Solutions to the Problems in Introduction to Commutative ...jtaylor/notes/atiyah... · Solutions to the Problems in Introduction to Commutative Algebra by M. F. Atiyah and I. G. MacDonald

Solutions to the Problems in

Introduction to Commutative Algebra

by M. F. Atiyah and I. G. MacDonald

J. David Taylor

October 20, 2018

Contents

1 Rings and Ideals 7

1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1

CONTENTS 2

1.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Modules 22

2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.27 Incomplete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 Rings and Modules of Fractions 34

3.1 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

CONTENTS 3

3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4 Primary Decomposition 46

4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

CONTENTS 4

4.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 Integral Dependence and Valuations 54

5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.19 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.22 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

CONTENTS 5

5.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.32 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.33 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.34 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.35 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6 Chain Conditions 64

6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

6.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

6.9 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

6.10 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

6.11 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

6.12 Undone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7 Noetherian Rings 67

7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8 Artin Rings 68

9 Discrete Valuation Rings and Dedekind Domains 69

10 Completions 70

11 Dimension Theory 71

12 Conventions 72

CONTENTS 6

13 Remarks 73

1 RINGS AND IDEALS 7

1 Rings and Ideals

1.1

Let A be a ring with nilpotent element x and unit element u. We will show that u + x is a unit. Take u = 1to see that 1 + x is a unit.

Constructive Proof: Let n > 0 be the least integer such that xn = 0 and define

y := u−1n−1∑i=0

(−u−1x)i.

Note that(−u−1x)n = (−u−1)nxn = 0.

As a consequence, we see that

y(u+ x) =

(n−1∑i=0

(−u−1x)i

)(1 + u−1x) = 1− (−u−1x)n = 1.

So u+ x is a unit with inverse y.

Non-Constructive Proof: By Proposition 1.8, the nilradical of A is the intersection of all prime ideals ofA. Since every maximal ideal of A is prime, the nilradical of A is contained in the Jacobson radical. Therefore, xis contained in the Jacobson radical of A. Apply Proposition 1.9 with y = −u−1 to seee that, 1 + u−1x is a unit.Since the units of A form a group,

u(1 + u−1x) = u+ x

is a unit as well.

1.2

Let f = a0 + a1x+ . . .+ anxn and g = b0 + b1x+ . . .+ bmx

m be elements of A[x].

(i) Suppose that f is a unit with inverse g. Let p be a prime ideal of A. Since A/p is a domain and f · g = 1in (A/p)[x], f must have degree 0 when reduced mod p. Thus, a1, . . . , an are contained in every prime ideal of A.On the other hand, Proposition 1.8 tells us that the nilradical of A is the intersection of all the prime ideals of A.So a1, . . . , an are nilpotent. As a consequence, a0 = f + (a0 − f) is the sum of a unit and a nilpotent element,hence a unit by Exercise 1.1.

If a0 is a unit and a1, . . . , an are nilpotent, then f − a0 is nilpotent and f = a0 + (f − a0) is the sum of a unitand a nilpotent element, hence a unit by Exercise 1.1.

(ii) Let p be a prime ideal of A. If fk = 0, then fk = 0 in (A/p)[x]. Since A/p is a domain, f = 0 in (A/p)[x].Thus, a0, . . . , an are contained in every prime ideal of A, and are nilpotent as a consequence of Proposition 1.8.

If a0, . . . , an are in the nilradical of A[x], then the A[x]-linear combination f is also in the nilradical becausethe nilradical is an ideal by Proposition 1.7.

(iii) If f is a zero-divisor, let g be a non-zero annihilator of f in A[x] of least degree. If g has positive degree,then ang is an annihilator of f of degree less than g, hence ang = 0. If k is a positive integer such that ajg = 0for j ≥ k, then (f − (akx

k + . . . + anxn))g = 0 implies that ak−1g = 0. Therefore, the least such k is 0. That is,

a0g = 0. Furthermore, we have shown that akbm = 0 for all k. Therefore, bmf = 0. (In fact, we have implicitlyshown that g has degree 0)

(iv) Let a, b, c be the ideals generated by the coefficients of f, g, and fg, respectively. The inclusion c ⊆ abshows that f and g are primitive if fg is. On the other hand, if fg is not primitive, then c is contained in somemaximal ideal, m, of A. Therefore fg = 0 in (A/m)[x]. Since A/m is a field, f = 0 or g = 0 in (A/m)[x]. This inturn implies that a ⊆ m or b ⊆ m. In either case, at least one of f or g is not primitive.


1.3

Define Ar := A[x1, . . . , xr] for all positive integers r. Let f = a0 + . . .+ anxnr be an element of Ar = Ar−1[xr]

with a0, . . . , an elements of Ar−1.

Suppose that every unit of Ar−1, when viewed as a polynomial with coefficients in A, has unit constant termand all other coefficients nilpotent. Furthermore, suppose that every nilpotent element of Ar−1, when viewed as apolynomial with coefficients in A, has nilpotent coefficients.

If f is a unit, then by Exercise 1.2 (i), a0 is a unit and a1, . . . , an are nilpotent in Ar−1. By our hypotheses,the constant coefficient of f over A is a unit, and all other coefficients of f over A are nilpotent.

If f is nilpotent, then by Exercise 1.2 (ii), a0, . . . , an are nilpotent in Ar−1. By our hypothesis, all coefficientsof f over A are nilpotent.

Pick a monomial order on the monomials in the indeterminates x1, . . . , xr. If f is a zero-divisor, let g ∈ Ar bea non-zero annihilator of f with minumum leading term. Let a ∈ A be the lead coefficient of f over A. Then agis an annihilator of f of degree less than g, hence ag = 0. By successively subtracting off the leading terms of f , asimilar induction on the chosen monomial order as in Exercise 1.2 (iii) shows that g annihilates every coefficientof f over A. Therefore, bf = 0, where b is the lead coefficient of g over A. (Remark 1)

Let f, g ∈ Ar and let a, b, c be the ideals generated by the respective coefficients of f, g, and fg over A. Theinclusion c ⊆ ab shows that f and g are primitive if fg is. On the other hand, if fg is not primitive, then c iscontained in some maximal ideal, m, of A. Therefore fg = 0 in (A/m)[x1, . . . , xr]. Since A/m is a field, f = 0 org = 0 in (A/m)[x1, . . . , xr]. This in turn implies that a ⊆ m or b ⊆ m. In either case, at least one of f or g is notprimitive.

1.4

Since every maximal ideal is prime, the nilradical of any ring is contained in the Jacobson radical.

If f is in the Jacobson radical of A[x], then 1 − fx is a unit by Proposition 1.9. By Exercise 1.2 (i), thecoefficients of f are nilpotent. Therefore f is in the nilradical of A[x] by Exercise 1.2 (ii).

1.5

Let f =∑

n≥0 anxn be an element of AJxK.

(i) If f is a unit with inverse g =∑

n≥0 bnxn, then fg = 1. On the other hand, fg has constant term a0b0.

Therefore a0 is a unit.

Suppose a0 is a unit. Define h := 1− a−10 f . Define g := a−1

0

∑n≥0 h

n. Since h has positive order, each of thecoefficients in the definition of g is a finite polynomial in the coefficients of h. Thus g is well defined. Furthermore,

fg = a0(1− h)a−10

∑n≥0

hn =∑n≥0

hn −∑n≥0

hn+1 = h0 +∑n≥1

hn −∑n≥1

hn = 1.

Therefore, f is a unit.

(ii) Let k be the order of f as a power series and suppose that fN = 0 for some positive integer N . The leastdegree term of fN has coefficient aNk = 0. Thus ak is nilpotent. Since the nilradical is an ideal, f − akxk is alsonilpotent. By induction on k, we see that every coefficient of f is nilpotent.

The converse is false: Define

A := Fp[T, T p−1, T p

−2, T p

−3, . . .]/(T p).

Claim: For every n ≥ 0, the element T p−n

has nilpotency order pn+1 in A.


Proof: Clearly,(T p−n)pn

= T . On the other hand, suppose(T p−n)a

= T p · g(T p−m)

for some integers a,m ≥ 0. Let N := maxm,n. Take pN th powers to get

T apN−n

= T pN+1 · g

(T p

N−m)

in Fp[T ]. By comparing degrees we see thatapN−n ≥ pN+1.

In particular, a ≥ pn+1 as desired.

Claim: The element f :=∑

n≥0 Tp−n

xn of AJxK is not nilpotent.

Proof: It is easy to see that

f(x)p =

∑n≥0

T p−nxn

p

=∑n≥0

T p1−n

xpn =∑n≥0

T p−nxp(n+1) = xpf(xp).

By repeated application of this identity, we see that

f(x)pn

= xpnf(xp

n)for all n ≥ 0. Since the right hand side has the same coefficients as the left (albeit with new indices), no power off(x) is zero.

(iii) By Proposition 1.9, f is in the Jacobson radical of AJxK if and only if 1 − fg is a unit of AJxK for all gin AJxK. By (i), this latter condition is equivalent to requiring that 1 − a0b is a unit in A for all b in A. Again,Proposition 1.9 shows that this is equivalent to letting a0 be an element of the Jacobson radical of A.

(iv) Let m be a maximal ideal of AJxK. Since 0 is in the Jacobson radical of A, x is in the Jacobson radical ofAJxK by (iii). Thus mc consists of the constant terms of elements of m and m = mc + xAJxK. Furthermore,

AJxK/m ∼= (AJxK/xAJxK)/(m/xAJxK) ∼= A/mc

shows that A/mc is a field. As a consequence, mc is maximal.

(v) Let p be a prime ideal of A, and define

φ : AJxK→ A→ A/p

by following the evaluation at 0 map with reduction modulo p. This map is obviously surjective. Furthermore,the kernel of φ consists of power series with constant term in p. That is, ker(φ) = p + xAJxK. By the FirstIsomorphism Theorem, AJxK/(p+xAJxK) ∼= A/p. Since A/p is a domain, p+xAJxK is a prime ideal. On the otherhand, p = (p + xAJxK)c shows that p is the contraction of a prime ideal of AJxK.

1.6

Let A be a ring such that every ideal not contained in the nilradical contains a non-zero idempotent. Let e bean idempotent in the Jacobson radical. By Proposition 1.9, 1− e is a unit. Since e is idempotent,

e(1− e) = e− e2 = 0.

Thus e = 0. Since, the Jacobson radical contains no non-zero idempotents, it must be contained in the nilradical.


1.7

Let A be a ring such that every element is a root of a polynomial of the form xn− x with n a positive integer.Let p be a prime ideal and x an element of A not in p. For some positive n, xn − x = 0 is in p. Since p is a primeideal not containing x, we have the equality xn−1 = 1 in A/p. We have now shown that the non-zero elements ofA/p have multiplicative inverses. Therefore, A/p is a field and p is a maximal ideal.

1.8

Let Σ be the set of prime ideals of A. By Theorem 1.3, Σ is not empty. Order Σ by reverse inclusion and letpn∞n=1 be a chain. Define p :=

⋂∞n=1 pn. If x, y are not elements of p, then there are i, j such that x is not in pi

and y is not in pj . Let k be the greater of i, j. Then x and y are both not in pk. Since pk is a prime ideal, xy isnot in pk. Thus xy is not in p. In particular, p is a prime ideal. Since p is an upper bound for the chain pn∞n=1,Zorn’s Lemma implies that Σ has maximal elements. Therefore, A has minimal prime ideals.

1.9

Let a be a proper ideal of A. If a =√a, then a is the intersection of the prime ideals that contain it by

Proposition 1.14. On the other hand, let a be the intersection of some collection of prime ideals Σ and let Σ′ bethe set of all prime ideals containing a. Then the following inclusions

a =⋂p∈Σ

⊇⋂p∈Σ′

=√a ⊇ a

show that a =√a.

1.10

Let A be a ring with nilradical R.

(i)⇒(ii) Let p be the unique prime ideal of A. By Proposition 1.8, p = R. Furthermore, p is a maximal idealby Corollary 1.4. If x is a non-unit, then it is contained in p and is nilpotent as a consequence.

(i)⇒(iii) Let p be the unique prime ideal of A. By Proposition 1.8, p = R. Furthermore, p is a maximal idealby Corollary 1.4. Thus A/R is a field.

(ii)⇒(i) Since every element of A not in R is a unit, Proposition 1.6 (i) asserts that A is a local ring withunique maximal ideal R. On the other hand, Proposition 1.8 states that R is contained in every prime ideal of A.Therefore, R is the only prime ideal of A.

(ii)⇒(iii) Since every element of A not in R is a unit, Proposition 1.6 (i) asserts that A is a local ring withunique maximal ideal R. In particular, A/R is a field.

(iii)⇒(i) Note that every prime ideal of A contains R. By the Lattice Isomorphism Theorem, p is a primeideal of A if and only if p/R is a prime ideal of A/R. Since A/R is a field, every prime ideal of A is equal to R.Thus A has exactly one prime ideal.

(iii)⇒(ii) Let x be a non-nilpotent element of A. Let y be an element of A such that xy = 1 in A/R. In otherwords, xy − 1 is nilpotent. By Exercise 1.1, xy = 1 + (xy − 1) is a unit. So x is a unit with inverse y(xy)−1.

1.11

Let A be a Boolean ring.

(i) Given x in A,0 = (x+ 1)2 − (x+ 1) = x2 + 2x+ 1− x− 1 = 2x.


(ii) If x is not an element of a prime ideal p, then x(1 − x) = 0 implies that x = 1 in A/p. Thus A/p is thefield with two elements and p is a maximal ideal.

(iii) Let x, y be elements of an ideal a. Define z := x+ y + xy and note that

xz = x(x+ y + xy) = x2 + 2xy = x

andyz = y(x+ y + xy) = y2 + 2xy = y.

Therefore, (x, y) = (z). If a is a finitely generated ideal with more than one generator, we may apply this processto reduce the number of generators by one. By repeated application, we can find a principal generator fora.

1.12

Let A,m be a local ring with an idempotent element x. If x is in m, then 1 − x is a unit because m is theJacobson radical. Thus x(1 − x) = 0 implies that x = 0. If x is a unit, then x(1 − x) = 0 implies that x = 1.Therefore the only idempotents of A are 0 and 1.

1.13

Let K0 = K be a field. Given a non-negative integer n for which the field, Kn, is defined, let Σn be theset of monic irreducible elements of Kn[x] and let An be the polynomial ring over Kn generated by the set ofindeterminants xf |f ∈ Σn. Define an as the ideal of An generated by the set f(xf ) ∈ A|f ∈ Σn. Since Kn

is a field, An is a domain. Thus every element of an has positive degree and an does not contain 1. Let mn be amaximal ideal of An containing an and define Kn+1 = An/mn. The map

Kn → An → An/mn = Kn+1,

given by the composition of the canonical inclusion and quotient maps, is a field homomorphism. Thus it isinjective and we may identify Kn with a subfield of Kn+1.

Let K =⋃n≥0Kn (really, K = lim−→Kn). If x, y are in K, then they are contained in some subfields Kn,Km.

Letting k be the greater of m and n, x and y are in Kk. Therefore the sum, difference, and product of x, y are inK. If y is non-zero, then the quotient x/y is in K as well. Since 0, 1 are in K, and any field arithmetic of K canbe performed in a subfield, K is a field.

Let f be an irreducible monic polynomial in K[x]. Since f has only finitely many coefficients, there is some nsuch that f is an irreducible monic polynomial in Kn[x]. By construction, f has a root in Kn+1, hence in K. Bythe Euclidean division, f must have degree 1. Therefore, K is algebraically closed.

By construction, the field extension Kn+1/Kn is algebraic for every n. Since the class algebraic extensions is adistinguished class of field extensions (Lang’s Algebra), Kn/K is an algebraic extension for every n ≥ 0. Therefore,every element of K is algebraic over K, and K is the algebraic closure of K up to isomorphism.

1.14

Let Σ be the set of ideals of the ring A which consist of solely zero-divisors. This set is non-empty becauseit contains (0). Order Σ by inclusion and suppose that ann≥1 is a chain. Since a =

⋃n≥1 an is an ideal of A

consisting solely of zero divisors, it is an upper bound for ann≥1 in Σ. By Zorn’s Lemma, Σ has a maximalelement b.

If x, y are not elements of b, then (x)+b, (y)+b each contain non-zero-divisors, because b is a maximal elementof Σ. Since the product of two non-zero-divisors is a non-zero-divisor and ((x) + b)((y) + b) ⊆ (xy) + b, someelement of (xy) + b is not a zero-divisor. Thus xy is not an element of b. Therefore, b is a prime ideal of A.


If d is a zero-divisor, then (d) is an element of Σ. Thus (d) is contained in a maximal element of Σ, which inturn is a prime ideal of A. Therefore, the union of the ideals that are maximal in Σ is the set of zero-divisors, andeach of these is prime ideal.

1.15

Let X be the set of all prime ideals of the ring A. For any subset E of A, define V (E) as the set of primeideals containing E.

(i) If a is the ideal of A generated by E, then V (a) ⊆ V (E) by the definition of V . If p is a prime idealcontaining E, then p contains a because a is the smallest ideal containing E. Thus V (E) ⊆ V (a) as well.

Given that every ideal is contained in its radical, V(√

a)⊆ V (a). Since

√a is the intersection of all prime

ideals containing a, V (a) ⊆ V(√

a).

(ii) Since (0) is contained in every ideal of A, V (0) = X. On the other hand, no proper ideals of A contain 1,hence V (1) is empty.

(iii) Let Eii∈I be a family of subsets of A and let p be a prime ideal. If p contains⋃i∈I Ei, then p contains

each Ei. Thus V(⋃

i∈I Ei)⊆⋂i∈I V (Ei). If p contains each Ei, then clearly p contains their union. Thus⋂

i∈I V (Ei) ⊆ V(⋃

i∈I Ei)

as well.

(iv) If a, b are ideals of A, then

V (ab) = V(√

ab)

= V(√

a ∩ b)

= V (a ∩ b)

by Exercise 1.13 of the text.

Given that ab is contained in a and b, V (a)∪V (b) ⊆ V (ab). If p is a prime ideal containing ab, then p containsa or b, by Proposition 1.11. Thus V (ab) ⊆ V (a) ∪ V (b).

1.16

• Because Z is a principal ideal domain, the prime ideals of Z are of the form (p), where p is a prime number,and (0). Therefore, SpecZ is a curve through all the principal ideals of Z of the form (p), with p a primeideal number, and the point (0). For geometric reasons that will be clear later, draw (0)) off to the side ofthe curve.

Let U ⊆ SpecZ be an open set. By definition, U = V (a)c for some ideal a of Z. Since Z is a principal idealdomain, a = (a) for some integer a. If a = 0, then V (a) = V (0) = SpecZ and U = ∅. If U is non-empty,then a 6= 0. By the Fundamental Theorem of Arithmetic, a = ±

∏ki=1 p

αii for some prime numbers p1, . . . , pk

and positive integers α1, . . . , αk. Define√a as

∏ki=1 pi. Since non-zero prime ideals of Z are maximal, we

have

V (a) = V (a) = V (√a) =

k⋃i=1

V (pi) = (p1), . . . , (pk).

Therefore, U = (p1), . . . , (pk)c. In summary, non-empty open subsets of SpecZ are cofinite sets containing(0).

• Since R is a field, SpecR is the one point space containing (0) under the trivial topology.

• As before, C[x] is a principal ideal domain. Thus its prime ideals are of the form (f), where f is a monicirreducible or f = 0. Since C is algebraically closed, the monic irreducible elements of C[x] are of the formx − a, where a is a complex number. Therefore, SpecC[x] can be drawn as the complex plane, where thepoint (x− a) is identified with the complex number a, with an additional point corresponding to the primeideal (0).


Let U ⊆ SpecC[x] be an open set. By definition, U = V (a)c for some ideal a of C[x]. Since C[x] is a principalideal domain, a = (f) for some monic polynomial f ∈ C[x]. If f = 0, then V (a) = V (0) = SpecC[x] andU = ∅. If U is non-empty, then f 6= 0. By the Fundamental Theorem of Algebra, f =

∏ki=1(x − ai)αi

for some complex numbers a1, . . . , ak and positive integers α1, . . . , αk. Define√f as

∏ki=1(x − ai). Since

non-zero prime ideals of C[x] are maximal, we have

V (a) = V (f) = V(√

f)

=

k⋃i=1

V (x− ai) = (x− a1), . . . , (x− ak).

Therefore, U = (x− a1), . . . , (x− ak)c. In summary, non-empty open subsets of SpecC[x] are cofinite setscontaining (0).

• As before, R[x] is a principal ideal domain. Therefore, its prime ideals are of the form (f), where f is amonic irreducible polynomial or f = 0. Since C/R is a degree 2 extension and C is algebraically closed,the irreducible polynomials in R[x] are degree 1 or 2. Let (f) be a non-zero prime ideal of R[x] and letι : R[x] → C[x] be the inclusion map. If deg(f) = 1, then ι(f) = x − r for some real number r, andι∗((x − r)) = (f). If deg(f) = 2, then ι(f) = (x − a)(x − a) for some non-real complex number a, andι∗((x − a)) = ι∗((x − a)) = (f). By Exercise 1.21 (i), ι∗ : SpecC[x] → SpecR[x] is a continuous map. Wehave just shown that it is surjective and that SpecR[x] is a quotient of SpecC[x] by identifying conjugatepoints. So SpecR[x] can be drawn as a copy of the complex plane, folded over the real axis, with an additionalpoint corresponding to the prime ideal (0).

Let U ⊆ SpecR[x] be a non-empty open set. Then ι∗−1(U) is a non-empty open subset of SpecC[x]. By ourprevious arguments, ι∗−1(U) = (x − a1), . . . , (x − ak)c for some complex numbers a1, . . . , ak. Since ι∗ issurjective, we know that

U = ι∗(ι∗−1(U)) = ι∗((x− a1)), . . . , ι∗((x− ak))c.

In summary, non-empty open subsets of SpecR[x] are cofinite sets containing (0).

• Let ι : Z→ Z[x] be the inclusion map. By Exercise 1.21 (i), ι∗ : SpecZ[x]→ SpecZ is continuous.

Claim: The map ι∗ is surjective.

Proof: Let p be a prime number. The identification, Z[x]/pZ[x] ∼= (Z/pZ) [x] = Fp[x] shows that pZ[x] is aprime ideal of Z[x]. Clearly, ι∗(pZ[x]) = (p). On the other hand, ι∗((0)) = (0). So ι∗ is surjective.

Therefore, we can partition SpecZ[x] as a union of fibers of the map ι∗ via

SpecZ[x] = ι∗−1((0)) ∪

⋃p prime number

ι∗−1((p))

.

Let p be a non-zero prime ideal of Z[x]. Then the map Z/ι−1(p) → Z[x]/p obtained from ι : Z → Z[x] bypassing to quotients is injective. In particular, if ι∗(p) = (c) with c ≥ 0, then c is the characteristic of Z[x]/p.

Claim: The fiber ι∗−1((0)) consists of prime ideals of the form (f) where f is either an irreducible polynomialor 0.

Proof: If ι∗(p) = (0), then every non-zero element of p has positive degree. Suppose that p 6= (0). Let f ∈ pbe a irreducible polynomial with minimum degree. Let g be any element of p. Since Q[x] is a EuclideanDomain, there are q, r ∈ Q[x] such that g = qf + r and r has degree less than f . Since f, g are in p, r is inp as well. By our hypothesis on f , r is zero. By Gauss’ Lemma, q is in Z[x]. Therefore, p = (f).

Claim: Let p be a prime number. The fiber ι∗−1((p)) consists of prime ideals of the form (p, f) where f isa polynomial that is irreducible mod p or f is 0.

Proof: If ι∗(p) = (p), then the quotient map Z[x] → Z[x]/p factors through the quotient Z[x] → Fp[x]. SoZ[x]/p is a quotient of Fp[x] by some prime ideal q. Since Fp[x] is a Principal Ideal Domain, q =

(f)

for


some element f of Fp[x] which is irreducible or 0. By the Lattice Isomorphism Theorem, q lifts to the ideal(p, f) of Z[x], where f is any lift of f .

In summary, the prime ideals of Z[x] break into four types:

(f), where f ∈ Z[x] is an irreducible polynomial of positive degree,

(p), where p ∈ Z is a prime number,

(p, f), where p ∈ Z is a prime number and f ∈ Z[x] is irreducible modulo p, and

(0).

Therefore, SpecZ[x] can be drawn as a grid. The bottom of the grid consists of ideals (c) where c is either aprime number or 0. To maintain consistency with the description of SpecZ given above, draw (0) off to theside from the other ideals of this form.

Above each ideal on the bottom row, place the ideals (c, f), where f is irreducible mod c. Each such fis irreducible, so you may as well put the ideal (p, f) in the same row as the ideal f . (Though f won’tnecessarily be irreducible mod every prime. For instance, x2 + x+ 1 is irreducible mod 2 but not mod 3.)

Claim: Every proper closed subset of SpecZ[x] is a finite union of sets of the form V (g1, . . . , gn), whereg1, . . . , gn are irreducible elements of Z[x].

Proof: Since Z is a Euclidean Domain, it is Noetherian. By Theorem 7.5, Z[x] is Noetherian. By Exercises6.7 and 6.8, V (a) is a finite union of irreducible closed subspaces of SpecZ[x]. If g1, g2 are elements ofZ[x], then Exercise 1.15 (iv) implies V (g1g2) = V (g1) ∪ V (g2). Therefore, if V (g) is irreducible, then g isirreducible. By Exercise 1.18 (ii) and our taxonomy of the prime ideals in Z[x], the converse holds as well.By Exercise 1.17 (i), the sets Xg form a basis for the topology on X = SpecZ[x]. By Exercise 6.6, every opensubset of SpecZ[x] is quasi-compact. If g has factorization into irreducibles

∏ni=1 gi, then Xg =

⋂ni=1Xgi .

By De Morgan’s Laws and the preceding facts, every closed subset of SpecZ[x] is a finite union of sets ofthe form V (g1, . . . , gn), where g1, . . . , gn are irreducible elements of Z[x].

Claim: Every proper closed subset of SpecZ[x] is a finite union of sets V (p), V (f), V (p, f) for varying p, f ,where p is a prime number and f is either irreducible (for (f)) or irreducible mod p (for (p, f)). Note thatirreducible mod p implies irreducible.

Proof: Let g1, . . . , gn be distinct irreducible elements of Z[x]. Let p be a minimal prime ideal containing theideal (g1, . . . , gn). If p = (f) for some irreducible f , then all the gi’s are multiples of f . If this case, g1 = fand n = 1. If n 6= 1, then (g1, . . . , gn) has only finitely many minimal primes, and each must be of the form(p, f) for p a prime number and f irreducible mod p.

Note that V (p), for p a prime number, consists of (p) and all prime ideals (p, f) with f irreducible mod p.Furthermore, V (f), for f an irreducible polynomial of positive degree, consists of (f) and all prime ideals ofthe form (p, f) where p is a prime number such that f is irreducible mod p.

Conclusion: Every non-empty open subset of SpecZ[x] can be formed by puncturing finitely many closedpoints and removing finitely many horizontal and vertical grid lines from the total space, and every non-emptyopen set contains (0).

1.17

For each f in A, define Xf = V (f)c in X = SpecA. Let U be an open set containing the point p of X. Bydefinition of the Zariski topology, there is an ideal a such that U c = V (a). Let f be an element of a not containedin p. Then U c = V (a) ⊆ V (f) implies that Xf ⊆ U and p is in Xf . Thus the sets Xf form a basis for the Zariskitopology.

(i) By Exercise 1.15 (iv),

Xf ∩Xg = V (f)c ∩ V (g)c = (V (f) ∪ V (g))c = V (fg)c = Xfg.

(ii) An element of A is nilpotent if and only if it is contained in every prime ideal. Since Xf is the set of primeideals not containing f , the equivalence follows.


(iii) By Corollary 1.5, every non-unit is contained in a maximal ideal. Conversely, no prime ideal contains aunit. Therefore, f is contained in no prime ideals precisely when f is a unit.

(iv) Every prime ideal containing g contains f if and only if√

(g) ⊇√

(f). The first condition is equivalentto Xf ⊆ Xg. Swapping the roles of f and g gives the desired equivalence.

(v) Let Xff∈E be an open cover of X. Taking complements shows that V (E) is empty. Therefore, 1 isin the ideal generated by E. This in turn implies that there are f1, . . . , fn in E and a1, . . . , an in A such that1 =

∑ni=1 aifi. Thus V (f1, . . . , fn) is empty. Taking complements again shows that Xfini=1 is an open cover of

X.

(vi) Let Xgg∈E be an open cover of Xf . Taking complements shows that V (f) ⊇ V (E). Therefore, f isin the radical ideal generated by E. This in turn implies that there are g1, . . . , gn in E, a1, . . . , an in A, and apositive integer m such that fm =

∑ni=1 aigi. Thus V (f) ⊇ V (g1, . . . , gn). Taking complements again shows that

Xgini=1 is an open cover of Xf .

(vii) Let U be a quasi-compact open set of X. Since U is open, it is the union of a collection of basic opensets Xff∈E . On the other hand, only finitely many such sets are needed to cover U because it is quasi-compact.

Conversely, finite unions of quasi-compact sets are quasi-compact.

1.18

(i) By (ii), the closure of x is the set of prime ideals containing px. Thus x is closed if and only if px isnot properly contained in any other prime ideals. In other words, x is closed if and only if px is maximal.

(ii) The closure of x is the intersection of all closed sets containing x. Furthermore, x is in V (E) if and onlyif E is a subset of px. Thus every closed set containing x contains V (px). On the other hand, V (px) is a closedset containing x.

(iii) By (ii), y is in the closure of x if and only if py ⊇ px.

(iv) If x, y are distinct points in X such that y is in the closure of x, then V (py) ⊆ V (px) by (iii). Since xand y are distinct, x is not in V (py). Therefore, x is in the complement of the closure of y.

1.19

If SpecA is irreducible and fg is nilpotent, then Xf ∩Xg = Xfg is empty by Exercise 1.17. Thus Xf or Xg isempty, and correspondingly f or g is nilpotent.

Suppose that the nilradical of A is prime ideal. Let Xf , Xg be non-empty open sets. Then f, g are not nilpotentand neither is fg. Thus Xf ∩Xg = Xfg is non-empty.

1.20

Let X be a topological space.

(i) Let U, V be open subsets of X such that U ∩ Y and V ∩ Y are non-empty. By the definition of the closure,U ∩ Y and V ∩ Y are non-empty open sets of Y . Since Y is irreducible, (U ∩ Y ) ∩ (V ∩ Y ) is a non-empty subsetof (U ∩ Y ) ∩ (V ∩ Y ). Thus Y is irreducible.

(ii) Note that every point is irreducible. If Y is an irreducible subspace of X, then let Σ be the set of irreduciblesubspaces of X containing Y , ordered by inclusion. Let Znn≥1 be a chain in Σ and define Z =

⋃n≥1 Zn. Suppose

that U, V are open sets intersecting Z. Then there are positive integers i, j such that U ∩ Zi and V ∩ Zj are non-empty. If k is the greater of i, j, then both U and V intersect Zk. Since Zk is irreducible, U ∩ V contains a pointof Zk. Thus U ∩ V contains a point of Z. Therefore, Z is an upper bound for Znn≥1 in Σ. By Zorn’s Lemma,there is a maximal irreducible subspace of X containing Y .


(iii) If Y is a maximal irreducible subspace of X, then it must equal its closure by (i). As in (ii), note thatpoints are irreducible subspaces of X. By (ii), the maximal irreducible subspaces cover X. In a Hausdorff space,any subspace with more than one point has disjoint non-empty open sets, and is thus not irreducible.

(iv) Let X = SpecA for some ring A. If Y is a closed irreducible subspace of X, then there is a radical ideala such that Y = V (a). If f, g are not in a, then Xf , Xg both meet Y , because a is its own radical. Since Y isirreducible, Xf ∩Xg = Xfg intersects Y . Thus fg is also not in a, and we see that a is a prime ideal.

If p is a prime ideal, Y = V (p), and Xf , Xg intersect Y , then f, g are not in p. Thus fg is not in p andXf ∩Xg = Xfg intersects Y . Therefore, Y is irreducible.

For any prime ideals p, q, V (p) ⊆ V (q) if and only if p ⊇ q. Thus maximal irreducible components of Xcorrespond to minimal prime ideals.

1.21

Let φ : A → B be a ring homomorphism, X = SpecA, and Y = SpecB. If q is a prime ideal of B, thenA/φ−1(q) → B/q is injective by the First Isomorphism Theorem. Thus φ−1(q) is a prime ideal of A. Defineφ∗ : Y → X by φ∗(q) = φ−1(q).

(i) If f is an element of A, then φ−1(q) is contained in Xf if and only if q is contained in Yφ(f). Therefore,φ∗−1 (Xf ) = Yφ(f).

(ii) If φ−1(q) ⊇ a, then q ⊇ qce ⊇ ae. Conversely, q ⊇ ae implies that φ−1(q) ⊇ aec ⊇ a. Therefore,φ∗−1 (V (a)) = V (ae).

(iii) Let q be a prime ideal containing b and define p = φ−1(q). Then bc is contained in p. Thus φ∗ (V (b)) ⊆V (bc).

Without loss of generality, b =√b. Let p ∈ V (bc) and suppose f /∈ p. By the transitivity of containment,

φ(f) /∈ b =√b. By Proposition 1.14, there exists q ∈ V (b) such that φ(f) /∈ q. Therefore,

φ∗(q) ∈ φ∗(Yφ(f) ∩ V (b)

)⊆ Xf ∩ φ∗(V (b)).

Hence, V (bc) ⊆ φ∗(V (b)).

(iv) If φ is surjective, then φ : A/ kerφ → B is an isomorphism by the First Isomorphism Theorem. By theLattice Isomorphism Theorem, φ∗ : Y → V (kerφ) is bijective and continuous. Since φ is surjective, every basicopen set of Y can be put in the form Yφ(f) for some f in A. By (i), φ∗

(Yφ(f)

)= Xf ∩ V (kerφ). Therefore, φ∗ is

a homeomorphism of Y onto its image.

(v) Suppose that kerφ is the in the nilradical of A and let Xf be a non-empty open set of X. Since f is not inthe nilradical of A, φ(f) is not nilpotent. Therefore, Yφ(f) contains some point q. By (i), φ∗(q) is in Xf . Therefore,φ∗(Y ) is dense in X.

Suppose that φ∗(Y ) is dense in X and let f be an element of kerφ. Then Yφ(f) = φ∗−1(Xf ) is empty. ThusXf is empty and f is nilpotent.

(vi) If ψ : B → C is another ring homomorphism and r is a prime ideal of C, then

(ψ φ)∗(r) = (ψ φ)−1(r) = φ−1(ψ−1(r)) = (φ∗ ψ∗)(r).

(vii) Let A be a domain with exactly one non-zero prime ideal p, and let K be the field of fractions of A.Define B = (A/p)×K and φ : A→ B by φ(x) = (x, x).

The prime ideals of B are (A/p)× (0) and (0)×K. Furthermore, φ∗ is bijective due to φ∗((A/p)× (0)) = (0)and φ∗((0)×K) = p. On the other hand, SpecB has the discrete topology, but the closure of (0) in SpecA isthe whole space.


1.22

Let A :=∏ni=1Ai be a direct product of the rings Ai, let πi : A → Ai be the canonical projections, let

ai := kerπi, and define Xi := V (ai). By the Chinese Remainder Theorem, ψ : A→∏ni=1A/ai is an isomorphism

that matches a prime ideal p to the product∏ni=1(p + ai)/ai. Therefore,

A/p ∼=n∏i=1

A/(p + ai)

is a domain. Thus all but one factor ring must be zero. Calling the remainding index j, A/p ∼= A/(p + aj) by thenatural projection. So p must contain aj . We have now shown that SpecA is the disjoint union of the closed setsXi. Since the complement of each Xi is a finite union of closed sets, each Xi is open. By Exercise 1.21 (iv), π∗igives a homeomorphism from SpecAi to Xi.

(i)⇒(ii) If X = SpecA is disconnected, then there are proper non-zero ideals a, b such that X = V (a)∪V (b) =V (ab) and ∅ = V (a) ∩ V (b) = V (a + b). By Exercise 1.15, there are f, g in a, b and a positive integer n suchthat f + g = 1 and (fg)n = 0. Since (f, g) ⊆

√(fn, gn) and V (f, g) is empty, V (fn, gn) is also empty. Thus

(fn) + (gn) = (1) and the Chinese Remainder Theorem implies that A→ (A/(fn))× (A/(gn)) is an isomorphism.Neither of f, g is a unit, because they are elements of the proper ideals a, b.

(i)⇒(iii) Suppose that A is reduced. If X = SpecA is disconnected, then there are proper non-zero ideals a, bsuch that X = V (a)∪V (b) = V (ab) and ∅ = V (a)∩V (b) = V (a+ b). By Exercise 1.15, there are f, g in a, b suchthat f + g = 1 and fg = 0. Therefore,

f2 = f(1− g) = f − fg = f.

Furthermore, f 6= 0, 1 because a, b are both proper ideals. (Remark 2)

(ii)⇒(i) This was shown above.

(ii)⇒(iii) Let φ : A→ A1×A2 be an isomorphism, where neither of A1, A2 are the zero ring. Then φ−1((1, 0))is a non-trivial idempotent of A.

(iii)⇒(i) If e is a non-trivial idempotent of A, then

V (e) ∩ V (1− e) = V (1) = ∅

andV (e) ∪ V (1− e) = V (e− e2) = V (0) = SpecA

is a separation of SpecA.

(iii)⇒(ii) Let e be a non-trivial idempotent of A. Then (e) + (1 − e) = (1) and (e)(1 − e) = (0). By theChinese Remainder Theorem, A ∼= A/(e)×A/(1− e).

(iii)⇒(ii) (Alternative) Let e be a non-trivial idempotent of A. The ideal Ae is a ring with identity e. Since

(1− e)2 = 1− 2e+ e2 = 1− e,

1− e is also a non-trivial idempotent. Let φ : A→ Ae×A(1− e) be the ring homomorphism given by

φ(x) = (xe, x(1− e)).

If x ∈ kerφ, then x = xe+ x(1− e) = 0. So φ is injective. On the other hand, let y, z be any elements of A. Thenφ(ye+ z(1− e)) = (ye, z(1− e)). So φ is surjective and an isomorphism.

Let A be a local ring. By Exercise 1.12, A contains no non-trivial idempotent. Therefore, SpecA is connectedby (iii).


1.23

Let A be a Boolean ring and X = SpecA.

(i) If f is an element of A, then Xf ∩X(1−f) = X0 is empty and

Xf ∪X(1−f) = (V (f) ∩ V (1− f))c = V ((f) + (1− f))c = V (1)c = X.

Therefore, Xf = Xc(1−f) is closed.

(ii) If f1, . . . , fn are elements of A, then by (i)

n⋃i=1

Xfi =n⋃i=1

V (1− fi) = V

(n∏i=1

(1− fi)

)= Xf ,

where f = 1−∏ni=1(1− fi).

(iii) Let Y be a closed and open subset of X. Because Y is open, it is a union of basic open sets. Since Y isclosed and X is quasi-compact, Y is also quasi-compact. Thus Y is a finite union of basic open sets. By (ii), Y isa basic open set.

(iv) If x, y are distinct points in X, then, without loss of generality, there is some f in A such that Xf containsx but not y, by Exercise 1.18 (iv). By (i), y is in X(1−f) and the two sets are disjoint. X is quasi-compact byExercise 1.17 (v).

1.24

Let L be a Boolean algebra. Then L has a greatest element 1 and a least element 0, each of ∨,∧ distributeover each other, and each element a has a unique complement a′.

Let A(L) be the set of L under the operations

a+ b = (a ∧ b′) ∨ (a′ ∧ b), a · b = a ∧ b.

Let a, b, c be elements of A(L). First, +, · are commutative because ∨,∧ are. The additive identity is 0 by

a+ 0 = (a ∧ 1) ∨ (a′ ∧ 0) = a ∨ 0 = a.

The multiplicative identity is given by 1. The additive inverse of a is a by

a+ a = (a ∧ a′) ∨ (a′ ∧ a) = 0 ∨ 0 = 0.

Addition is associative by

(a+ b) + c =[((a ∧ b′) ∨ (a′ ∧ b)) ∧ c′

]∨[((a′ ∨ b) ∧ (a ∨ b′)) ∧ c

]= (a ∧ b′ ∧ c′) ∨ (a′ ∧ b ∧ c′) ∨ (a′ ∧ b′ ∧ c) ∨ (a ∧ b ∧ c)= a+ (b+ c),

where the second line follows from the first by distributing c, c′ and further shows that the expression is symmetricin a, b, and c. Multiplication is associative because ∧ is. Finally, the distributive law holds:

a(b+ c) = a ∧((b ∧ c′) ∨ (b′ ∧ c)

)= [(a ∧ b) ∧ c′] ∨ [b′ ∧ (a ∧ c)]= [(a ∧ b) ∧ (a′ ∨ c′)] ∨ [(a′ ∨ b′) ∧ (a ∧ c)]= ab+ ac.

Therefore, A(L) is a Boolean ring.


Let A be a Boolean ring with elements a, b and define a ∨ b = a+ b+ ab, a ∧ b = ab, and a ≤ b if and only ifab = a.

First, we show that ≤ is a partial order. If a ≤ b and b ≤ a, then a = ab = ba = b. So ≤ is anti-symmetric. Ifa ≤ b ≤ c, then ac = abc = ab = a, and a ≤ c. So ≤ is transitive.

Second, we show that A is a lattice under this ordering.

a(a ∨ b) = a(a+ b+ ab) = a and a(a ∧ b) = aab = a ∧ b

shows that a ∧ b ≤ a ≤ a ∨ b. By symmetry, a ∧ b ≤ b ≤ a ∨ b as well. If c ≤ a, b, then

c(a ∧ b) = cab = c

shows that c ≤ a ∧ b. Thus a ∧ b is the infimum of a and b. If a, b ≤ c, then

c(a ∨ b) = c(a+ b+ ab) = a ∨ b

shows that a ∨ b ≤ c. Thus a ∨ b is the supremum of a and b.

Third, we show that this is a Boolean lattice. In general,

a ∨ 1 = a+ 1 + a · 1 = 1

anda ∧ 0 = a · 0 = 0.

For distributivity,

(a ∧ b) ∨ c = ab+ c+ abc = (a+ c+ ac)(b+ c+ bc) = (a ∧ c) ∨ (b ∧ c)

and(a ∨ b) ∧ c = ac+ bc+ abc = ac+ bc+ acbc = (a ∧ c) ∨ (b ∧ c).

Finally, if we define a′ = 1 + a, thena ∧ a′ = a(1 + a) = a+ a = 0

anda ∨ a′ = a+ (1 + a) + a(1 + a) = 1.

Furthermore, if x is a complement of a, then

0 = a ∧ x = ax

and1 = a ∨ x = a+ x+ ax.

These equations imply that x = 1 + a = a′. So complements are unique. Thus A,≤ is a Boolean lattice.

1.25

Given a Boolean lattice L, define

φ : L→ P(SpecA(L)) : f 7→ Xf .

By Exercise 1.24, f ≤ g in L if and only if fg = f . This in turn implies that Xf ∩Xg = Xfg = Xf , which yieldsXf ⊆ Xg.

If Xf = Xg, thenXf = Xg = Xc

(1+g)

by the solution to Exercise 1.23 (i). So Xf ∩ X(1+g) = Xf(1+g) is empty. Therefore, f(1 + g) is nilpotent. It iseasy to see that every Boolean ring is reduced. So f(1 + g) = 0. In particular, f = fg. So f ≤ g. Swap the rolesof f and g to see that g ≤ f . Since the ordering on L is reflexive, f = g and φ is injective.

On the other hand, the image of φ is precisely the class of open-and-closed subspaces of the compact Hausdorffspace SpecA(L) by Exercise 1.23.


1.26

We will use the notation of the problem.

(i) Shown in the text.

(ii) Shown in the text.

(iii) Since µ : X → X is bijective,

µ(Uf ) = µ(x ∈ X : f(x) 6= 0)= µ(x) : x ∈ X and f(x) 6= 0= mx : x ∈ X and f /∈ mx

= m ∈ X : f /∈ m

= Uf .

Let U be an open subset of X. If x ∈ U , then x and U c are disjoint closed sets. By Urysohn’s Lemma thereis a continuous function f : X → R such that f(x) = 1 and f(y) = 0 for all y /∈ U . Therefore, x ∈ Uf ⊆ U . So thesets Uf form a basis for the topology on X.

Let U be an open subset of X. Since X ⊆ SpecC(X) has the subspace topology, there is an open set Wof SpecC(X) such that U = X ∩ W . Let m ∈ U ⊆ W . By Exercise 1.17, there is f ∈ C(X) such thatm ∈ (SpecC(X))f ⊆ W . Note that X ∩ (SpecC(X))f = Uf . Therefore,

m ∈ Uf = X ∩ (SpecC(X))f ⊆ X ∩ W = U .

So the sets Uf form a basis for the topology on X.

1.27

We will use the notation of the problem.

We will prove that the map µ : X → X is surjective. If m is a maximal ideal of P (X), then P (X)/m is afinitely generated k-algebra that is also a field. By Corollary 5.24, P (X)/m is a finite field extension of k. Sincek is algebraically closed, k → P (X)/m is an isomorphism of k-algebras. For each i, let xi be the image of ξi in k.Then the elements ξ1 − x1, . . . , ξn − xn each map to 0 in k. Therefore, m = (ξ1 − x1, . . . , ξn − xn) = mx, and µ issurjective.

1.28

Let X,Y be affine algebraic varieties in kn, km respectively, let φ : X → Y be a regular map, let P (X) =k[t1, . . . , tn]/I(X) and P (Y ) = k[s1, . . . , sm]/I(Y ). Furthermore, let ξ1, . . . , ξn and η1, . . . , ηm be the respectiveimages of t1, . . . , tn and s1, . . . , sm in P (X) and P (Y ).

(Inverse Functor) Given a k-algebra homomorphism φ∗ : P (Y ) → P (X), the elements φ∗η1, . . . , φ∗ηm lift

to polynomials g1, . . . , gm in k[t1, . . . , tn]. Let ψ : X → km be the regular morphism with coordinate functionsg1, . . . , gm. Since all lifts of g1, . . . , gm are congruent mod I(X), the regular morphism ψ only depends on φ∗ andnot on the choice of g1, . . . , gm.

Suppose that h ∈ I(Y ). Then

0 = φ∗(h(η1, . . . , ηm)) = h(g1, . . . , gm) mod I(X) = (h ψ)(ξ1, . . . , ξn).

Therefore, imψ ⊆ Y . By abuse of notation, we call the corestricted morphism ψ : X → Y .

(φ 7→ φ∗ is injective) If φ∗ : P (Y )→ P (X) is induced by some regular morphism φ : X → Y , then

fj mod I(X) = φ∗ηj = gj mod I(X)


for all j. So ψ = φ as functions from X to Y .

(φ 7→ φ∗ is surjective) Furthermore, ψ∗ηj = gj mod I(X). So ψ∗ = φ∗ as k-algebra homomorphisms.

2 MODULES 22

2 Modules

2.1

Direct Proof: If m,n are coprime, then there are integers x, y such that mx+ny = 1. For any simple tensor,f ⊗ g, we get

f ⊗ g = (mx+ ny)(f ⊗ g) = x(mf ⊗ g) + y(f ⊗ ng) = 0.

Thus Z/mZ⊗Z Z/nZ = 0.

Proof via Localization: (Remark 3) If p is a prime not dividing m, then the Euclidean algorithm providesintegers q, r such that p = qm+ r. Since any common factor of m and r divides p, r is coprime to m. Therefore,p is a unit in Z/mZ. In particular, (Z/mZ)(p) = 0.

Let X = SpecZ. If m,n are coprime, then (m) + (n) = (1). Therefore,

Xm ∪Xn = V (m,n)c = V (1)c = X.

Let p be a prime. If (p) ∈ Xm, then

(Z/mZ⊗ Z/nZ)(p) = (Z/mZ)(p) ⊗ (Z/nZ)(p) = 0⊗ (Z/nZ)(p) = 0.

Swap the roles of m and n to see that Z/mZ⊗Z/nZ localizes to 0 at every closed point of X. By Proposition 3.8,Z/mZ⊗ Z/nZ = 0.

Proof via Deferment: Set A = Z, a = (m), and M = Z/nZ. Apply Exercise 2.2 and note that aM = M inthis case.

2.2

If a is an ideal of A and M is an A-module, then

0→ a→ A→ A/a→ 0

is a short exact sequence of A-modules. Tensoring with M yields

a⊗AM → A⊗AM → (A/a)⊗AM → 0,

another exact sequence of A-modules, since the tensor product is right exact. If we define f : a ×M → aM byf(a,m) = am, then f is obviously A-bilinear and surjective. By the universal property of the tensor product,there is a unique A-linear map f⊗ : a ⊗A M → aM that satisfies f⊗(a ⊗ m) = am. Furthermore, defineg : (A/a) ×M → M/aM by g(a,m) = am. This map is obviously A-bilinear and surjective as well. By theuniversal property of the tensor product, there is a unique A-linear map g⊗ : (A/a) ⊗A M → M/aM such thatg⊗(a⊗m) = am.

On the other hand,0→ aM →M →M/aM → 0

is a short exact sequence of A-modules. Putting this together gives the following commutative diagram with exactrows, where central vertical arrow is the canonical isomorphism.

a⊗AM A⊗AM (A/a)⊗AM 0

0 aM M M/aM 0

f⊗ g⊗

2 MODULES 23

Applying the Snake Lemma to this diagram yields another exact sequence,

0→ ker g⊗ → coker f⊗ = 0.

Therefore, g⊗ is injective. Given that g⊗ is surjective as well,

g⊗ : (A/a)⊗AM →M/aM

is an isomorphism of A-modules.

2.3

Let A be a local ring with maximal ideal m and finitely generated A-modules M,N such that M ⊗A N = 0.Additionally, let κ = A/m be the residue field of A. By Proposition 2.14 and Exercise 2.15,

Mκ ⊗κ Nκ = (κ⊗AM)⊗κ (κ⊗A N)

= [(κ⊗AM)⊗κ κ]⊗A N= (κ⊗AM)⊗A N= κ⊗A (M ⊗A N)

= (M ⊗A N)κ = 0,

where the equalities are the canonical identifications. Since M,N are finitely generated A-modules, Mκ, Nκ arefinitely generated κ-vector spaces by Proposition 2.17. If d, e are the respective dimensions of Mκ, Nκ, thenMκ ⊗κ Nκ is isomorphic to κd ⊗κ κe, which can further be identified with κde by Proposition 2.14. This showsthat d or e must by zero, because Mκ ⊗κ Nκ = 0. Without loss of generality, d = 0 and Mκ = 0. On the otherhand, Mκ = M/mM by Exercise 2.2. Finally, Nakayama’s Lemma implies that M = 0.

2.4

Let Mii∈I be a family of A-modules with direct sum M .

Claim: The module M ⊗AN is canonically isomorphic to⊕

i∈I(Mi ⊗AN) for regardless of the A-module N .

Proof 1: (Construct Maps) By the universal property of the tensor product, the inclusions ιi : Mi →M inducemaps Mi⊗AN →M ⊗AN . These in turn induce a unique A-linear map f :

⊕i∈I(Mi⊗AN)→M ⊗AN such that

f(mj ⊗ n) = mj ⊗ n. On the other hand, the map M ×N →⊕

i∈I(Mi ⊗A N) given by (∑

imi, n) 7→∑

imi ⊗ nis A-bilinear. By the universal property of the tensor product, it induces a unique A-linear map g : M ⊗A N →⊕

i∈I(Mi⊗AN) such that g(∑

imi⊗n) =∑

imi⊗n. By construction, f g = idM⊗AN and g f = id⊕i∈I(Mi⊗AN)

and M ⊗A N is canonically isomorphic to⊕

i∈I(Mi ⊗A N).

Proof 2: (Yoneda) We will make use of the adjunction preceding Proposition 2.18 and the following charac-terization of the direct sum:

“There are morphisms ιi : Mi →M such that for any A-module N and any family of A-linear mapsfi : Mi → Ni∈I , there is a unique A-linear map f : M → N such that f ιi = fi for all i ∈ I.”

In particular, the map

homA(M,N)→∏i∈I

homA(Mi, N) : f 7→ (f ιi)i∈I

is a bijection, natural in N .

2 MODULES 24

Let N,P be any A-modules. We have the following natural isomorphisms,

homA(M ⊗A N,P ) = homA(M,homA(N,P ))

=∏i∈I

homA(Mi, homA(N,P ))

=∏i∈I

homA(Mi ⊗A N,P )

= homA

(⊕i∈I

(Mi ⊗A N) , P

)

By Yoneda’s Lemma, M ⊗A N is naturally isomorphic to⊕

i∈I (Mi ⊗A N).

(Remark 4)

Solution of Problem 1: (Element Chase) Let φ : N ′ → N be injective, let M be a flat A-module, and let ηbe an element of ker (idMi ⊗ φ). Then

0 = (ιi ⊗ idN )(idMi ⊗ φ)(η) = (idM ⊗ φ)(ιi ⊗ idN ′)(η).

Since M is flat, idM ⊗ φ is injective. Furthermore, for each i ∈ I, the map ιi ⊗ idN ′ is injective because it isinclusion into the ith coordinate of the direct sum M ⊗A N ′. Therefore, η = 0 and idMi ⊗ φ is injective. ByProposition 2.19, the A-modules Mi are all flat.

Now suppose that every Mi is a flat A-module and let∑

i θi be an element of ker (idM ⊗ φ). Then

(ιi ⊗ idN )(idMi ⊗ φ)(θi) = (idM ⊗ φ)(ιi ⊗ idN ′)(θi) = 0

for every i. Since idMi ⊗ φ and ιi⊗ idN are injective, θi = 0. Thus∑

i θi = 0, idM ⊗ φ is injective, and M is a flatA-module.

Solution of Problem 2: (Exactness) Let f : N ′ → N be injective. The map

idM ⊗f : M ⊗A N ′ →M ⊗A N

has the property that for each i ∈ I, this map takes the ith coordinate into the ith coordinate. In particular, it isinjective if and only if it is injective in each coordinate. By Proposition 2.19, the equivalence follows.

2.5

Define f :⊕

n≥0A→ A[x] by f(∑

n≥0 anen) =∑

n≥0 anxn. The map f is obviously an A-linear isomorphism.

By Exercise 2.4,⊕

n≥0A is a flat A-module due to each A being a flat A-module. Therefore, A[x] is a flat A-algebrabecause it is A-linearly isomorphic to a flat A-module.

2.6

Let M be an A module and define M [x] as the space of polynomials with coefficients in M . Let E be theendomorphism ring of M and let E′ be the endomorphism ring of M [x] (as abelian groups). Letting E act onM [x] diagonally gives an injective ring homomorphism E → E′. Since M is an A-module, we already have a ringhomomorphism A → E. Therefore, M [x] is an A-module via the composition A → E → E′. Extend A → E′

to A[x] → E′ be letting x ·mxn := mxn+1 and extending by linearity. By the Universal Property of PolynomialRings, this map is uniquely determined – and it makes M [x] into an A[x]-module.

We have the following identifications

A[x]⊗AM =

⊕n≥0

Axn

⊗AM =⊕n≥0

(Axn ⊗AM) =⊕n≥0

Mxn = M [x].

2 MODULES 25

2.7

If p is a prime ideal of A, then (A/p)[x] is a domain. By Exercise 2.6, this is isomorphic to A[x] ⊗A (A/p)as an A[x]-module. Furthermore, Exercise 2.2 shows that this is isomorphic to A[x]/p[x] as an A-module. Thecomposite map f : (A/p)[x]→ A[x]/p[x] is an A-linear map such that f(

∑n≥0 anx

n) =∑

n≥0 anxn. Thus f is an

A[x]-algebra isomorphism, A[x]/p[x] is a domain, and p[x] is a prime ideal of A[x].

On the other hand, (2) is a maximal ideal of Z, but

Z[x]/(2)[x] = (Z/2Z) [x] = F2[x]

is not a field (x is not a unit).

2.8

(i) Let 0→ P ′ → P be exact, and let M,N be flat A-modules. Tensoring with M shows that

0→ P ′ ⊗AM → P ⊗AM

is exact. Now tensor with N to show that

0→ (P ′ ⊗AM)⊗A N → (P ⊗AM)⊗A N

is exact. By the Proposition 2.14 (ii),

0→ P ′ ⊗A (M ⊗A N)→ P ⊗A (M ⊗A N)

is exact. Thus M ⊗A N is a flat A-module.

(ii) Let 0 → P ′ → P be an exact sequence of A-modules, let B be a flat A-algebra, and let N be a flatB-module. Extend scalars by B to get the exact sequence of B-modules

0→ B ⊗A P ′ → B ⊗A P.

Tensoring with N yields another exact sequence of B-modules,

0→ N ⊗B (B ⊗A P ′)→ N ⊗B (B ⊗A P ).

By Exercise 2.15,0→ (N ⊗B B)⊗A P ′ → (N ⊗B B)⊗A P

is an exact sequence of A-modules. Thus N = N ⊗B B is a flat A-module.

2.9

Let0→M ′

f−→Mg−→M ′′ → 0

be an exact sequence of A-modules with M ′,M ′′ finitely generated.

Constructive Proof: Let x1, . . . , xn be generators for M ′ and g(y1), . . . , g(ym) be generators for M ′′. Givenz in M , there are b1, . . . , bm in A such that g(z) =

∑mi=1 big(yi). Thus z −

∑mi=1 biyi is in the kernel of g.

By exactness of the sequence, there are a1, . . . , an in A such that z =∑n

j=1 ajf(xj) +∑m

i=1 biyi. Therefore,f(x1), . . . , f(xn), y1, . . . , ym is a finite generating set for M .

Non-Constructive Proof: Let α : A⊕n → M ′ and β : A⊕m → M ′′ be surjective A-linear maps. Since A⊕m

is a free A-module, it is projective. So there is an A-linear map γ : A⊕m →M such that g γ = β. Define

δ = f α⊕ γ : A⊕(n+m) = A⊕n ⊕A⊕m →M

via the universal property of the direct limit. Then we have the following commutative diagram with exact rows:

2 MODULES 26

0 A⊕n A⊕(n+m) A⊕m 0

0 M ′ M M ′′ 0

α δ β

Apply the Snake Lemma to get the exact sequence

0 = cokerα→ coker δ → cokerβ = 0.

Therefore, δ is surjective.

2.10

Let A be a ring, let a be an ideal in its Jacobson radical, and let u : M → N be a homomorphism of A-modulessuch that N is finitely generated. Let v, w be the maps induced from u by restriction and passing to the quotient.Then the diagram

0 aM M M/aM 0

0 aN N N/aN 0

v u w

is commutative and has exact rows. Applying the Snake Lemma yields an exact sequence

coker vf−→ cokeru→ cokerw.

Note that im f = a cokeru.

Suppose that w is surjective. Then f is surjective,

cokeru = im f = a cokeru,

and cokeru is finitely generated because it is a quotient of N . By Nakayama’s Lemma, cokeru = 0. Therefore, uis surjective.

2.11

Let A be a non-zero ring and let φ : A⊕m → A⊕n be a homomorphism of free A-modules.

If φ is surjective, then the sequence of A-modules,

A⊕mφ−→ A⊕n → 0

is exact. Tensor this sequence with κ = A/m for some maximal ideal m of A to get an exact sequence

κ⊕m → κ⊕n → 0

of κ-vector spaces. Since m,n are the respective dimensions of κ⊕m and κ⊕n, m ≥ n.

On the other hand, suppose that m > n and embed A⊕n as the submodule of the first n coordinates of A⊕m.The containment,

φ(A⊕m) ⊆ A⊕n ( A⊕m,

yields the polynomial equation,φd + a1φ

d−1 + . . .+ ad−1φ+ ad = 0,

2 MODULES 27

by the Cayley-Hamilton Theorem. Let d be the minimal degree for such an equation and let π : A⊕m → A⊕(m−n)

be the projection onto the last m− n coordinates. Then

0 = π(φd + a1φd−1 + . . .+ ad−1φ+ ad) = adπ

implies that ad = 0. Since d is minimal, there is an x in A⊕m such that

y = (φd−1 + a1φd−2 + . . .+ ad−1)(x)

is non-zero, but φ(y) = 0. Therefore, if φ is injective, then m ≤ n.

Putting the above results together shows that if φ is an isomorphism, then m = n.

2.12

Let M be a finitely generated A-module, φ : M → An a surjective A-module homomorphism, and e1, . . . , enbe the standard basis of An. Let f1, . . . , fn be lifts of e1, . . . , en to M via φ, and define the A-linear mapψ : An →M : ei 7→ fi. Since φ ψ = idAn , M is isomorphic to im(ψ φ)⊕ ker(φ) by the map

m 7→ ((ψ φ)(m),m− (ψ φ)(m)).

Therefore, kerφ is a quotient of the finitely generate A-module M , and is itself finitely generated as a consequence.(Remark 5)

2.13

Let f : A → B be a ring homomorphism and let N be a B-module. Give N an A-module structure byrestriction of scalars, define

g : N → NB : n 7→ 1⊗ n,

and definep : NB → N : b⊗ y 7→ by.

Since p g = idN , g is injective and NB is isomorphic to im(g)⊕ ker(p) by the map

b⊗ y 7→ ((g p)(b⊗ y), b⊗ y − (g p)(b⊗ y)).

(Remark 5)

2.14

We need only show that µi = µj µij . If xi ∈Mi, then

(µj µij)(xi) = µij(xi) +D = µij(xi) + (xi − µij(xi)) +D = xi +D = µi(xi).

(Remark 6)

2.15

Let x ∈M . By the construction of M , x =∑

i∈I xi +D for some xi ∈Mi, and I is finite. Since I is finite, itselements have a common upper bound, k. Then

x =∑i∈I

xi +D =∑i∈I

xi −∑i∈I

(xi − µik(xi)) +D =∑i∈I

µik(xi) +D = µk

(∑i∈I

µik(xi)

).

2 MODULES 28

If µi(xi) = 0, then xi ∈ D. Let xi =∑n

l=1(xjl − µjlkl(xjl)) be a representation of xi as a sum of elements of Dwith jl < kl for each l. Without loss of generality, j1 is minimal. If j1 = j2, then

(xj1 − µj1k1(xj1)) + (xj2 − µj2k2(xj2)) = ((xj1 + xj2)− µj1k1(xj1 + xj2)) + µj1k1(xj2)− µj2k2(xj2).

That is, µj1k1(xj2) − µj2k2(xj2) ∈ D and is 0 in coordinate j1. So we may assume that j1 < jl for all l > 1. Ifj1 6= i, then xj1 = 0 by the independence of coordinates in the direct sum.

Without loss generality, j2 is minimal in I \ i. If j2 6= k1, then xj2 = 0 by the independence of coordinates.Therefore, xj2 = µik1(xi), and

xi =n∑l=1

(xjl − µjlkl(xjl))

= xi − µik1(xi) + xj2 − µj2k2(xj2) +

n∑l=3


= xi − µk1k2(µik1(xi)) +

n∑l=3


= xi − µik2(xi) +

n∑l=3

(xjl − µjlkl(xjl)).

By repeated application of the above argument we arrive at

xi = xi − µikn(xi).

Therefore, µikn(xi) = 0 in Mkn . (Remark 7)

Alternative Proof: We will use the construction of the direct limit in Remark 6. The first statement followsfrom the construction of M as a quotient of the disjoint union

∐i∈IMi. The second statement follows directly

from the definition of ∼.

2.16

Let αi : Mi → N be a family of A-module homomorphisms such that

αj µij = αi

for i ≤ j, and let M = lim−→i∈IMi. If i, j ≤ k such that µi(xi) = µj(xj), then

0 = µi(xi)− µj(xj) = µk(µik(xi))− µk(µjk(xj)) = µk(µik(xi)− µjk(xj)).

By Exercise 2.15 there is l ≥ k such that µil(xi) = µjl(xj). Thus

αi(xi) = αl(µil(xi)) = αl(µjl(xj)) = αj(xj).

Therefore, we may define α : M → N by α(µi(xi)) = αi(xi). The preceding calculations have show that this iswell defined and satisfies the requirement that α µi = αi. Furthermore, this requirement forces the map to beunique. (Remark 8)

Alternative Proof: We will use the construction of the direct limit in Remark 6. Define

α : M → N : [(x, i)] 7→ αi(x).

Suppose that µik(x) = µjk(y). Then

αi(x) = αk(µik(x)) = αk(µjk(y)) = αj(y).

So α is well-defined. It is easy to verify that α is A-linear. Additionally, α(µi(x)) = αi(x) by construction.Therefore, this condition uniquely determines α.

2 MODULES 29

2.17

Let Mii∈I be a family of submodules of an A-module, such that for each i, j there exists k such thatMi +Mj ⊆Mk. Define i ≤ j to mean Mi ⊆Mj and define µij : Mi →Mj as the inclusion map.

Since Mj ⊆⋃i∈IMi for every j,

∑Mi ⊆

⋃Mi. On the other hand,

⋃Mi is a union of subsets of

∑Mi. Thus⋃

Mi ⊆∑Mi.

For each j, let αj : Mj →⋃Mi be the inclusion map, and let α : lim−→Mi →

⋃Mi be the unique induced

A-module homomorphism. If x ∈⋃Mi, then there is j such that x ∈ Mj . Thus, α(µj(x)) = αj(x) = x, and α is

surjective. On the other hand, α(µj(xj)) = αj(xj) = xj shows that α is injective. Therefore, α is an isomorphism.

Take Mii∈I to be the family of finitely generated submodules of M . Order I by i ≤ j, if Mi ⊆Mj . Then

M =⋃i∈I

Mi = lim−→Mi.

2.18

Let M = (Mi, µij) and N = (Ni, νij) be direct systems of A modules over the same directed set, I. Let M,Nbe the respective direct limits and µi : Mi → M,νi : Ni → N be associated homomorphisms. A homomorphismΦ : M→ N of direct systems is defined to be a collection of A-module homomorphisms φi : Mi → Ni such that

φj µij = νij φi

for all i ≤ j.

Let Φ : M→ N be a homomorphism of direct systems. For each i, define the map αi = νi φi : Mi → N . Wecheck that the collection of maps αi are compatible by this derivation:

αj µij = νj φj µij = νj νij φi = νi φi = αi.

By the universal property of the direct limit there is a unique map φ : M → N such that φ µi = αi = νi φi forevery i.

2.19

Let Mf−→ N

g−→ P be an exact sequence of direct systems over a common index set I. Let M,N,P be theirrespective direct limits and let f : M → N, g : N → P be the unique maps obtained from f ,g as in Exercise 2.18.Then we have

g f µi = g νi fi = ρi gi fi = 0

for every i. By Exercise 2.15, the images of the maps µi cover M . Therefore f g = 0.

Let n ∈ ker(g). By Exercise 2.15 there is ni ∈ Ni such that νi(ni) = n. Thus, by Exercise 2.18,

0 = g(n) = (g νi)(ni) = (ρi gi)(ni).

We again cite Exercise 2.15 to obtain j ≥ i such that

0 = (ρij gi)(ni) = (gj νij)(ni).

Since νij(ni) ∈ ker(gj) and Mf−→ N

g−→ P is exact, there is mj ∈Mj such that fj(mj) = νij(ni). Let m = µj(mj).We have obtained the following:

f(m) = (f µj)(mj) = (νj fj)(mj) = (νj νij)(ni) = νi(ni) = n.

Therefore, ker(g) ≤ im(f).

2 MODULES 30

2.20

Let M,D be as in the statement of Exercise 2.14, and let M ′, D′ be their analogues for the system (Mi ⊗AN,µij ⊗ idN ). So we have two short exact sequences of A-modules

0→ D →M → lim−→Mi → 0 (1)

and0→ D′ →M ′ → P → 0. (2)

Note that

M ⊗A N =

(⊕i∈I

Mi

)⊗A N =

⊕i∈I

(Mi ⊗A N) = M ′,

andD ⊗A N → D′ : (xi − µij(xi))⊗ n 7→ xi ⊗ n− (µij ⊗ idN ) (xi ⊗ n)

is clearly surjective.

Tensor (1) with N to obtain the following commutative diagram with exact rows (The right vertical arrow isinduced from the middle vertical arrow by passing to the quotient):

D ⊗A N M ⊗A N(lim−→Mi

)⊗A N 0

0 D′ M ′ P 0

Apply the Snake Lemma and recall that the middle arrow is an isomorphism and the left arrow is surjectiveto see that the rightmost vertical arrow has trivial kernel and cokernel.

2.21

Let i ∈ I. For each j ≥ i, ring homomorphism αij : Ai → Aj makes Aj into an Ai-algebra. Therefore (Remark9),

lim−→j∈Ji

Aj = lim−→j∈I

Aj = A

has a canonical Ai-module structure, and this is true for every i ∈ I. Let αi(x), αj(y) ∈ A. Define

αi(x) · αj(y) := x · αj(y),

where the operation on the right is scalar multiplication of A as an Ai-module. The associative and distributiveproperties are inherited from the Ai-module axioms. Let us check that this is well-defined and commutative. Letk ≥ i, j. Then αi(x) = αk(αik(x)) and αj(y) = αk(αjk(y)). Furthermore, the definition of scalar multiplicationyields

x · αj(y) = x · αk(αjk(y)) = αk(αik(x)αjk(y)).

The expression on the far right is obviously commutative and independent of the choice of x, y as representatives.By construction, the maps Ai → A are ring homomorphisms.

Suppose that A = 0. Let i ∈ I. Then αi(1) = 0. By Exercise 2.15, there is j ≥ i such that 1 = µij(1) = 0.Therefore, Aj = 0.

2 MODULES 31

2.22

Let “ρ” be the symbol for the structure maps on the system of Ri’s. Let i, j ∈ I with i ≤ j. Since Rj is anAj-module and Aj is an Ai-algebra, we can view Rj as an Ai-module via restriction of scalars. Therefore,

lim−→j∈Ji

Rj = lim−→i∈I

Ri =: R

is an Ai-module. Furthermore, ιj : Rj → Aj is Ai-linear for all j ≥ i. By Exercise 2.18, there is a unique Ai-linearmap ι : R→ A such that ιρi = ιi for all i ∈ I. These Ai-linear maps are compatible for varying i ∈ I. Therefore,R→ A is A-linear. By Exercise 2.19, ι is injective. So R may be canonically identified with an ideal of A.

By Exercise 2.15, every element of R is nilpotent, so R is contained in the nilradical of A. Suppose that x ∈ Aand xn = 0 for some n ≥ 0. Let x = αi(ξ) for some i ∈ I. Let j ≥ i such that 0 = αij(ξ

n) = αij(ξ)n. Then

αij(ξ) = ιj(y) for some y ∈ Rj . Then we have

x = αj(αij(ξ)) = αj(ιj(y)) = ι(ρj(y)).

So every element of the nilradical of A is in the image of ι.

Suppose that Ai is an integral domain for each i ∈ I. Let x, y ∈ A such that xy = 0. By Exercise 2.16, x, y liftto a common index k such that xy = 0 in Ak. If x 6= 0 in A, then x 6= 0 in Ak. Since Ak is an integral domain,y = 0 in Ak; and y = 0 in A as a consequence. Therefore, A is an integral domain.

2.23

This Exercise asks no questions. I’m suppose the reader is expected to verify the assertions. These follow fromExercises 2.14, 2.15, 2.16, 2.18, 2.19, and 2.21.

2.24

(Remark 10)

(i)⇒(ii) Let N be an A-module. Let F∗ be a free resolution of N . Since M is flat, the functor M⊗A is exact.So the complex M ⊗A F∗ is exact. Therefore, TorAn (M,N) = Hn(M ⊗A F∗) = 0.

(ii)⇒(iii) This is immediate.

(iii)⇒(i) Let0→ N ′ → N → N ′′ → 0

be a short exact sequence of A-modules. Apply the functor TorA∗ (M, ·) to get the exact sequence

0 = TorA1 (M,N ′′)→M ⊗A N ′ →M ⊗A N →M ⊗A N ′′ → 0.

By Proposition 2.19, M is flat.

2.25

Let0→ N ′ → N → N ′′ → 0

be a short exact sequence of A-modules, let N ′′ be flat, and let f : M ′ → M be an injective A-linear map. TheTor long exact sequence yields the following commutative diagram with exact rows:

0 = TorA1 (M ′, N ′′) M ′ ⊗A N ′ M ′ ⊗A N M ′ ⊗A N ′′ 0

0 = TorA1 (M,N ′′) M ⊗A N ′ M ⊗A N M ⊗A N ′′ 0

f⊗idN′ f⊗idN f⊗idN′′

2 MODULES 32

By the Snake Lemma, the sequence

0→ ker(f ⊗ idN ′)→ ker(f ⊗ idN )→ ker(f ⊗ idN ′′) = 0

is exact. Therefore, the map ker(f ⊗ idN ′)→ ker(f ⊗ idN ) is an isomorphism and N is flat if and only if N ′ is flat.

2.26

(⇒) This follows from Exercise 2.24.

(⇐) Let (Mi)i∈I be a directed system of A-modules with direct limit M and let N be an A-modules. LetF∗ → N be a flat resolution of N . By Exercises 2.19 and 2.20,

TorAn (M,N) = Hn(M ⊗A F∗) = Hn

(lim−→ (Mi ⊗A F∗)

)= lim−→Hn(Mi ⊗A F∗) = lim−→TorAn (Mi, N). (3)

Suppose that N is an A-module such that TorA1 (A/a, N) = 0 for all finitely generated ideals a of A. Let b bean ideal of A and let Σ be the directed family of all finitely generated subideals of b (as in Exercise 2.17). By theexactness of direct limits and (3),

TorA1 (A/b, N) = TorA1

(lim−→a∈Σ

A/a, N

)= lim−→

a∈Σ

TorA1 (A/a, N) = 0.

So TorA1 (·, N) vanishes on cyclic A-modules.

Let M =∑n

i=1Ami be a finitely generated A-module, and define Mk :=∑k

i=1Ami for 1 ≤ k ≤ n. Note thatA→Mk/Mk−1 : 1 7→ mk is surjective, so Mk/Mk−1 is cyclic. The Tor functor yields exact sequences

TorA1 (Mk−1, N)→ TorA1 (Mk, N)→ TorA1 (Mk/Mk−1, N) = 0,

where the equality is due to Mk/Mk−1 being a cyclic A-module. By induction on k, there is a surjective map fromTorA1 (M1, N) onto TorA1 (Mn, N) = TorA1 (M,N). Since M1 is a cyclic A-module, TorA1 (M,N) = 0.

Let M be any A-module and let (Mi)i∈I be the directed family of its finitely generated submodules. By (3),TorA1 (M,N) = 0. Since M was arbitrary, Exercise 2.24 implies that N is flat. (Remark 2.26)

2.27 Incomplete

(i)⇒(ii) Let A be absolutely flat with element a. Since the inclusion (a) → A is injective and A/(a) is flat,the tensored map,

(a)/(a)2 = (a)⊗A A/(a)→ A⊗A A/(a) = A/(a)

is also injective. However, xa 7→ 0 for all x ∈ A. Therefore, (a)/(a)2 = 0.

(ii)⇒(iii) If a is an element of A, then (a) = (a)2 implies that there is an element b such that a = a2b.Furthermore, (ab)2 = a2b · b = ab shows that ab is idempotent, and a · ab = a2b = a shows that (a) = (ab). Thusevery principal ideal is generated by an idempotent element of A.

If e, f are idempotent elements of A, then

e(e+ f − ef) = e+ ef − ef = e

and(e+ f − ef)f = ef + f − ef = f

show that (e, f) = (e+ f − ef). Additionally,

(e+ f − ef)2 = e(e+ f − ef) + f(e+ f − ef)− ef(e+ f − ef) = e+ f − ef.

2 MODULES 33

By induction, every finitely generated ideal is principal, generated by an idempotent element. Thus, if a is afinitely generated ideal, there is an idempotent g such that a = (g). Then A = (g) ⊕ (1 − g) shows that a is adirect summand of A.

(iii)⇒(i) Let N be an A-module, a a finitely generated ideal, and A = a ⊕ b. Additionally, let ι : a → A bethe natural inclusion and let π : A→ a be the natural projection. Then π ⊗ idN is a left inverse for ι⊗ idN , andι⊗ idN is injective. Thus TorA1 (A/a, N) = 0. By Exercise 2.26, N is flat and A is absolutely flat.

(Remark 12)

2.28

If a is an element in a boolean ring A, then a2 = a implies that (a) = (a)2. Thus A is absolutely flat byExercise 2.27.

Let A be a ring with a map n : A → Z such that n(x) ≥ 2 and xn(x) = x for every x. Then xn(x)−2 · x2 = ximplies that (x) ⊆ (x)2 ⊆ (x). Thus A is absolutely flat.

Let f : A → B be a ring homomorphism with A absolutely flat. If y is in the image of f , then there are a, xin A such that x = ax2 and

y = f(x) = f(ax2) = f(a)y2.

Thus (y) = (y)2 and the image of f is absolutely flat.

Let A be an absolutely flat local ring with maximal ideal m. Let x be an element of m and a an element ofA such that x = ax2. Then ax is a non-unit idempotent in a local ring. By Exercise 1.12, ax is zero. Therefore,x = ax · x = 0. As a consequence, m is the zero ideal and A is a field.

Let A be absolutely flat with x = ax2. Then x(1− ax) = 0. If x is not a zero-divisor, then 1 = ax and x is aunit.

3 RINGS AND MODULES OF FRACTIONS 34

3 Rings and Modules of Fractions

(Remark 13)

3.1 Current

Let S be a multiplicatively closed subset of the ring A and let M =∑k

i=1Axi be an A-module. By Corollary3.4, S−1M = 0 if and only if S−1Axj = 0 for every j. This is equivalent to the requirement that S ∩AnnA(Axj) isnon-empty for each j. On the other hand, the existence of s in S such that sM = 0 is equivalent to the conditionthat S ∩AnnA(M) is non-empty.

The following inclusions show that S ∩ AnnA(M) is non-empty precisely when every S ∩ AnnA(Axj) is non-empty.

k∏i=1

(S ∩AnnA(Axi)) ⊆k⋂j=1

(S ∩AnnA(Axj))

= S ∩

k⋂j=1

AnnA(Axj)

= S ∩AnnA(M)

⊆ S ∩AnnA(Axj),

where the step from the second to third line uses Exercise 2.2 (i).

3.2

Let a be an ideal of A, and define S = 1 + a. If a1+b is an element of S−1a, then

1− a

1 + b=

1 + b− a1 + b

is a unit in S−1A. Thus S−1a is in the Jacobson radical of S−1A.

Let M be a finitely generated A-module such that aM = M . Then (S−1a)(S−1M) = S−1M and S−1a is inthe Jacobson radical of S−1A. By Nakayama’s Lemma, S−1M = 0. By Exercise 3.1 there is some s in S = 1 + asuch that sM = 0.

3.3

Let S, T be multiplicatively closed subsets of A, let f : A→ S−1A be the natural map, and let U = f(T ). Notethat ST is a multiplicatively closed subset of A and let g : A→ (ST )−1A be the natural map. Since g(s) = s

1 is aunit in (ST )−1A for each s in S, g has a unique lift h : S−1A→ (ST )−1A such that h f = g by Proposition 3.1.

If u = f(t) is in U = f(T ), then

h(u) = h(f(t)) = g(t) =t

1

is a unit in (ST )−1A. If 0 = h(a/s) = g(a)g(s)−1 = a/s in (ST )−1A, then there are elements s0, t0 in S, T suchthat s0t0a = 0 in A. Thus

0 =1

s0· f(s0t0a) = f(t0)

a

1

in S−1A. Finally, for every element ast of (ST )−1A,

a

st= g(a)g(st)−1 = (g(a)g(s)−1)g(t)−1 = h(a/s)h(f(t))−1.

By Corollary 3.2, h lifts to an isomorphism from U−1(S−1A) to (ST )−1A.


3.4

Let f : A → B be a ring homomorphism, S a multiplicatively closed subset of A, and T = f(S). Consider Bas an A-module by restriction of scalars, and form the S−1A-module S−1B. Let b

s = b′

s′ in S−1B. Then

f(s′′)(f(s′)b− f(s)b′) = s′′ · (s′ · b− s · b′) = 0

in B. Thus, the map g : S−1B → T−1B : bs 7→

bf(s) is well defined and bijective. We show that it is additive and

S−1A-homogeneous of degree 1.

The calculation,

g

(b

s+b′

s′

)= g

(s′ · b+ s · b′

ss′

)=s′ · b+ s · b′

f(ss′)=f(s′)b+ f(s)b′

f(s)f(s′)

=b

f(s)+

b′

f(s′)= g

(b

s

)+ g

(b′

s′

),

shows that g is additive.

On the other hand,

g

(a

s· bs′

)= g

(a · bss′

)=

a · bf(ss′)

=f(a)

f(s)

b

f(s′)= (S−1f)

(as

)g

(b

s′

)=a

s· g(b

s′

)shows that g is S−1A-homogeneous of degree 1. Therefore, g is an isomorphism of S−1A-modules.

3.5

By Corollary 3.12, localization commutes with taking nilradicals. If A is a ring such that (R(A))p = R(Ap) = 0for every prime ideal p, then R(A) = 0 by Proposition 3.8.

Let A = Z/6Z. Then (2), (3) are maximal ideals, and 2 · 3 = 0 shows that every prime ideal contains 2 or 3.Thus SpecA = (2), (3). Furthermore, 2 · 3 = 0 implies that A(2) = F2 and A(3) = F3. Yet, A is not a domain.

3.6

Let A be a non-zero ring and let Σ be the set of multiplicatively closed subsets of A that do not contain 0.Since A is not the zero ring, 1 is an element of Σ. Let Sii∈I be a chain in Σ and define S =

⋃i∈I Si. Then

1 ∈ S and if x, y ∈ S, then there is some j ∈ I such that x, y ∈ Sj . Thus xy ∈ Sj ⊆ S. Furthermore, if 0 were inS, then it would have to be in some Si. Since S is a multiplicatively closed subset of A not containing 0, it is anelement of Σ. By Zorn’s Lemma, Σ has maximal elements.

Let S be a maximal element of Σ and let x be an element of A. Since S is maximal in Σ, x is not in S ifand only if the multiplicative sub-monoid of A generated by x and S contains 0. This in turn is equivalent to thecondition that x is a nilpotent element of S−1A. Therefore, A \ S is the contraction of the nilradical of S−1A,hence an ideal. Since S is multiplicatively closed, p = A \ S is a prime ideal not meeting S. By Proposition 3.11(iv), S−1p = R(Ap) is a minimal prime ideal, and thus p is a minimal prime ideal.

Let p be a minimal prime ideal of A, x an element, and define S = A \ p. Then S−1p = R(Ap). Thus x isnot in S if and only if x is nilpotent in Ap. This in turn is equivalent to the condition that the multiplicativesub-monoid of A generated by x and S contains 0. Thus S is a maximal element of Σ.

3.7

Let S be a multiplicatively closed subset of the ring A,


(i) If S is saturated, then A× ⊆ S because 1 is in S. If 0 is in S then S = A, and A \ S is the empty union ofprime ideals. Suppose that 0 is not in S, and let mii∈I be the set of maximal ideals of S−1A. If a/s is a unit inS−1A, then there is some s′ in S such that s′(a− s) = 0 in A. In other words, s′a = s′s is in S. Thus, a is in S,because S is saturated. Therefore, A \ S ⊆

⋃i∈I mi

c. Conversely,⋃i∈I mi

c ⊆ A \ S by Proposition 3.11 (iv).

Now, suppose that A \ S =⋃i∈I pi where the pi are prime ideals. If xy is in S, then xy is in the complement

of each prime ideal pi. By primality, x, y are each in the complement of each prime ideal pi. Thus x, y are both inS, and S is saturated.

(ii) Let Σ be the set of X such that S ⊆ X ⊆ A and X is saturated. Since A is in Σ, it is non-empty. Order Σby reverse inclusion, let Xii∈I be a chain, and let X =

⋂i∈I Xi. Then X is obviously a multiplicatively closed

set containing S. Furthermore, A \ X =⋃i∈I A \Xi is a union of prime ideals by (i). Thus X is saturated and an

element of Σ. By Zorn’s Lemma, Σ has minimal elements. If S1, S2 are elements of Σ, then S1 ∩ S2 is in Σ. Thusthere is a unique minimal element, S in Σ.

Let mii∈I be the set of maximal ideals in S−1A. Then A \(⋃

i∈I mci

)is a saturated subset of A, by (i), that

contains S. Thus A \(⋃

i∈I mci

)is in Σ and S ⊆ A \

(⋃i∈I m

ci

). Since S is saturated, A \ S =

⋃j∈J qj where the

qj are prime ideals of A not meeting S. Therefore,⋃j∈J qj ⊇

⋃i∈I m

ci .

On the other hand, if p is a prime ideal of A not meeting S, then p does not meet S, and S−1p is a prime idealof S−1A. Thus

⋃j∈J qj ⊆

⋃i∈I m

ci holds as well.

Let a be an ideal of A and let S = 1 + a. Note that every factor of an element of S is in S, and that thecollection of these factors is a multiplicatively closed subset of A. Thus S is the set of factors of elements of S.We make this more concrete as follows:

If x is in S, then there is y in S such that xy ≡ 1 mod a. Conversely, if xy ≡ 1 mod a, then there is a in asuch that xy = 1 + a, and x, y are in S. Thus

S = π−1((A/a)×

),

where π : A→ A/a is the natural projection.

3.8

Let S ⊂ T be multiplicatively closed subsets of the ring A, and let

φ : S−1A→ T−1A

be the lift of f : A→ T−1A as given by Proposition 3.1.

(i)⇒(ii) If t be an element of T , then φ−1(

1t

)· t1 = φ−1

(1t ·

t1

)= 1 in S−1A.

(ii)⇒(iii) Let xs be the inverse of t

1 in S−1A. The equation xts = 1

1 in S−1A implies that there is s′ in S suchthat s′(xt− s) = 0 in A. Hence, (s′x)t = ss′ is in S.

(iii)⇒(iv) For each t in T there is x in A such that xt is in S. Thus t is in S.

(iv)⇒(v) Since T ⊆ S, A \ T contains every prime ideal not meeting S, by Exercise 3.7. Thus every primeideal meeting T also meets S.

(v)⇒(i) Note that φ : S−1A → T−1A is a homomorphism of S−1A-modules. If p is a prime ideal of S−1A,then S ⊆ T ⊆ A \ pc. By Exercise 3.3, (S−1A)p = Apc = (T−1A)p, and φp is the identity on Ap, hence bijective.By Proposition 3.9, φ is bijective.

3.9

Let S0 be the set of all non-zero-divisors in the non-zero ring A. Then S0 is obviously a saturated multiplica-tively closed subset and its complement D is a union of prime ideals by Exercise 3.7. If p is a minimal prime ideal,


then S = A \ p is a saturated, maximal multiplicatively closed set not containing 0, by Problems 3.6 and 3.7. Ifx is not in S, then 0 is in the multiplicative monoid generated by S and x. Therefore x is a zero divisor, andp = A \ S ⊆ D.

(i) Let S be a multiplicatively closed subset of A containing a non-zero, zero-divisor, d, annihiliated by c. Then

c

1=cd

d=

0

d= 0

shows that A → S−1A is not injective. On the other hand, if a/1 = 0 in S−10 A, then there is s in S0 such that

sa = 0 in A. Since s is not a zero-divisor, a = 0. Thus, the canonical map A→ S−10 A is injective.

(ii) If d is a zero-divisor of A, annihilated by c, then ds ·

c1 = cd

s = 0 for every s in S0. Therefore, if a/s is not azero-divisor, then a is in S0 = A \D, and a/s is a unit with inverse s/a.

(iii) If A = A× ∪D, then S0 = A×. Let f : A→ S−10 A be the canonical map. The f is surjective, since

a

s=as−1

ss−1=as−1

1= f(as−1),

and f is injective, since 0 = f(a) = a1 implies that there is s in S such that as = 0 in A. Thus 0 = ass−1 = a.

Therefore, f is an isomorphism.

3.10

Let A be a ring.

(i) Suppose A is absolutely flat, S is a multiplicatively closed subset, and N is an S−1A-module. Then Nis a flat A-module by restriction of scalars. Now extend scalars to get NS−1A = S−1A ⊗A N , which is a flatS−1A-module by Exercise 2.20. Furthermore, NS−1A = S−1N by Proposition 3.5. Finally, S−1N = N because Nis an S−1A-module. Thus N is a flat S−1A-module, and S−1A is absolutely flat.

(ii) If A is absolutely flat, then Am is an absolutely flat local ring for each maximal ideal m, hence a field.

On the other hand, let A be a ring that localizes to a field at each maximal ideal, m, and let N be an A-module. Then Nm is an Am-vector space, hence a free Am-module. Therefore, Nm is a flat Am-module. Since mwas arbitrary, N is a flat A-module by Proposition 3.10.

3.11

Let A be a ring.

(iv)⇒(iii) Every Hausdorff space is T1.

(iii)⇒(ii) Closed points of Spec(A) correspond to maximal ideals by Exercise 1.18 (i).

(ii)⇒(i) By the Lattice Isomorphism Theorem, we may identify the maximal ideals of A and A/R(A). Let mbe such an ideal. Then

(A/R(A))m = Am/R(Am) = Am/m

is a field, and A/R(A) is absolutely flat.

(i)⇒(iv) By Exercise 1.21 (iv), Spec(A) and Spec(A/R(A)) are naturally homeomorphic. Without loss ofgenerality, R(A) = 0 and A is absolutely flat. Let p be a prime ideal of A contained in a maximal ideal m. SinceAm is a field, it has only one prime ideal, the zero ideal. Thus p = m by Proposition 3.11 (iv), and every primeideal of A is maximal.

Suppose that m, n are distinct maximal ideals. Then there is f ∈ m and g ∈ n such that f + g = 1. Onthe other hand, the elements of m vanish in Am. So there is s not in m such that sf = 0. Together, we haves = sf + sg = sg is an element of n and f = 1 − g is not. Therefore, m ⊆ Xs, n ⊆ Xf and Xs ∩Xf = Xsf = ∅.So Spec(A) is Hausdorff.


In general, Spec(A) is quasi-compact by Exercise 1.17 (v). If it is Hausdorff, then it is compact. Let X be asubset of Spec(A) containing distinct points x, y, and let Xf contain x but not y. Since Xf is a quasi-compactsubset of a Hausdorff space, it is closed. Thus Xf ∩X,X \Xf is a disjoint cover of X by non-empty, relativelyopen sets, and X is disconnected. Therefore, the only connected subsets of Spec(A) are singletons.

3.12

Let A be a domain, M an A-module, and T (M) the set of torsion elements of M . If x, y ∈ T (M), a ∈ A, andb, c are respectively non-zero elements of AnnA(x),AnnA(y), then

bc(x+ ay) = c(bx) + ab(cy) = 0,

and bc 6= 0 because A is a domain. Thus T (M) is a submodule of M .

(i) Let x ∈ M, 0 6= a ∈ A such that ax ∈ T (M). Then there is some non-zero b ∈ AnnA(ax), and (ba)x =b(ax) = 0. Since A is a domain, ba is a non-zero element of AnnA(x), and x ∈ T (M). Thus T (M/T (M)) = 0,and M/T (M) is torsion-free.

(ii) Let f : M → N be an A-module homomorphism, x ∈ T (M), and a a non-zero element of AnnA(x). Thenaf(x) = f(ax) = f(0) = 0. Thus f(T (M)) ⊆ T (N).

(iii) Suppose

0→M ′f−→M

g−→M ′′

is an exact sequence of A-modules, and consider the induced sequence

0→ T (M ′)T (f)−−−→ T (M)

T (g)−−−→ T (M ′′).

Since it is the restriction of an injective function, T (f) is injective. Similarly, T (g) T (f) = 0, because g f = 0.If x ∈ T (M) such that T (g)(x) = 0, then g(x) = 0. Since the first sequence is exact, and T (M) is a submoduleof M , there is y ∈ M ′ such that f(y) = x. If a is a non-zero element of AnnA(x), then f(ay) = af(y) = ax = 0.Since f is injective, ay = 0. Thus y ∈ T (M ′) and T (f)(y) = x. Therefore the induced sequence on torsion isexact.

(iv) Let K be the field of fractions of A and define φ : M → K ⊗AM by f(x) = 1⊗ x. If x is in T (M), thenit is annihilated by some a in A. Thus φ(x) = 1 ⊗ x = a−1 ⊗ ax = 0, and x is in the kernel of φ. The followingdemonstrates the converse.

The family Aξξ∈K is a family of A-submodules of K. Given ξ, η ∈ K, there are non-zero a, b in A such thataξ, bη are in A. Thus Aξ +Aη ⊆ A 1

ab . By Exercise 2.17,

K =⋃ξ∈K

Aξ = lim−→ξ∈K

Aξ.

By Exercise 2.20, we may identify K ⊗AM with lim−→ξ∈K(Aξ ⊗AM).

If φ(x) = 0, then there is some ξ in K such that 1⊗ x = 0 in Aξ⊗AM , by Exercise 2.15. Since 1 is in Aξ, ξ−1

is in A. Furthermore,A→ Aξ : a 7→ aξ

is an isomorphism of A-modules because K is a field. Therefore,

ψ : Aξ ⊗AM → A⊗AM →M : aξ ⊗ x 7→ ax

is an A-module isomorphism. Finally, we look at the composition

0 = ψ(0) = (ψ φ)(x) = ψ(1⊗ x) = ξ−1x.

Thus x is in T (M).


3.13

Let S be a multiplicatively closed subset of the domain A. For any x in M and s in S,

Ann(x/s) = Ann(S−1Ax) = S−1 Ann(Ax) = S−1 Ann(x)

by Proposition 3.14. Thus xs is in T (S−1M) if and only if it is in S−1T (M).

In particular, T (Mp) = T (M)p for all prime ideals p of A. By Proposition 3.8, (i), (ii), and (iii) are equivalent.

3.14

Let a be an ideal of A and let M be an A-module such that Mm = 0 for all maximal ideals m containing a.Since localization is exact, M/aM is an A/a-module such that (M/aM)m = Mm/(aM)m = 0 for all maximal idealsm of A/aA. By Proposition 3.8, M/aM = 0. Thus M = aM .

3.15

Let F = An for some ring A. Let x1, . . . , xn be a set of generators for F , let e1, . . . , en be the canonical basis,and let φ : F → F be the map given by φ(ei) = xi. The map φ is surjective because x1, . . . , xn generate F . Thefollowing demonstrates that φ is injective.

Without loss of generality, we may assume A is a local ring by Proposition 3.9. Let m be the maximal ideal ofA, let k be its residue field, and let N be the kernel of φ. Then the sequence

0→ N → Fφ−→ F → 0

is exact. Since F is a flat A-module,

0→ k ⊗A N → k ⊗A Fφ−→ k ⊗A F → 0

is exact. Furthermore, k⊗A F is naturally isomorphic to kn. Thus φ is a surjective k-linear map of n-dimensionalvector spaces, and bijective as a consequence. By Exercise 2.12, N is a finitely generated A-module. Sincek ⊗A N = 0, Exercise 2.3 implies that N = 0.

3.16

Let f : A→ B be a flat A-algebra.

(i)⇒(ii) Let p be a point in Spec(A). By Proposition 3.16 there is q in Spec(B) such that f∗(q) = qc = p.

(ii)⇒(iii) Let m be a maximal ideal of A and let q be a prime ideal of B such that f∗(q) = m. Thenme = qce ⊆ q 6= (1).

(iii)⇒(iv) Let x be a non-zero element of M and let a = Ann(x). Since x is non-zero, a is a proper ideal of A.By hypothesis, ae is then a proper ideal of B. By Exercise 2.2 and the First Isomorphism Theorem, we have thefollowing isomorphisms of B-modules

B/ae ∼= B ⊗A (A/a) ∼= B ⊗A Ax.

Since B is flat, the map B ⊗A Ax → B ⊗A M induced by the inclusion Ax → M is injective. Thus MB is anon-zero B-module.

(iv)⇒(v) Let M ′ be the kernel of the natural map M → B ⊗A M so that we have a short exact sequence ofA-modules

0→M ′ →M →MB → 0.


Since B is flat, we may tensor with B ⊗A − to get a short exact sequence of B-modules

0→M ′B →MB → (MB)B → 0.

By Exercise 2.13, the map on the right is injective. Therefore M ′B = 0. By hypothesis, this implies that M ′ = 0.Thus M → B ⊗AM is injective.

(v)⇒(i) By hypothesis, the natural map

A/a→ (A/a)B ∼= B/ae

is injective. Thus a ⊆ aec ⊆ a.

3.17

Let Af−→ B

g−→ C be a pair of ring homomorphisms such that g f is flat and g is faithfully flat, and leth : M → N be an injective homomorphism of A-modules. Let K be the kernel of the extension hB : MB → NB.Now tensor with C to get the commutative diagram

0 K ⊗B C (M ⊗A B)⊗B C (N ⊗A B)⊗B C

0 M ⊗A C N ⊗A C,

where the right and center vertical arrows are the canonical isomorphisms. The bottom row is exact because g fis flat, and the top row is exact because g is flat. Thus K ⊗B C = 0. Since g is faithfully flat, K = 0 by Exercise3.16 (iv). Therefore, f : A→ B is a flat A-algebra.

3.18

Let f : A → B be a flat homomorphism of rings, let q be a prime ideal of B, and let p = qc. By Proposition3.10, Bp is flat over Ap and Bq is flat over Bp. By Exercise 2.8 (ii), Bq is flat over Ap. Since pe = qce ⊆ q,

f : Ap → Bq satisfies the conditions of Exercise 3.16 (iii). Therefore, Bq is a faithfully flat Ap-algebra, andf∗ : Spec(Bq)→ Spec(Ap) is surjective.

3.19

Let M be an A-module and define the support of M by

Supp(M) = p ∈ Spec(A)|Mp 6= 0.

We will use the result that S−1A = 0 if and only if 0 ∈ S. In particular, S−1(A/a) = 0 if and only if S∩a 6= ∅.

(i) My Proposition 3.8, M = 0 if and only if Mp = 0 for all prime ideals p. This is equivalent to Supp(M) = ∅.

(ii) A given prime ideal p contains a if and only if (A \ p)∩ a = ∅. Thus p is in V (a) if and only if (A/a)p 6= ∅.

(iii) Let0→M ′ →M →M ′′ → 0

be an exact sequence of A-modules. By Proposition 3.3,

0→M ′p →Mp →M ′′p → 0

is an exact sequence of Ap-modules for every prime ideal p in A. Thus Mp = 0 if and only if M ′p = M ′′p = 0.


(iv) By (ii), Supp(Mi) ⊆ Supp(M) for each i. Conversely, suppose that x/s is a non-zero element of Mp. Then

x/s ∈

k∑j=1

Mij

p

=k∑j=1

(Mij )p

for some indices i1, . . . , ik. Therefore, (Mi)p 6= 0 for some i, and Supp(M) ⊆⋃i∈I Supp(Mi) as a consequence.

(v) A prime ideal p contains Ann(M) if and only if (A \ p)∩Ann(M) = ∅. This is equivalent to the conditionthat (A/Ann(M))p = Ap/Ann(M)p is non-zero. Since Mp = 0 if and only if Ap = Ann(Mp), we are done.

(vi) By Proposition 3.7, (M ⊗A N)p = Mp ⊗Ap Np as Ap-modules. If Mp = 0 or Np = 0, then Mp ⊗ap Np = 0.Thus

Supp(M ⊗A N) ⊆ Supp(M) ∩ Supp(N).

By Corollary 3.4, Mp, Np are finitely generated Ap-modules. Since Ap is a local ring, if Mp ⊗Ap Np = 0, thenMp = 0 or Np = 0 by Exercise 2.3. Therefore,

Supp(M ⊗A N) ⊇ Supp(M) ∩ Supp(N).

(vii) By Exercise 2.2, M/a = (A/a)⊗AM . Thus, parts (ii), (v), and (vi) imply that

Supp(M/aM) = Supp((A/a)⊗AM)

= Supp(A/a) ∩ Supp(M)

= V (a) ∩ V (Ann(M))

= V (a + Ann(M)).

By Proposition 3.3 and Exercise 2.2,

(M/aM)p = Mp/apMp = (Ap/ap)⊗Ap Mp = (A/a)p ⊗Ap Mp.

By Corollary 3.4, Mp is a finitely generated Ap-module. Therefore, (M/aM)p = 0 if and only if (A/a)p = 0 orMp = 0.

(viii) Let q be a point in Spec(B) and let p = f∗(q). By Proposition 3.5,

(B ⊗AM)q = Bq ⊗B (B ⊗AM)

= Bq ⊗AM= (Bq ⊗Ap Ap)⊗AM= Bq ⊗Ap Mp.

Therefore, Supp(B ⊗AM) ⊆ f∗−1(Supp(M)).

Let κp = Ap/pAp and κq = Bq/qBq be the respective residue fields of p and q, and assume that 0 = (B ⊗AM)q.Tensor with κq over Bq to get

0 = κq ⊗Bq (B ⊗AM)q = κq ⊗Bq Bq ⊗Ap Mp = κq ⊗Ap Mp.

Since the tensor product is right exact, κq⊗Ap Mp surjects onto κq⊗Ap (Mp/pMp). Therefore, the κp-vector spaceκq ⊗Ap (Mp/pMp) is zero. Since κq is a field, Mp/pMp = 0. Furthermore, Mp is a finitely generated Ap-moduleand p is the Jacobson radical of Ap. By Nakayama’s Lemma, Mp = 0. Thus Supp(B ⊗AM) ⊇ f∗−1(Supp(M)).

3.20

Let f : A→ B be a ring homomorphism.


(i) If f∗ : Spec(B) → Spec(A) is surjective, then every prime ideal of A is a contracted ideal. Conversely, letp = bc for some ideal b in B then

pec = bcec = bc = p.

By Proposition 3.16, p is the contraction of a prime ideal of B, and f∗ is surjective.

(ii) Suppose that every prime ideal of B is an extended ideal, and let q1 = ae1, q2 = ae2 be prime ideals of B. Iff∗(q1) = f∗(q2), then

q1 = ae1 = aece1 = f∗(q1)e = f∗(q2)e = aece2 = ae2 = q2,

and f∗ is injective.

For a counterexample to the converse, let

f : A = C→ C[x]/(x2) = B

be the natural inclusion map. Since x2 = 0 in B, every prime ideal of B contains x. On the other hand, (x) is amaximal ideal, so B has exactly one prime ideal. Similarly, A has only the prime ideal (0) because it is a field.Thus

f∗ : Spec(B)→ Spec(A)

is the injective map between one point sets. Finally,

(x)ce = (0A)e = (0B)

shows that (x) cannot be an extended ideal.

3.21

(i) Let S be a multiplicatively closed subset of the ring A, and let φ : A→ S−1A be the canonical homomor-phism.

By Proposition 3.11 (iv), φ∗ : Spec(S−1A) → Spec(A) = X is bijective onto its image, S−1X. Furthermore,S−1X consists of prime ideals of A not meeting S. Thus S−1X =

⋂s∈S Xs.

Now we will show that φ∗ is an open map. Let Yf/g be an open set in Spec(S−1A). Exercise 1.21 (i) allows usto calculate that

Yf/g = V (f/g)c = V (fs)c = Yfs = Yφ(fs) = φ∗−1(Xfs)

= φ∗−1(Xf ∩Xs) = φ∗−1(Xf ) ∩ φ∗−1(Xs),

because gs is a unit in S−1A, for every s in S. Thus

Yf/g =⋂s∈S

(φ∗−1(Xf ) ∩ φ∗−1(Xs))

= φ∗−1(Xf ) ∩ φ∗−1

(⋂s∈S

Xs

)= φ∗−1(Xf ) ∩ φ∗−1(S−1X).

Since φ∗ : Spec(S−1A) → S−1X is bijective, φ∗(Yf/g) = Xf ∩ S−1X, and φ∗ is an open map. Therefore, φ∗ is ahomeomorphism onto its image.

If S is the monoid generated by f , then S−1X =⋂n≥0Xfn = Xf .

(ii) Let f : A → B be a ring homomorphism, X = Spec(A), Y = Spec(B), and f∗ : Y → X the associatedmapping. The diagram,


A B

S−1A S−1B,

f

S−1f

obviously commutes. By Proposition 3.11 (iv), we may identify the prime ideals of A and B not meeting S withtheir extensions to S−1A and S−1B, and

(S−1f)−1(S−1q) = S−1p

if and only iff−1(q) = p,

by the commutativity of the diagram. Thus the restriction of f∗ to S−1Y has image in S−1X, and can be identifiedwith the map

(S−1f)∗ : Spec(S−1B)→ Spec(S−1A).

If f∗(q) = p, then S−1(A/p) → S−1(B/q) is injective, because localization is exact. Thus, p must meet S,if q does. Consequently, the only points of Y mapping into S−1X by f∗ are those of S−1Y . In other words,S−1Y = f∗−1(S−1X).

(iii) Let a be an ideal of A with extension b to B, and let f : A/a→ B/b be the map obtained from f : A→ Bby passing to the quotient. If q is a prime ideal of B containing b, then qc ⊇ bc = aec ⊇ a. Thus f∗ restricted toV (b) has image in V (a).

By Exercise 1.21 (iv), the natural quotient maps A → A/a, B → B/b induce homeomorphisms Spec(A/a) →V (a), Spec(B/b)→ V (b). We must show that the diagram,

Spec(B/b) Spec(A/a)

V (b) V (a),

f∗

f∗

commutes. But the diagram,

A B

A/a B/b,

f

f

already commutes. Pulling back a prime ideal from the lower right corner to the upper left along the two routesshows that the previous diagram of spectra commutes.

(iv) Let p be a prime ideal of A, let S = A \ p, let f : A→ B be a ring homomorphism, and X = Spec(A), Y =Spec(B). Since V (p) is the set of prime ideals containing p and S−1X is the set of prime ideals contained in p,p = V (p) ∩ S−1X. Therefore,

f∗−1(p) = f∗−1(V (p) ∩ S−1X) = V (pB) ∩ S−1Y,

by Exercise 1.21 (ii), and (iii) above. By (iii), we may identify S−1Y with Spec(Bp). Under this identification,V (pB) ∩ S−1Y = V (pBp). The latter space is canonically homeomorphic to Spec(Bp/pBp) by Exercise 1.21 (iv).On the other hand,

Bp/pBp = Ap/pAp ⊗Ap Bp

= κp ⊗Ap Ap ⊗A B= κp ⊗A B.

Therefore, f∗−1(p) = Spec(κp ⊗A B).


3.22

Let p be a prime ideal of the ring A. Every open neighborhood U of p contains a basic open set Xs with s inS = A \ p. Thus,

Spec(Ap) = S−1X =⋂s∈S

Xs =⋂p∈U

U,

by Exercise 3.21 (i).

3.23

Let X = Spec(A) for some ring A and let U = Xf for some f in A.

(i) Suppose that U = Xg. By Exercise 1.17 (iv),√

(f) =√

(g). Thus there are positive integers m,n andelements a, b of A such that fm = ag and gn = bf . Let φ : A→ Af be the canonical map.

We will use Corollary 3.2 to get our desired canonical isomorphism. First,

fmn = (ag)n = anbf

shows that a, b are units in Af , and moreover ag = fm shows that g is a unit in Af . Second, φ(x) = 0 impliesthat fkx = 0 for some k ≥ 0. Multiply through by bk to get 0 = bkfkx = gnkx. Third,

x

fk= φ(x)φ(fk)−1 = φ(x)φ(bk)φ(bkfk)−1 = φ(xbk)φ(gnk)−1.

Therefore, there is a unique isomorphism Ag → Af lifting the canonical homomorphism A → Af . Flipping theroles of f and g yields the inverse map. Thus A(U) is a well defined ring.

(ii) Now let Xg = U ′ ⊆ Xf = U . By Exercise 1.17 (iv), g ∈√

(f). Therefore, there is an integer n ≥ 0 anda in A such that gn = af . As a consequence, f is a unit in A(U ′). By Proposition 3.1, there is a unique ringhomomorphism ρ : A(U)→ A(U ′) lifting the canonical map A→ A(U ′).

(iii) If U = U ′, then the uniqueness of ρ implies that it is the identity map.

(iv) The three arrows are all unique maps lifting the identity by Proposition 3.1. The uniqueness forces thecomposition of the lower to maps to equal the top map.

(v) Let x = p be a point of X. If x ∈ U and U = Xf , then f ∈ A \ p. Thus there is a unique mapµU : A(U)→ Ap lifting the canonical homomorphism A→ Ap. Furthermore, µU = µU ′ ρUU ′ by uniqueness. ByExercise 2.16 there is a unique map ρ : lim−→x∈U A(U)→ Ap lifting the maps µU .

Given xf ∈ Ap. We have that x

f is in the image of µU for U = Xf . Thus ρ is surjective.

If ρ(a) = 0, then there is f ∈ A \ p such that fa = 0 in A. Let U = Xf , and note that this implies that a = 0in A(U), and thus in lim−→x∈U A(U) as well. Therefore, ρ is injective.

3.24

Let Uii∈I be a covering of X by basic open sets and let si ∈ A(Ui) for each i such that si = sj in A(Ui ∩Uj)for each i, j. For each i, let Ui = Xfi and si = xi

fmii

. Since si = sj in A(Ui ∩ Uj), there are integers nij ≥ 0

such that (fifj)nij (xif

mj

j − xjfmii ) = 0 in A for all i, j. On the other hand, there are finitely many f ’s such that

(f1, . . . , fk) = (1) because X is quasi-compact. Raising the ideal (f1, . . . , fk) to a sufficiently large power yields∑ki=1 aif

mii = 1 for some a1, . . . , ak in A such that ai ∈ (fi)

maxi 6=j(nij). Define s =∑k

i=1 aixi. Then, in A(Uj), we


have

s =k∑i=1

aixi

= ajsjfmj

j +∑i 6=j

aixi

= sj

1−∑i 6=j

aifmii

+∑i 6=j

aixi

= sj +∑i 6=j

ai(xi − sjfmii )

= sj +1

fmf

j

∑i 6=j

ai(xifmj

j − xjfmii )

= sj ,

for j = 1, . . . , k.

Now, suppose that i ∈ I is arbitrary. For each j = 1, . . . , k, s = sj = si in A(Ui ∩ Uj). Therefore, there isn ≥ 0 such that (fmi

i s− xi)(fifj)n = 0 in A for each j. Taking powers of (f1, . . . , fk) as before, we may obtain an

A-linear combination∑k

j=1 bjfnj = 1. Summing the equations

bj(fmii s− xi)(fifj)n = 0

in A yields (fmii s− xi)(fi)n = 0. Thus s = si in A(Ui).

Now, suppose that s′ ∈ A also has image si in A(Ui) for every i ∈ I. Given the f1, . . . , fk as before, we getthe systems of equations (s− s′)fnj = 0 for j = 1, . . . , k. Raising (f1, . . . , fk) to a sufficiently high power and thensumming our equations yields s = s′.

3.25

Let f : A → B, g : A → C, h : A → B ⊗A C be ring homomorphisms such that h(x) = f(x) ⊗ g(x), and letX,Y, Z, T be the respective prime ideal spectra of A,B,C,B ⊗A C. We will make multiple uses of Exercise 3.21(iv). For any prime ideal p of A, p ∈ h∗(T ) if and only if

h∗−1(p) = Spec(κp ⊗ (B ⊗A C)) = Spec((κp ⊗A B)⊗κp (κp ⊗A C))

is non-empty. This is equivalent to the condition that the κp-vector space (κp ⊗A B) ⊗κp (κp ⊗A C) is non-zero.In turn, this is equivalent to the condition that both κp ⊗A B and κp ⊗A C are non-zero κp-vector spaces. Wemay flip this again, to the condition that f∗−1(p) = Spec(κp ⊗A B) and g∗−1(p) = Spec(κp ⊗A C) are non-empty.Finally, this is equivalent to p ∈ f∗(Y ) ∩ g∗(Z).

3.26

Let B be the direct limit of the direct system of rings Bα, gαβ, let fα : A → Bα be A-algebras such thatgαβ fα = fβ for all α ≤ β, and let f : A→ B be the induced map. By Exercise 3.21 (iv), f∗−1(p) = Spec(κp⊗AB)is empty if and only if κp ⊗A B = 0. On the other hand,

κp ⊗A B = lim−→α

(κp ⊗A Bα)

by Exercise 2.20. By Exercise 2.15, a direct limit is 0 if and only if every element is eventually zero. In the caseof a ring, this is equivalent to 1 = 0 for some α. Now, κp ⊗A Bα = 0 if and only if f∗−1

α (p) = Spec(κp ⊗A Bα) isempty. In summary, p is in the image of f∗ if and only if it is in the image of f∗α for every α.

4 PRIMARY DECOMPOSITION 46

4 Primary Decomposition

4.1

Suppose a is a decomposable ideal of A. The irreducible components of Spec(A/a) are the sets V (p), where pis a minimal prime ideal over a, by Exercise 1.20 (iv). By Proposition 4.6, these are the isolated prime ideals of a.Furthermore, Theorem 4.5 tells us that there are only finitely many such prime ideals.

4.2

Suppose a is a decomposable ideal of A such that a =√a. Let

⋂ni=1 qi be an irredundant primary decompo-

sition of a with associated prime ideals p1, . . . , pn. Then a =√a =

⋂ni=1

√qi =

⋂ni=1 pi yields another primary

decomposition of a. Since the initial decomposition was assumed to be irredundant, this decomposition is as well.Therefore a has no embedded prime ideals.

4.3

Suppose A is absolutely flat and q is a p-primary ideal. By Exercise 2.28, A/q is absolutely flat, and non-unitsare zero-divisors. Since q is primary, non-units are nilpotent. Therefore, A/q is an absolutely flat local ring. ByExercise 2.28, A/q is a field, and q is a maximal ideal as a consequence.

4.4

Define m = (2, t), q = (4, t) as ideals of Z[t]. Since Z[t]/m ∼= F2, m is a maximal ideal. Note that 22, t ∈ q.Thus m ⊆ √q, and we have equality because m is maximal. By Proposition 4.2, q is m-primary.

On the other hand, t ∈ q \m2. So mk ⊆ m2 ⊂ q ⊂ m for all integers k ≥ 2. Therefore, q is not a power of m.

4.5

Define p1 = (x, y), p2 = (x, z),m = (x, y, z) in the polynomial ring K[x, y, z], where K is a field and x, y, z areindeterminates. Since K[x, y, z]/p1

∼= K[z], K[x, y, z]/p2∼= K[y], and K[x, y, z]/m ∼= K, we know that p1, p2 are

prime ideal and m is maximal.

Define a = p1p2 = (x2, xy, xz, yz). Clearly p1 and p2 are respectively p1-primary and p2-primary. By Proposi-tion 4.2, m2 is m-primary. Since p1 ∩ p2 = (x, yz) and m2 = (x2, xy, xz, yz, y2, z2), we know that

p1 ∩ p2 ∩m2 = a

is a primary decomposition. Furthermore, the primary components p1, p2,m2 have distinct associated prime ideals,

andx ∈ p1 ∩ p2 \m2, y2 ∈ p1 ∩m2 \ p2, z

2 ∈ p2 ∩m2 \ p1

shows that the decomposition is irredundant.

Let p be a prime ideal containing a, then x ∈ p because x2 ∈ a ⊆ p. Similarly, yz ∈ p implies that y ∈ p orz ∈ p. Thus p1, p2 are the isolated prime ideals of a and m is an embedded prime ideal.

4.6

Let X be a compact Hausdorff space where (0) is a decomposable ideal of C(X), and let (0) =⋂ni=1 qi be an

irredundant primary decomposition of (0).


Suppose a is an ideal contained in distinct maximal ideals mx,my. Since X is Hausdorff, x and y have disjointopen neighborhoods U, V . By Urysohn’s Lemma, there are f, g ∈ C(X) such that f(x) = g(y) = 1, f |X\U = 0,and g|X\V = 0. Thus, fg = 0 ∈ a, but f /∈ my and g /∈ mx. Therefore, each primary ideal of C(X) is contained ina unique maximal ideal.

For each i, let mi be the maximal ideal over qi. By Exercise 1.26, X is homeomorphic to

Max(C(X)) = Max(C(X)) ∩ V (0) = Max(C(X)) ∩

(n⋃i=1

V (qi)

)= m1, . . . ,mn.

Therefore, X must be finite.

4.7

Let A[x] be the ring of polynomials in one indeterminate over the ring A, and let a[x] be the ideal of polynomialswith coefficients in the ideal a of A.

(i) Clearly, ae ⊆ a[x]. On the other hand, axn ⊆ ae for every non-negative integer n. Thus, ae = a[x].

(ii) Lift the natural map A→ A/p to obtain a map A[x]→ (A/p)[x] with kernel p[x]. Therefore, A[x]/p[x] ∼=(A/p)[x], and p[x] is a prime ideal of A[x] whenever p is a prime ideal of A.

(iii) Let q be a p-primary ideal of A. As in (ii), (A/q)[x] ∼= A[x]/q[x]. Let f ∈ (A/q)[x] be a zero-divisor. ByExercise 1.2, there is b ∈ A \ q such that bf ∈ q. Therefore, the coefficients of f are zero-divisors in A/q. Since qis primary, the coefficients of f are nilpotent in A/q. Again we cite Exercise 1.2 to see that f is nilpotent. Thusq[x] is a primary ideal of A[x]. Furthermore, fk ∈ q[x] for some positive integer k if and only if the coefficients off are in

√q = p. So

√q[x] = p[x], and q[x] is in fact p[x]-primary.

(iv) Let a =⋂ni=1 qi be an irredundant primary decomposition of an ideal a of A. The prime ideals (pi[x])ni=1

of A[x] are distinct because they have distinct contractions to A. Furthermore, for each j we have

⋂i 6=j

(qi[x]) =

⋂i 6=j

qi

[x] * qj [x]

because⋂i 6=j qi * qj . Therefore,

a =

(n⋂i=1

qi

)[x] =

n⋂i=1

qi[x]

is an irredundant primary decomposition of a.

(v) Suppose that p is an isolated prime ideal of a and let p′ be a prime ideal of A[x] such that a[x] ⊆ p′ ⊆ p[x].Taking contractions yields a ⊆ (p′)c ⊆ p. Since p is minimal, (p′)c = p. On the other hand, p′ ⊇ (p′)ce = p[x].Thus p′ = p[x]. So p[x] is an isolated prime ideal of a[x].

4.8

Let k be a field and define ideals pi = (x1, . . . , xi) of k[x1, . . . , xn] for each i = 1, . . . , n. The map k[x1, . . . , xn]→k[xi+1, . . . , xn] obtained by evaluating each xj at 0 for 1 ≤ j ≤ i is surjective with kernel pi. Since k[xi+1, . . . , xn]is a domain, pi is prime ideal. If i = n, then pi is a maximal ideal. By Proposition 4.2, the powers of pi arepi-primary in k[x1, . . . , xi]. By Exercise 4.7 (iii), the powers of pi are pi-primary in k[x1, . . . , xn] as well.

4.9

Let A be a ring and define

D(A) = p ∈ Spec(A) : p ⊇ (0 : a) is a minimal for some a ∈ A.


If x ∈ A is a non-zero, zero-divisor, then x ∈ (0 : a) for some non-zero a ∈ A. If p is a minimal prime idealabove (0 : a), then x ∈ p and p ∈ D(A).

Let x be a non-zero element of A and suppose that there is a prime ideal p such that x ∈ p ∈ D(A). Bythe definition of D(A), there is a ∈ A such that p is a minimal prime ideal above (0 : a). Thus x is nilpotent in(A/(0 : a))p. In other words, there is s ∈ A \ p and a positive integer n such that xns ∈ (0 : a). Since s /∈ p, we

know that sa is non-zero. If n is the smallest such integer, then xn−1sa is non-zero, but x ·xn−1sa = 0. Therefore,x is a zero-divisor.

Let S be a multiplicatively closed subset of A. By definition, S−1p ∈ D(S−1A) if and only if S−1p is a minimalprime ideal above (0 : a/s) for some a ∈ A and s ∈ S. By Proposition 3.11 (ii), (0 : a/s) = S−1(0 : a). Therefore,by Proposition 3.11 (iv), this is equivalent to the condition that p is a minimal prime ideal above (0 : a) for somea ∈ A and S ∩ p = ∅. In turn, this is holds precisely when p ∈ D(A) ∩ Spec(S−1A).

Suppose that (0) is a decomposable ideal and p is one of its associated prime ideals. By Theorem 4.5, p =√(0 : a) for some a ∈ A. Thus p ∈ D(A).

On the other hand, suppose that p ∈ D(A) is a minimal prime ideal above (0 : a) for some a ∈ A. Thenpp ∈ D(Ap). So pp is a minimal prime ideal above (0 : a)p. Since Ap is local, pp =

√(0 : a)p. By Proposition 4.2,

(0 : a)p is pp-primary. Taking contractions yields

(0 : a) =(0 : (a)cp

)⊇ (0 : a)cp ⊇ (0 : a).

Thus (0 : a) is a primary ideal. Furthermore, by Proposition 4.8, (0 : a) is p-primary. So p is an associated primeideal of (0).

4.10

For every prime ideal p of A, let Sp(0) be the kernel of the natural map A→ Ap.

(i) Since (0)p ⊆ pp, Sp(0) ⊆ p by Proposition 3.11.

(ii) If√Sp(0) = p, then pp =

(√Sp(0)

)p

=√

(Sp(0))p =√

0 in Ap. Therefore, Ap has only one prime ideal,and p is a minimal prime ideal of A.

If p is a minimal prime ideal of A, then pp is the nilradical of Ap. By Exercise 1.18,√Sp(0) = p.

(iii) If p ⊇ p′, then the universal property of localization gives a unique lift of A→ Ap′ to Ap → Ap′ such thatthe following diagram commutes:

A Ap

Ap′

Therefore, Sp(0) ⊆ Sp′(0).

(iv) Suppose x ∈⋂

p∈D(A) Sp(0), and p is a minimal prime ideal above (0 : x). By definition, p ∈ D(A). Thusx ∈ Sp(0) by hypothesis. If x 6= 0, then there is s /∈ p such that sx = 0. In turn, this implies that s ∈ (0 : x) ⊆ p,which is a contradiction. Therefore, x = 0.

4.11

Let p be a minimal prime ideal of A and let q be p-primary. Since p is minimal, pp is the unique prime ideal idealof Ap. Therefore, (0)p is pp-primary by Proposition 4.2, and its contraction, Sp(0), is p-primary by Proposition4.8. Furthermore, by Exercise 1.18 and Proposition 4.8, (0) ⊆ q implies that Sp(0) ⊆ q.

Let a be the intersection of all Sp(0) for which p is a minimal prime ideal of A. Then a ⊆ Sp(0) ⊆ p for eachminimal prime ideal. As a consequence, a ⊆

⋂p minimal p =

√(0).


Suppose (0) is a decomposable ideal of A. Then A has only finitely many minimal prime ideals p1, . . . , pn byCorollary 4.11.

If a = (0), then⋂ni=1 Spi(0) is a primary decomposition of (0). Note that the prime ideals p1, . . . , pn are distinct.

If Spj (0) ⊇⋂i 6=j Spi(0) for some j, then take radicals to obtain pj ⊇

⋂i 6=j pi. By Proposition 1.11 (ii), pj ⊇ pi

for some i. Since pj is a minimal prime ideal, we have equality. This contradicts the distinctness of the minimalprime ideals p1, . . . , pn. Therefore,

⋂ni=1 Spi(0) is an irredundant primary decomposition of (0). By Theorem 4.5,

p1, . . . , pn are the associated prime ideals of (0). So every associated prime ideal of (0) is isolated.

On the other hand, if every associated prime ideal of (0) is isolated, then (0) =⋂

p∈D(A) Sp(0) =⋂ni=1 Spi(0) = a

by Problems 4.9 and 4.10 (iv).

4.12

Let S be a multiplicatively closed subset of the ring A and for each ideal a of A define the saturation of a,S(a), as the contraction of S−1a to A.

(i) By Proposition 3.11 (v) and Exercise 1.18,

S(a) ∩ S(b) =(S−1a

)c ∩ (S−1b)c

=(S−1a ∩ S−1b

)c=(S−1 (a ∩ b)

)c= S(a ∩ b).

(ii) By Proposition 3.11 (iv) and Exercise 1.18,

S(√

a)

=(S−1√a)c

=(√

S−1a)c

=√

(S−1a)c =√S(a).

(iii) By Proposition 1.17, S(a) = (1) if and only if S−1a = S−1A. By Proposition 3.11 (ii), this is equivalentto S ∩ a 6= ∅.

(iv) Let f : A → S−1A be the natural map and note that x ∈ S(a) if and only if f(x) ∈ S−1a. This isequivalent to the existence of s ∈ S such that xs ∈ a. Therefore,

x ∈ S1(S2(a)) if and only if xs1 ∈ S2(a) for some s1 ∈ S1

if and only if xs1s2 ∈ a for some s1 ∈ S1, s2 ∈ S2

if and only if x ∈ (S1S2)(a).

Suppose a =⋂ni=1 qi is an irredundant primary decomposition. By Proposition 4.9, S(a) =

⋂mj=1 qij is a

irredundant primary decomposition for some 1 ≤ i1, . . . , im ≤ n. Since there are only finitely many subsets of1, . . . , n, there are only finitely many ideals S(a).

4.13

Let p be a prime ideal in a ring A, and define the nth symbolic power of p as p(n) = Sp(pn), where Sp = A \ p.

(i) By Proposition 3.11 (v),√S−1p pn = S−1

p√pn = S−1

p p. Thus S−1p pn is an S−1

p p-primary ideal by Proposition

4.2. Since p(n) is the contraction of S−1p pn, it is a p-primary ideal by Proposition 4.8 (ii).

(ii) Suppose that pn =⋂mi=1 qi is an irredundant primary decomposition and J is the set of indices j for which

qj ⊆ p. By Proposition 4.9, p(n) = Sp(pn) =

⋂j∈J qj . By (i), p =

√p(n) ⊆ √qj for each j ∈ J . Therefore, qj is

p-primary for each j ∈ J . Since the decomposition is irredundant by hypothesis, there is only one index j in J ,and qj = p(n). Thus p(n) is the p-primary component of pn.

(iii) Suppose that p(m)p(n) =⋂ki=1 qi is an irredundant primary decomposition and J is the set of indices j for

which qj ⊆ p. Since S−1p p(m)p(n) = S−1

p pm+n, Proposition 4.8 (ii) implies that Sp(p(m)p(n)

)= p(m+n). On the

other hand, Proposition 4.9 tells us that Sp(p(m)p(n)

)=⋂j∈J qj . By (i), p =

√p(m+n) ⊆ √qj for each j ∈ J .


Therefore, qj is p-primary for each j ∈ J . Since the decomposition is irredundant by hypothesis, there is only oneindex j in J , and qj = p(m+n). Therefore, p(m+n) is the p-primary component of p(m)p(n).

(iv) If p(n) = pn, then pn is p-primary because p(n) is by (i). If pn is p-primary, then pn = p(n) by (ii).

4.14

Let a be a decomposable ideal in a ring A and let p be a maximal element of the set of ideals of the form (a : x)where x ∈ A \ a. If yz ∈ p and y /∈ p, then z ∈ (a : xy) and p ⊆ (a : xy). Since p is maximal, z ∈ (a : xy) = p. Sop is a prime ideal and p =

√(a : x) is an associated prime ideal of a by Theorem 4.5.

4.15

Let a be a decomposable ideal in a ring A, let Σ be an isolated set of prime ideals associated to a, let qΣ bethe intersection of the corresponding primary components. Suppose that there is f ∈ A such that f ∈ p if andonly p /∈ Σ, for each associated prime ideal p of A, and let Sf be the multiplicative monoid generated by f .

By Proposition 4.9, Sf (a) =⋂mi=1 qi = qΣ, where p1, . . . , pm = Σ. By Proposition 3.11 (ii), Sf (a) =⋃

j≥0(a : f j). Since f ∈ pi =√qi for i = m + 1, . . . , n, there is some k such that fk ∈

⋂ni=m+1 qi. Therefore,

Sf (a) ⊆ (a : fk) ⊆⋃j≥0(a : f j) = Sf (a).

4.16

Let A be a ring in which every ideal is decomposable, let S be a multiplicatively closed subset, and let b be anideal of S−1A. By Proposition 3.11 (i), b = S−1a for some ideal a of A. Let

⋂ni=1 qi be an irredundant primary

decomposition of a. By Proposition 4.9, S−1a =⋂mi=1 S

−1qi is an irredundant primary decomposition of b.

4.17

(L1) For every proper ideal a and every prime ideal p, there exists x /∈ p such that Sp(a) = (a : x).

Let A be a ring with the property (L1), and suppose a is a proper ideal of A. Let p1 be a minimal prime idealabove a. The ideal q1 = Sp1(a) is the contraction of the p1-primary ideal Sp1(0) ⊆ A/a to A, so q1 is p1-primaryby Exercise 4.11. By (L1), q1 = (a : x) for some x /∈ p1. Clearly, a ⊆ q1 ∩ (a + (x)). If f = a+ bx ∈ q1 ∩ (a + (x)),then fx − ax = bx2 ∈ a, because f ∈ q1 = (a : x) and a ∈ a. Thus b ∈ Sp1(a) = q1 = (a : x). So f ∈ a and wehave a = q1 ∩ (a + (x)). Let a1 be an ideal maximal among those for which a = q1 ∩ a1 and x ∈ a1. Now we havea ( a1 * p1.

Given a = q1 ∩ · · · ∩ qn ∩ an where aj−1 ( aj * pj for each j = 1, . . . , n, if an is a proper ideal, then we mayrepeat the process of the previous paragraph to extend the strictly increasing sequence of ideals a ( a1 ( . . . ( an.

4.18

(L2) For every ideal a and every descending chain S1 ⊇ S2 ⊇ S3 ⊇ . . . of multiplicatively closed subsets, thereexists a positive integer N such that Sn(a) = SN (a) for every n ≥ N .

(i)⇒(ii) Let a be an ideal of A, let p be a prime ideal, and let a =⋂ni=1 qi be an irredundant primary

decomposition of a with associated prime ideals p1, . . . , pn. Let Σp be the set of associated prime ideals of acontained in p. Clearly, Σp is an isolated set of prime ideals. Without loss of generality, p1, . . . , pm = Σp andp + pm+1 · · · pn. Thus there is f ∈ pm+1 · · · pn \ p. By Exercise 4.15 and Proposition 4.9,

Sp(a) = qΣp = (a : fk)

for some positive integer k. Therefore, A satisfies (L1).


Let S1 ⊇ S2 ⊇ . . . be a descending chain of multiplicatively closed subsets of A. Without loss of generality,the first mj associated prime ideals of a meet Sj , for every j ≥ 1. Since m1,m2,m3, . . . is a decreasing sequence ofpositive integers, it has a least element mN . By Proposition 4.9, if j ≥ N , then Sj(a) =

⋂ni=mj

qi =⋂ni=mN

qi =

SN (a). Therefore, A satisfies (L2).

(ii)⇒(i) Let a be an ideal of A. Using the property (L1), obtain the prime ideals pi and ideals ai, where we areusing the notation of Exercise 4.17. Now, define Sn = Sp1 ∩ · · · ∩ Spn for each positive integer n. By (L2), thereexists a positive integer N such that n ≥ N implies that Sn(a) = SN (a).

4.19

Let p be a prime ideal in a ring A, and let q be p-primary. Since (0) ⊆ q, Sp(0) ⊆ Sp(q) = q by Proposition4.8.

Suppose that for every prime ideal p of A, Sp(0) is the intersection of all p-primary ideals of A, and let p1, . . . , pnbe distinct non-minimal prime ideals of A. If n = 1, then a = p1 has p1 as its only associated prime ideal. Supposethat pn is maximal among the pi, and that there is a decomposable ideal b with irredundant primary decomposition⋂n−1i=1 qi, where each qi is pi-primary. Let p be a minimal prime ideal of A contained in pn. If b ⊆ Spn(0), then

b ⊆ Spn(0) ⊆ Sp(0) ⊆ p by Exercise 4.10 (iii). By taking radicals, this implies that pi ⊆ p for some i. Thiscontradicts our hypothesis that the pi’s are non-minimal prime ideals. By our assumption on A, this implies thatthere is a pn-primary ideal, qn, such that b * qn. Since p1, . . . , pn were assumed to be distinct, we have thata = p1 ∩ · · · ∩ pn is an irredundant primary decomposition.

4.20

Let N be a submodule of an A-module, M . Define the M -radical of N as

M√N = x ∈ A : xqM ⊆ N for some q > 0.

It follows directly from the definition that, M√N =

√(N : M) =

√Ann (M/N).

Analogues of Exercise 1.13: Let N,N ′ be submodules of M and let a be an ideal of A. It is unclear whatanalogues the author is requesting. The following are what come to my mind.

(i) M√N ⊇ (N : M) (The analogy is that a is an annihilator of A/a)

(ii)√

M√N ⊇ M

√N

(iii) I will break this into two formulas. The first deals with intersections of submodules. The second dealswith the product of an ideal and a submodule.

M√N ∩N ′ =

√(N ∩N ′ : M)

=√

(N : M) ∩ (N ′ : M)

=√

(N : M) ∩√

(N ′ : M)

=M√N ∩ M

√N ′

Let x ∈ M√aM ∩ M

√N , then there are positive integers k1, k2 such that xk1 ∈ (aM : M) and xk2 ∈ (N : M). If

m ∈M , then xk1m =∑k

i=1 aimi for some a1, . . . , ak ∈ a and m1, . . . ,mk ∈M . Thus,

xk1+k2m = xk2k∑i=1

aimi =

k∑i=1

ai(xk2mi) ∈ aN.

Therefore, M√aM ∩ M

√N ⊆ M

√aN .


On the other hand,M√aN =

√(aN : M) ⊆

√(aM ∩N : M) =

M√aM ∩ M

√N.

Given the previous paragraph, we have establishedM√aN =

M√aM ∩ M

√N.

(iv) M√N = (1) if and only if 1 ∈ Ann (M/N).

(v) Suppose xk = y + z ∈ M√N + M

√N ′, and get k1, k2 so that yk1 ∈ (N : M) and zk2 ∈ (N ′ : M). Then

xk(k1+k2) ∈ (N +N ′ : M). Therefore,√

M√N + M

√N ′ ⊆ M

√N +N ′.

(vi) If p is a prime ideal, then

M√pnM = M

√p (pn−1M) = M

√pM ∩ M

√pn−1M = M

√pn−1M

for every positive integer n. By induction, p ⊆ M√pM = M

√pnM . Now suppose that M/pM and M

√pM are finitely

generated A-modules. By Propositions 3.11 (v) and 3.14, we have(

M√pM)p

=

√Ann

((M/pM)p

)= (0) because

(M/pM)p is a κp-vector space. By Exercise 3.1, there is s /∈ p such that s · M√pM = (0) ⊆ p. Since p is prime

ideal, we have M√pnM = M

√pM ⊆ p.

In particular, if A is Noetherian, and M/pM is a finitely generated A-module, then M√pnM = p.

4.21

A proper submodule Q of M is primary if and only if every zero-divisor of M/Q is nilpotent. Suppose that Qis a primary submodule of M and that xy ∈ (Q : M), but y /∈ (Q : M). Let m ∈ M such that ym /∈ Q. Then xis a zero-divisor of M/Q. Since Q is primary, xn ∈ (Q : M) for some positive integer n. Therefore, (Q : M) is aprimary ideal of A.

Analogue of Lemma 4.3: Let Q1, . . . , Qn be p-primary submodules of M and let Q =⋂ni=1Qi. By Lemma 4.3,

(Q : M) =⋂ni=1(Qi : M) is a p-primary ideal. Therefore, Q is a p-primary submodule of M .

Analogue of Lemma 4.4: Let Q be a p-primary submodule of M .

(i) If m ∈ Q, then (Q : m) = (1) because Q is an A-module.

(ii) If m /∈ Q and xy ∈ (Q : m) with y /∈ (Q : m), then xym ∈ Q and ym /∈ Q. Since Q is primary, thereis a positive integer n such that xn ∈ (Q : M) ⊆ (Q : m). Letting y = 1, we see that (Q : M) ⊆ (Q : m) ⊆√

(Q : M) = p. Therefore, (Q : m) is p-primary.

(iii) If x /∈ p and m ∈ (Q : x), then xm ∈ Q, but x is not nilpotent in M/Q. Therefore, m ∈ Q, and we seethat (Q : x) = Q.

4.22

Analogue of Theorem 4.5: Let N =⋂ni=1Qi be an irredundant primary decomposition. For every m ∈M ,√

(N : m) =

n⋂i=1

√(Qi : m) =

⋂m/∈Qj

pj .

If√

(N : m) is prime ideal, then it equals pj for some Qj not containing m, by Proposition 1.11 (ii). On theother hand, since the decomposition is irredundant, we may choose mj ∈

⋂i 6=j Qi \ Qj for each index j. Thus√

(N : mj) = pj for each j.

By the Lattice Isomorphism theorem, the natural map M →M/N yields an order preserving bijection betweenthe submodules of M containing N and the submodules of M/N . In particular, sums, intersections, inclusions,and quotients are preserved. Thus the associated prime ideals of N in M are the associated prime ideals of 0 inM/N .


4.23

By the Lattice Isomorphism Theorem, we may assume that N is the zero submodule.

Analogue of Proposition 4.6: Suppose 0 is a decomposable submodule of M , let 0 =⋂ni=1Qi be an irredundant

primary decomposition, and let p be a prime ideal of A. If p ⊇ Ann(M) =⋂ni=1(Qi : M), then taking radicals

shows that p ⊇⋂ni=1 pi. By Proposition 1.11 (ii), p contains an associated prime ideal of 0. Therefore, every

minimal prime ideal of A/Ann(M) is an isolated prime ideal of 0 ≤M .

Analogue of Proposition 4.7: Suppose 0 is a decomposable submodule of M , and let D be the set of zero-divisors of M . Then x ∈ D if and only if there exists a non-zero m ∈ M such that xm = 0. In particular,D =

⋃0 6=m∈M

√(0 : m). If p is an associated prime ideal of 0, then Exercise 4.22 implies that p =

√(0 : m) for

some m ∈ M . Therefore, p ⊆ D. On the other hand, for every non-zero m ∈ M , there is some associated primeideal, p, of 0 such that √

(0 : m) =⋂

m/∈Qj

√(Qj : m) =

⋂m/∈Qj

pj ⊂ p.

Thus D ⊆⋃ni=1 pi, and we see that the set of zero-divisors of M is the union of the associated prime ideals of 0 in

M .

Analogue of Proposition 4.8: Let S be a multiplicatively closed subset of A and let Q be a p-primary submoduleof M .

(i) If s ∈ S ∩ p, then sn ∈ S ∩ (Q : M) for some positive integer n. Therefore, S−1M/S−1Q = S−1(M/Q) = 0.

(ii)

Analogue of Proposition 4.9:

Analogue of Theorem 4.10:

Analogue of Corollary 4.11:

5 INTEGRAL DEPENDENCE AND VALUATIONS 54

5 Integral Dependence and Valuations

5.1

Let f : A → B be an integral map of rings and let b be an ideal of B. By Exercise 1.21 (iii), f∗(V (b)) ⊆V (f−1(b)). Let p ∈ V (f−1(b)). By Proposition 5.6 (i), B/b is integral over A/f−1(b). By Proposition 5.10, thereis q ∈ Spec(B/b) = V (b) such that f∗(q) = p. Therefore, f∗(V (b)) = V (f−1(b)), and f∗ is a closed map.

5.2

Let B/A be an integral extension of rings, let Ω be an algebraically closed field, and let f : A → Ω be a ringmap. Let m = ker(f). By Proposition 5.10, there is a prime ideal n of B over m. Since Ω is a field, m is maximal.By Corollary 5.8, n is maximal. By Proposition 5.6 (i), κn/κm is an integral, hence algebraic, extension of fields.Therefore, f : κm → Ω lifts to some g : κn → Ω. Precompose g with the natural map B → κn to get g : B → Ω.

Our construction yields the following diagram:

A κm Ω

B κn Ω,

f

id

g

where the vertical maps are the extension maps A→ B, κm → κn, and the identity map. The horizontal compositesare f and g, respectively, and the right square commutes by construction. On the other hand, the left squarecommutes since the central map is the quotient of the map A→ B. Therefore the entire diagram commutes andwe see that g is an extension of f .

5.3

Let f : B → B′ be a map of A-algebras and let C be an A-algebra. Suppose f is integral. If x ∈ B′, thenB[x] = ⊕n−1

i=0 Bxi as a B-module, for some non-negative integer n. Therefore,

(B ⊗A C)[x⊗ 1] = B[x]⊗A C= ⊕n−1

i=0 (Bxi ⊗A C)

= ⊕n−1i=0 (B ⊗A C)(x⊗ 1)i

is a finitely generated B ⊗A C-module. Thus x⊗ 1 is integral over B ⊗A C by Proposition 5.1. Since B′ ⊗A C isthe B ⊗A C-span of x⊗ 1x∈B′ , f ⊗ idC is an integral map of A-algebras.

5.4

Let A = k[y], B = k[x], where k is a field and x is a root of t2 − (y + 1). Note that t2 − (y + 1) ∈ A[t] isirreducible by Eisenstein’s criterion. Let n = (x− 1) and m = (y) be maximals ideal so B and A respectively, andnote that n ∩A = m. Additionally, let z = 1

x+1 ∈ Bn and let K = A(0).

Since (x + 1)z = 1, we have yz2 + 2z − 1 = 0. Therefore, mzK(t) = t2 + 2

y t −1y . Let g(t) = yt2 + 2t − 1, and

suppose z is integral over Am. Then f(z) = 0 for some f(t) = c0tn + . . . + cn ∈ A[t] with c0 /∈ m. Without loss

of generality, f is primitive of minimum positive degree. Since f ∈ K[t] and f(z) = 0, f(t) = mzK(t)q(t) for some

q(t) ∈ K[t]. Thus yf(t) = g(t)q(t). By Gauss’ Lemma, q(t) ∈ A[t] and y divides q(t). Therefore, f(t) = g(t)h(t)for some h(t) ∈ A[t]. However, this implies that y divides c0, contrary to hypothesis.


5.5

Let B/A be an integral extension of rings.

(i) Suppose x ∈ A and x−1 ∈ B. Then x−n+a1xn−1+. . .+an = 0 implies x−1 = −(a1+a2x+. . .+anx

n−1) ∈ A.

(ii) By Corollary 5.8, every maximal ideal of B contracts to a maximal ideal of A. By Theorem 5.10, everymaximal ideal of A is the contraction of a prime ideal of B, which is maximal by Corollary 5.8. Thus the contractionof the Jacobson radical of B is the Jacobson radical of A.

5.6

Let B1, . . . , Bn be integral A-algebras and let B =∏ni=1Bi. Then B is clearly an A-algebra by the diagonal

map A → B. Let x = (x1, . . . , xn) ∈ B and let mi(t) be the minimal polynomial of xi over A for each i. Definep(t) =

∏ni=1mi(t) ∈ A[t]. By construction, p(t) is monic and

p(x) = (p(x1), . . . , p(xn)) = 0.

Thus x is integral over A.

5.7

Let A be a subring of B such that B \A is closed under multiplication, and let x ∈ B be integral over A withminimal polynomial tn + a1t

n−1 + . . .+ an. If x /∈ A, then

x(xn−1 + a1xn−2 + . . .+ an−1) = −an ∈ A

implies that xn−1 + a1xn−2 + . . .+ an−1 = a′ ∈ A. Thus

xn−1 + a1xn−2 + . . .+ an−1 − a′ = 0.

Since n was assumed to be minimal, we must have n = 1 and the minimal polynomial of x is t + a1. So x ∈ A,contrary to hypothesis. Therefore, x ∈ A.

5.8

Let A be a subring of B, and let C be the integral closure of A in B.

(i) Suppose B is a domain, let f, g ∈ B[x] be monic such that fg ∈ C[x], and let K be a splitting field of f, gover B(0). Then we may factor as f(x) =

∏ni=1(x− ξi) and g(x) =

∏mj=1(x− ηj). Since fg ∈ C[x], the ξ’s and η’s

are integral over C. Therefore coefficients of f, g ∈ B[x] are polynomials in elements of K, integral over C. SinceC is integrally closed in B, we have f, g ∈ C[x].

(ii) Let h ∈ B[x] be monic, let B′ = B[x]/hB[x], let i : B → B[x] be the inclusion, and let q : B[x] → B′ bethe quotient map. We claim that q i is injective. Note that a ∈ hB[x] if and only if there exists some b ∈ B[x]such that a = bh. Since h is monic, a has positive degree. Since the domain of q i contains no elements of B[x]of positive degree, q i is injective. Therefore, B′ is a finite B-algebra with an injection B → B′. Furthermore,h(t) = (t− x)h1(t) over B′, for some h1(t) ∈ B′[t] (it is here that we are using the fact that a non-unique form ofEuclidean division lets us divide by a monic polynomial over an arbitrary ring).

Let f, g ∈ B[x] be monic such that fg ∈ C[x]. Repeat the preceding construction with f and g to obtain afinite B-algebra B′ such that f and g split over B′. The same argument as in (i) concerning polynomials in theroots of f and g establishes that f, g ∈ C[x].


5.9

Let A be a subring of B, and let C be the integral closure of A in B. Let f ∈ B[x] be integral over A[x]with minimal polynomial p(t) = tn + a1t

n−1 + . . .+ an, where a1, . . . , an ∈ A[x]. Let r be an integer greater thandeg a1, . . . ,deg an, deg f , and n. If f0 = f − xr, then

0 = p(f) = p(f0 + xr) = fn0 + h1fn−10 + . . .+ hn−1f0 + p(xr).

Therefore,f0, f

n−10 + h1f

n−20 + . . .+ hn−1 ∈ B[x]

andf0(fn−1

0 + h1fn−20 + . . .+ hn−1) = −p(xr) ∈ C[x].

By Exercise 5.8, f0 ∈ C[x].

5.10

Let f∗ : SpecB → SpecA be the map associated to a ring homomorphism f : A→ B.

(i) (a)⇒(b) Suppose that f∗ is closed, p1 ⊆ p2 are prime ideals of A, and q1 is a prime ideal of B over p1. Then

f∗(V (q1)) = V (f∗(q1)) = V (p1)

and p2 ∈ V (p1). Thus there is q2 ∈ V (q1) such that f∗(q2) = p2.

(b)⇔(c) The map f∗

: Spec(B/q)→ Spec(A/p) is surjective if and only if every prime ideal of A containing phas a lift to a prime ideal of B over q.

(ii) (a)⇒(b) Suppose that f∗ is open, p2 ⊆ p1 are prime ideals of A, and q1 is a prime ideal of B over p1. LetU be an open neighborhood of q1 in SpecB. Since f∗(q1) = p1 and f∗(U) is open, there is s ∈ A \ p1 such thatXs is an open neighborhood of p1 in f∗(U). Since p2 ⊆ p1, we also have p2 ∈ Xs ⊆ f∗(U).

By Exercise 3.26,

f∗(SpecBq1) =⋂t/∈q1

f∗(SpecBt) =⋂t/∈q1

f∗(Yt).

Since p2 ∈ f∗(Yt) for all t /∈ q1, there is some q2 ⊆ q1 over p2.

(b)⇔(c) The map f∗ : SpecBq → SpecAp is surjective if and only if every prime ideal of A contained in p hasa lift to a prime ideal of B contained in q.

5.11

Let f : A→ B be a flat ring homomorphism. By Exercise 3.18, f∗ : SpecBq → SpecAp is surjective wheneverq is a prime ideal of B and p = qc. By Exercise 5.10 (ii), f has the going-down property.

5.12

Let G be a finite group of automorphisms of a ring A, let AG be the ring of G-invariants, and let x ∈ A.Define p(t) =

∏σ∈G(t− σ(x)) ∈ A[t]. Since the coefficients of p(t) are symmetric polynomials in the G-orbit of x,

p(t) ∈ AG[t]. Furthermore, p(x) = 0. So x is integral over AG.

Let S be a G-stable multiplicatively closed subset of A and let SG = S ∩ AG. For any σ ∈ G, the universalproperty of localization lifts the map A

σ−→ A→ S−1A uniquely to a map σ : S−1A→ S−1A such that σ(as

)= σ(a)

σ(s) .This is the desired group action.


Since AG ⊆ A and SG ⊆ S, the universal property of localization yields a unique map f : (SG)−1AG → S−1Aextending the inclusion AG → A such that f

(as

)= a

s . Given as ∈ (SG)−1AG and σ ∈ G,

σ(f(as

))= σ

(as

)=σ(a)

σ(s)=a

s= f

(as

).

So the image of f is in (S−1A)G.

Let as ∈ (S−1A)G, and define s′ =

∏σ∈G σ(s) ∈ SG.

5.13

Let p be a prime ideal of AG and let P be the set of prime ideals of A over p. Suppose that p1, p2 ∈ P andx ∈ p1. Then we have ∏

σ∈Gσ(x) ∈ p1 ∩AG = p ⊆ p2.

Since p2 is prime ideal, there is σ ∈ G such that x ∈ σ(p2). Therefore, p1 ⊆⋃σ∈G σ(p2). By the prime ideal

Avoidance Lemma, p1 ⊆ p2 for some σ. Since A is integral over AG and p1 ∩ AG = p2 ∩ AG, we know thatp1 = σ(p2).

5.14

Let A be an integrally closed domain with K = A(0), L/K a finite, separable, normal field extension, andlet B be the integral closure of A in L, and let G = Gal(L/K). Suppose b ∈ B has minimal polynomialp(x) = xn + a1x

n−1 + . . . + an. Given σ ∈ G, 0 = σ(0) = σ(p(b)) = p(σ(b)), because G fixes A. Since B is theintegral closure of A in L, B is G-stable. Furthermore, we have shown that σ permutes the roots p(x). Therefore,there is b′ ∈ B such that σ(b′) = b.

If b ∈ BG, then b ∈ K because it is fixed, and b is integral over A because it is in B. Since A is integrallyclosed, b ∈ A. Therefore, BG ⊆ A.

5.15

Let A be an integrally closed domain with K = A(0), L/K a finite extension of fields, and let B be the integralclosure of A in L.

5.16

Let k be an infinite field and let A be a finitely generated k-algebra. Then A = k[x1, . . . , xn]/a for some ideala of k[x1, . . . , xn]. If a = (0) or n = 0, then the lemma holds trivially. Suppose that a 6= (0), n > 0, and thatthe lemma holds for any number of generators less than n. Let f ∈ a be non-zero. Without loss of generality, xnis one of the variables appearing in f . Let F ∈ k[x1, . . . , xn] be the homogeneous component of f of maximumdegree, d. Since k is infinite, there are scalars λ1, . . . , λn−1 such that F (λ1, . . . , λn−1, 1) 6= 0. Define the variables,x′i = xi − λixn for 1 ≤ i < n. For an arbitrary monomial, we have

n∏i=1

xνin = xνnn

n−1∏i=1

(x′i + λixn)νi =

(n−1∏i=1

λνii

)xdn + a1x

d−1n + . . .+ ad,

where a1, . . . , ad ∈ k[x′1, . . . , x′n−1]. Therefore, by change of coordinates, we may write

f = F (λ1, . . . , λn−1, 1)xdn + b1xd−1n + . . .+ bd,


with b1, . . . , bd ∈ k[x′1, . . . , x′n−1]. Let a′ be the kernel of the natural map k[x′1, . . . , x

′n−1] → A, and let A′ =

k[x′1, . . . , x′n−1]/a′. Note that the map A′ → A is injective. By induction, there y1, . . . , yr ∈ A′ that are alge-

braically independent over k, and k[y1, . . . , yr]→ A′ is finite, injective, and y1, . . . , yr are k-linear combinations ofx′1, . . . , x

′n−1. By Proposition 2.16, k[y1, . . . , yr] → A is finite, injective, and y1, . . . , yr are k-linear combinations

of x1, . . . , xn. (Practically, the process of proof can be repeated until a = 0. At that point we have obtained thedesired algebraically independent set of generators.)

5.17

Let k be an algebraically closed field and let X ⊆ kn be an affine algebraic set with ideal I(X) 6= (1). LetP (X) = k[t1, . . . , tn]/I(X) be the coordinate ring of X. Since I(X) 6= (1), we know that P (X) is not the zeroring. By Exercise 1.27, we know that any maximal ideal m of P (X) has the form (t1 − a1, . . . , tn − an) forsome a1, . . . , an ∈ k. Let f ∈ I(X). Since m is a maximal ideal of P (X), we know that f ∈ m. Therefore,f(a1, . . . , an) = 0. Since this is true for every f ∈ I(X), the point (a1, . . . , an) is in X.

5.18

Let k be a field and let B be a finitely generated k-algebra that is also a field. Let B be generated by x1, . . . , xnover k. If n = 1, then

1

x1= xn1 + a1x

n−11 + . . .+ an

for some a1, . . . , an ∈ k shows that x1 is integral over k. Therefore, B/k is a finite field extension. Therefore, wemay assume that n > 1.

Let A = k[x1] and let K = A(0). By the inductive hypothesis, B/K is a finite field extension. Therefore,x2, . . . , xn are integral over K, hence algebraic over A. Let f be the product of the lead coefficients of the minimalpolynomials of x2, . . . , xn over A. Then x2, . . . , xn are integral over Af . By the generation assumption on B,this implies that B, hence K is integral over Af . If x1 is transcendental over k, then A is a unique factorizationdomain, hence integrally closed. By Proposition 5.12, Af is integrally closed. Since K/Af is integral, K = Af .

5.19 Undone

5.20

Let B be a finitely generated faithful A-algebra that is a domain, and let S = A \ (0). Then S−1B is afinitely generated K-algebra, where K = S−1A is the field of fractions of A. By Noether’s Normalization Lemma,there are x1, . . . , xn ∈ S−1B, algebraically independent over K, such that S−1B is integral over K[x1, . . . , xn].Let z1, . . . , zm generate B as an A-algebra, and let mi(t) = tdi + ai1

si1tdi−1 + . . . +

aidisidi

be the minimal polynomial

of zi over K[x1, . . . , xn] for each 1 ≤ i ≤ m. For each 1 ≤ k ≤ n, let sk ∈ S be such that skxk ∈ B. Define

s =(∏

i,j sij

)(∏k sk), define yk = sxk ∈ B, and let mi(t) ∈ A[x1, . . . , xn][t] be monic polynomials such that

mi(st) = smi(t) for each i. Then y1, . . . , yn are algebraically independent over K, hence over A.

Let B′ = A[y1, . . . , yn]. For each i, polynomial mi(t) gives an equation of integrality for szi over A[x1, . . . , xn].Since yi = xi

s , zi ∈ Bs is integral over B′s. Since Bs is generated by z1, . . . , zn over B′s, the ring extension is integral.

5.21

Let B be a finitely generated faithful A-algebra that is a domain. By Exercise 5.20, there exist y1, . . . , yn ∈ Bthat are algebraically independent over A and a non-zero s ∈ A such that Bs is integral over A[y1, . . . , yn]s. LetΩ be an algebraically closed field and let f : A → Ω be a ring homomorphism for which f(s) 6= 0. Since they1, . . . , yn are algebraically independent over A, f has an extension to g : A[y1, . . . , yn] → Ω given by mapping


each yi to 0. Furthermore, f(s) 6= 0 implies that g(s) is a unit. By the universal property of localization, g hasa lift to a map g : A[y1, . . . , yn]s → Ω. Since Bs is finitely generated and integral over A[y1, . . . , yn]s, it is a finiteA[y1, . . . , yn]s-module. Let z1, . . . , zm be a list of generators. We can successively extend g by mapping each zi toa root of its minimal polynomial in Ω. Since there are only finitely many zi’s, this process terminates in a mapg : B → Ω.

5.22 Undone

Let B be a finitely generated faithful A-algebra that is a domain.

5.23

Let A be a ring.

(i)⇒(ii) By the First Isomorphism Theorem, homomorphic images of A may be identified with quotients.Let a be an ideal of A. The nilradical of A/a is the intersection of all prime ideals of A/a. By the LatticeIsomorphism Theorem we may identify ideals of A containing a with ideals of A/a in such a way that containment,primality, maximality, and intersections are preserved. If p is a prime ideal of A containing a, then there is afamily of maximal ideals mii∈Ip such that p =

⋂i∈Ipmi. By the lattice isomorphism theorem, this intersection

is preserved on passing to the quotient. Therefore the nilradical of A/a is an intersection of maximal ideals. Sincethe Jacobson radical is the intersection of all maximal ideals, the Jacobson radical is contained in the nilradical ofA/a. The reverse containment holds in general, so the Jacobson radical and nilradical of A/a are the same.

(ii)⇒(iii) Let p be a non-maximal prime ideal of A. We apply the Lattice Isomorphism Theorem. The nilradicalof A/p corresponds to the ideal p. The Jacobson radical of A/p corresponds to the intersection of all maximal idealscontaining p. By hypothesis, these are equal. If q is a prime ideal of A strictly containing p, then q is containedin a maximal ideal. Therefore, the intersection of all prime ideals strictly containing p is an ideal containing thenilradical of A/p and contained in the Jacobson radical of A/p. Since these are equal, the intersection of all primeideals strictly containing p is p.

(iii)⇒(i) Suppose A has a prime ideal ideal which is not an intersection of maximal ideals. By passing to thequotient we may assume that A is a domain whose Jacobson radical is non-zero. Let f be a non-zero element ofthe Jacobson radical of A. Since A is a domain, Af 6= 0. Let p be the contraction of a maximal ideal of Af toA. Then p is a prime ideal of A not containing f and is maximal with respect to this property. Therefore, theintersection of all prime ideals strictly containing p contains f , and is not equal to p.

5.24

Let A be a Jacobson ring and let B be an A-algebra.

(i) Suppose that B is integral over A. Let b be an ideal of B. We want to show that√b is the intersection

of maximal ideals of B containing b. Without loss of generality, let b =√b, and let a be the contraction of b to

A. Since taking the radical commutes with taking the contraction, a =√a as well. By Proposition 5.6 (i), B/b is

integral over A/a. Let J be the Jacobson radical of B/b. By Exercise 5.5 (ii), Jc is the Jacobson radical of A/a.Since A is a Jacobson ring, Jc = (0).

Suppose there is f ∈ J that is not nilpotent and f has minimal polynomial tn + a1tn−1 + . . . + an over A/a.

Since fn + a1fn−1 + . . . + an = 0, an ∈ J , hence an ∈ Jc = (0). Since f is not nilpotent, f /∈ q for some prime

ideal q of B/b, thenf · (fn−1 + a1f

n−2 + . . .+ an−1) = 0 ∈ q

implies that fn−1 +a1fn−2 + . . .+an−1 ∈ q. Let p be the contraction of q to A/a. Passing to the quotient gives B/q

integral over A/p, f is in the Jacobson radical of B/q with minimal polynomial of degree n − 1, the contractionof the J to A/p is the Jacobson radical of A/p and is (0), and f is again not nilpotent. As long as n > 1 we mayrepeat this process until we arrive at the case that f has minimal polynomial t + a1. Since f is in the Jacobson


radical of B/q and f = −a1 ∈ A/p, f is in the Jacobson radical of A/p. Therefore, f ∈ p ⊆ q. This contradictsthe hypothesis on q. Therefore, the Jacobson radical of B/b is equal to the nilradical, and B is a Jacobson ring.

(ii) Suppose that B is finitely generated over A. Let q be a non-maximal prime ideal of B and let p be thecontraction of q to A. By passing to the quotient we may assume that B is a finitely generated faithful A-algebrawhich is also a domain and that the Jacobson radical of A is (0). By Exercise 5.22, we know that the Jacobsonradical of B is (0). Since every non-trivial prime ideal of B is contained in a maximal ideal, the intersection of allnon-trivial prime ideals is contained in the Jacobson radical of B, and is therefore (0). This establishes criterion(iii) of Exercise 5.23.

Since Z is a Jacobson ring and every field is a Jacobson ring, finitely generated rings and finitely generatedk-algebras are Jacobson rings.

5.25

Let A be a ring.

(i)⇒(ii) Let A be a Jacobson ring and let B be a field which is a finitely generated A-algebra. Since B is afield, the kernel of A→ B is a prime ideal ideal. Passing to the quotient, we may assume that A is a domain whoseJacobson radical is (0), and B is a faithful A-algebra. Let s ∈ A be as in Exercise 5.21. Since s 6= 0, there is amaximal ideal m of A not containing s. Let k = A/m and let Ω be the algebraic closure of k. Compose the naturalmaps A→ k and k → Ω to obtain a map f : A→ Ω such that f(s) 6= 0. By Exercise 5.21, f has an extension tosome g : B → Ω. Since B is a field, g is injective. Therefore, B is an algebraic extension of k. Since B is finitelygenerated over k, B/k is a finite field extension. Given x ∈ B, let m(t) = tn + a1t

n−1 + . . . + an ∈ (A/m)[t] bethe minimal polynomial of x over k, where a1, . . . , an ∈ A. Then m(t) lifts to some monic m(t) ∈ A[x] such thatm(x) = a′ ∈ m. Thus x satisfies the polynomial m(t)−a′ and is integral over A. Therefore, B is a finite A-algebra.

(ii)⇒(i) Let p be a non-maximal prime ideal of A and let B = A/p. Let f be a non-zero element of B. ThenBf is a finitely generated A-algebra. Suppose that Bf is a field. By hypothesis, it is a finite A-algebra. Thus it isa finite B-algebra, hence integral over B. By Proposition 5.7, B is a field, in contradiction to our hypothesis onp. Therefore, Bf is not a field. Contract a maximal ideal of Bf to B to obtain a prime ideal q not containing fand maximal with this property. Therefore, for every f ∈ \p, there is a prime ideal strictly containing p but notf . So p is the intersection of all prime ideals strictly containing it.

5.26

Let X be a topological space. Suppose Uopen, Cclosed ⊆ X. We claim that U ∩ C is open in its closure. SinceU and C are closed, and U ∩ C ⊆ U ∩ C, we have U ∩ C ⊆ U ∩ C. Thus

U ∩ C ∩ Cc ⊆ U ∩ C ∩ Cc = ∅.

Therefore,

U ∩ C \ (U ∩ C) = U ∩ C ∩ (U c ∪ Cc)= (U ∩ C ∩ U c) ∪ (U ∩ C ∩ Cc)= (U ∩ C ∩ U c),

which is closed. Thus U ∩ C is open in U ∩ C.

On the other hand, suppose that S ⊆ X is open in S. Then there is Uopen ⊆ X such that S = U ∩ S, and wesee that S is the intersection of an open and a closed set.

Let X0 be a subset of X.

(1)⇒(2) Suppose that every non-empty locally closed subset of X meets X0, and let Eclosed ⊆ X. Let x ∈ Eand let U be a neighborhood of x. Since U ∩ E is locally closed, U ∩ E ∩Xo 6= ∅. Therefore, x is in the closureof E ∩X0. The reverse containment holds in general.


(2)⇒(3) Suppose that E ∩X0 = E for every Eclosed ⊆ X, and let U, V be open subsets of X such thatU ∩X0 = V ∩X0. Taking X0-complements yields U c ∩X0 = V c ∩X0. Now take closures in X to obtain

U c = U c ∩X0 = V c ∩X0 = V c.

Thus U = V and the correspondence is injective. On the other hand, the correspondence is surjective by thedefinition of the subspace topology.

(3)⇒(1) Suppose that the correspondence U 7→ U∩X0 from the topology of X to that of X0 is bijective, and letS be a non-empty locally closed subset of X. By definition, there are Uopen, Cclosed ⊆ X such that S = U∩C. Since∅ 7→ ∅ and the correspondence is bijective, U ∩X0 6= ∅. Since S is non-empty, X 7→ X0, and the correspondenceis bijective, Cc ∩X0 6= X0. Therefore, S ∩X0 6= ∅.

Let A be a ring, let X = SpecA, and let X0 be the set of maximal ideals of A.

(i)⇒(ii) Suppose that A is a Jacobson ring. Let a be an ideal of A and let V (b) = V (a) ∩X0. By construction,V (b) ⊆ V (a). On the other hand, V (a) ∩X0 ⊆ V (b) is the set of maximal ideals containing a. Therefore, a andb are contained in exactly the same maximal ideals. Since A is a Jacobson ring,

√a,√b are each the intersection

of the maximal ideals containing them, and are thus equal. In summary, V (a) ∩X0 = V (a), and V (a) was anarbitrary closed set. So X0 is very dense in X.

(ii)⇒(iii) Suppose that X0 is very dense in X and suppose that x is locally closed. By criterion (1), x∩X0

is non-empty. Therefore, x is a maximal ideal of A, and x is closed.

(iii)⇒(i) Suppose that every locally closed subset of X consisting of single point is closed. Let p be a non-maximal prime ideal of A and let f ∈ A \ p. Since Xf ∩ V (p) is locally closed and p is not closed, Xf ∩ V (p)contains a point q other than p, and f /∈ q. Since f was arbitrary, p is equal to the intersection of prime idealideals strictly containing it, and A is a Jacobson ring.

5.27

Let K be a field and let Σ be the set of all local subrings of K. Since K ∈ Σ, Σ is non-empty. Order Σ bydomination and let A1 ≤ A2 ≤ . . . be a chain. Let A =

⋃i≥1Ai and let m =

⋃i≥1 mi, where mi is the maximal

ideal of Ai for each i. Then A is a ring and m is an ideal. Suppose f ∈ A \m. Then f ∈ Ai \mi for some i. SinceAi is a local ring, f−1 ∈ Ai ⊆ A. Therefore, A is a local ring and m is its maximal ideal. By Zorn’s Lemma, Σhas maximal elements.

Let A,m be a maximal element of Σ and let f ∈ K \ A. Then A is a subring of A[f ]. If mA[f ] is a properideal, then it is contained in some maximal ideal n of A[f ]. Since we are working inside of an ambient field, A isa subring of the local ring (A[f ])n and m = n ∩ A. Since A is maximal, A = (A[f ])n, contrary to our assumptionon f . Therefore, 1 ∈ mA[f ], f−1 ∈ m ⊆ A, and we see that A is a valuation ring of K.

Suppose that A is a valuation ring of K. By Proposition 5.18 (i), A,m is a local ring. Suppose that B, n is alocal subring of K dominating A,m. If f ∈ B \ A, then f−1 ∈ A is a non-unit. So f−1 ∈ m ⊆ n. Thus, 1 ∈ n,contrary to hypothesis. Therefore, A,m is a maximal element of Σ.

5.28

Let A be a domain with fraction field K.

(i)⇒(ii) Suppose that A is a valuation ring of K with ideals a, b, let f ∈ b \ a, and let 0 6= g ∈ a. Since A isa valuation ring, f

g ∈ A or gf ∈ A. If f

g ∈ A, then f = fg · g ∈ a, contrary to hypothesis. Therefore, g

f ∈ A and

g = gf · f ∈ b. Thus a ⊆ b.

(ii)⇒(i) Suppose that the ideals of A are totally ordered by containment and let fg ∈ K with f, g ∈ A. Then

either (f) ⊆ (g) or (f) ⊇ (g). So f = ag for some a ∈ A or bf = g for some b ∈ A. Therefore, fg = a ∈ A or

gf = b ∈ A, and A is a valuation ring.


If A is a valuation ring with prime ideal p, then the ideals of A are totally ordered by containment. Since therespective lattices of ideals of Ap and A/p are sublattices of the lattice of ideals of A, they are also totally orderedby containment. Therefore, Ap and A/p are valuation rings of their fields of fractions.

5.29

Let A be a valuation ring of a field K and let B be a subring of K containing A. Since A is a valuation ringof K, B is also a valuation ring of K by Proposition 5.18 (ii). By Proposition 5.188 (i), B is a local ring withmaximal ideal n. Let p = n ∩ A. If s ∈ A \ p, then s ∈ B \ n. So s is a unit in B. By the universal property oflocalization, the inclusion A → B lifts uniquely to a map Ap → B. Furthermore, this map is injective because Ais a domain. If f ∈ B \A, then f−1 ∈ A because A is a valuation ring. Since f ∈ B, f−1 is not in n, hence not inp. Therefore, f = (f−1)−1 ∈ Ap. Therefore, the localization map is surjective as well.

5.30

Let A be a valuation ring of a field K, let U ≤ K∗ be the group of units of A, and let Γ = K∗/U . For ξ, η ∈ Γdefine ξ ≥ η by xy−1 ∈ A for some coset representatives x, y ∈ K∗ of ξ, η ∈ Γ. We claim that this is a well-definedtotal ordering on Γ that is compatible with the group structure.

(Well-Defined) Let a, b ∈ U and let x, y ∈ K∗ such that xy−1 ∈ A. Then

(ax)(by)−1 = ab−1xy−1 ∈ A.

So the relation is well-defined on elements of Γ.

(Total) Since A is a valuation ring, xy−1 ∈ A or yx−1 ∈ A for every x, y ∈ K∗.

(Reflexive) Since xx−1 = 1 ∈ A, the relation is reflexive.

(Anti-Symmetric) If xy−1, yx−1 ∈ A, then xy−1 ∈ U . Therefore, ξη−1 = 1, and ξ = η.

(Transitive) Let ζ ∈ Γ with coset representative z ∈ K∗, and suppose that xy−1, yz−1 ∈ A. Then xz−1 =xy−1 · yz−1 ∈ A.

(Compatible) If xy−1 ∈ A, then (xz)(yz)−1 = xy−1 ∈ A.

Let v : K∗ → Γ be the quotient map. Since A is a valuation ring, xy−1 ∈ A or yx−1 ∈ A. Without loss ofgenerality, xy−1 ∈ A. So min(v(x), v(y)) = v(y). Therefore,

(x+ y)y−1 = xy−1 + 1 ∈ A,

and we see that v(x+ y) ≥ min(v(x), v(y)).

5.31

Let Γ be a totally ordered abelian group (written additively) and let K be a field. A valuation of K withvalues in Γ is a map v : K∗ → Γ such that (i) v(xy) = v(x) + v(y) and (ii) v(x+ y) ≥ min(v(x), v(y)).

Let A ⊆ K be the set of x ∈ K such that x = 0 or v(x) ≥ 0. By (i) and (ii), A is closed under addition andmultiplication. By (i), v(1) = v(1) + v(1), which implies that v(1) = 0. Therefore, 1 ∈ A. On the other hand,0 = v(1) = v(−1) + v(−1) implies that v(−1) = 0 (even if K has characteristic 2). Thus v(−x) = v(x) for allx ∈ A by (i). So A is closed under the taking of additive inverses, and A is a ring.


5.32 Undone

5.33 Undone

5.34 Undone

5.35 Undone

6 CHAIN CONDITIONS 64

6 Chain Conditions

6.1

Let M be an A-module and let u : M →M be A-linear.

(i) Suppose that u is surjective and M is Noetherian. Then the ascending chain, 0 ≤ keru ≤ keru2 ≤ . . .,stabilizes at some index n ≥ 1. Since u is surjective, un is as well. If x ∈ keru and x = un(y) for some y ∈ M ,then

0 = u(x) = un+1(y).

Thus y ∈ kerun+1 = kerun. So x = un(y) = 0. Therefore, u is injective.

(ii) Suppose that u is injective and M is Artinian. Then the descending chain, 0 ≥ imu ≥ imu2 ≥ . . .,stabilizes at some index n ≥ 1. Since u is injective, it lifts to an injective map cokerui → cokerui+1 for each i. Ifx ∈ cokeru, then un(x) ∈ cokerun+1 = cokerun. Thus un(x) = 0. Therefore, x = 0, and u is surjective.

6.2

Let M be an A-module such that every non-empty set of finitely generated submodules has a maximal element,let N be a submodule of M , and let Σ be the set of finitely generated submodules of N . Since 0 ∈ Σ, it is non-empty. Let N ′ be a maximal element of Σ. Since N ′ ∈ Σ, N ′ is finitely generated. If x ∈ N , then Ax + N ′ isa finitely generated submodule of N containing N ′. Since N ′ is maximal, N ′ = Ax + N ′. Since x was arbitrary,N ′ = N . We have proven that every submodule of M is finitely generated. Therefore, M is Noetherian.

6.3

Let N1, N2 be submodules of an A-module M .

(Noetherian) Suppose M/N1,M/N2 are Noetherian, and let

M1 ≤M2 ≤M3 ≤ . . .

be an ascending chain of submodules of M containing N1 ∩N2. For j = 1, 2, adjoin Nj and pass to the quotientto obtain an ascending chain,

(M1 +Nj)/Nj ≤ (M2 +Nj)/Nj ≤ . . . ,

of submodules of M/Nj . Since M/Nj is Noetherian, the ascending chain terminates at some index nj . Letn = max(n1, n2). Then we have the two identities,

(Mn +N1)/N1 = (Mn+1 +N1)/N1

and(Mn +N2)/N2 = (Mn+1 +N2)/N2.

Let f ∈ Mn+1. By the first identity, there is f1 ∈ N1 such that f + f1 ∈ Mn. By the second identity, there isf2 ∈ N2 such that f + f1 = f + f2. Therefore, f1 = f2 ∈ N1 ∩N2, and f ∈Mn +N1 ∩N2. Since f was arbitrary,Mn+1 +N1 ∩N2 = Mn +N1 ∩N2. Therefore, the chain,

M1/(N1 ∩N2) ≤M2/(N1 ∩N2) ≤M3/(N1 ∩N2) ≤ . . . ,

of submodules of M/(N1 ∩N2) terminates. So M/(N1 ∩N2) is Noetherian.

(Artinian) If M/N1,M/N2 are Artinian, the same proof works with the reverse ordering.


6.4

Let M be a Noetherian A-module, let a = Ann(M), and suppose that a1 ≤ a2 ≤ . . . is an ascending chainof ideals of A containing a. Since M is Noetherian, it is finitely generated by some x1, . . . , xk. Again, since Mis Noetherian, for each i the ascending chain of submodules, a1xi ≤ a2xi ≤ . . ., stabilizes for some index ni. Letn = max(n1, . . . , nk). Then anxi = an+1xi for each i. Hence, anM = an+1M . Let a ∈ an+1. Since anM = an+1M ,the multiplication by a map M/anM →M/anM is the zero map. Since a ⊆ an, the annihilator of M/anM is an.Therefore, a ∈ an. Since a was arbitrary, an+1 ⊆ an, and the chain stabilizes. Therefore, A/a is a Noetherian ring.

Let M = Z2/Z as a Z-module (that is M consists of fractions whose denominators are powers of 2, modulo 1).Let N be a submodule of M . If n

2k∈ N is in lowest terms, then there are a, b ∈ Z such that

a · n2k

=an+ b2k

2k=

1

2k∈ N.

Therefore, (2−kZ)/Z ⊆ N . If N contains elements of arbitrarily large denominator, then N = M . Otherwise,N = (2−kZ)/Z for some k ≥ 0. Now, having classified all of the Z-submodules of M , we can see that M is Artinian.If n ∈ Z annihilates M , then 2k|n for all k ≥ 0. Thus Ann(M) = (0). However, Z/(0) ∼= Z is not Artinian because(2) ⊇ (4) ⊇ (8) ⊇ . . . is a non-terminating descending chain of ideals.

6.5

Let X be a Noetherian topological space, let Y be a subspace, and let U1 ⊆ U2 ⊆ . . . be an ascending chainof open subsets of Y . By the definition of the subspace topology, there is some open subset V1 of X such thatV1 ∩ Y = U1. Given open subsets V1 ⊆ V2 ⊆ . . . ⊆ Vn−1 of X such that Vi ∩ Y = Ui for each i, let V ′n be an opensubset of X such that V ′n ∩ Y = Un, and define Vn = Vn−1 ∪ V ′n. Then we have Vn−1 ⊆ Vn and

Vn ∩ Y = (Vn−1 ∩ Y ) ∪ (V ′n ∩ Y ) = Un−1 ∪ Un = Un,

so the inductive selection of the family of V ’s may continue. Since X is a Noetherian topological space, there isan index, N , at which the chain of V ’s stabilizes. Therefore,

UN = VN ∩ Y = Vn ∩ Y = Un,

for all n ≥ N . So Y is a Noetherian topological space.

Let Wii∈I be an open cover of X. Without loss of generality, I is an ordinal. For each i ∈ ω ∩ I, defineYi =

⋃j<iWj . Then Y1 ⊆ Y2 ⊆ . . . is an ascending chain of open subsets of X. Since X is Noetherian, this

chain stabilizes at some index n ∈ N. Since Wii∈I is an open cover, X ⊆ Yn =⋃j<nWj . Therefore, X is

quasi-compact.

6.6

Let X be a topological space.

(i)⇒(iii) If X is Noetherian, and Y is a subspace of X, then Y is Noetherian by Exercise 6.5. Furthermore,Exercise 6.5 implies that Y is quasi-compact.

(iii)⇒(ii) This is obvious.

(ii)⇒(i) Suppose that every open subspace of X is quasi-compact, let U1 ⊆ U2 ⊆ . . . be an ascending chain ofopen subsets of X, and define U =

⋃i≥1 Ui. Since U is a union of open subsets it is open, hence quasi-compact, and

Uii≥1 is an open cover of U . Therefore, there is some n such that U1, . . . , Un cover U , and the chain stabilizes.


6.7

6.8

Let A be a Noetherian ring, let X = SpecA, and let V1 ⊇ V2 ⊇ . . . be a descending chain of closed subsetsof X. By the definition of the Zariski topology, there are radical ideals ai of A such that Vi = V (ai) for each iand a1 ⊆ a2 ⊆ . . . is an ascending chain of ideals. Since A is Noetherian, this chain terminates at some index, n.Therefore, the descending chain of closed subspaces of X terminates at n as well. Therefore, X is Noetherian.

By our method of proof one can see that if X is Noetherian, then A satisfies the ACC on the set of radicalideals.

6.9 Undone

6.10 Undone

6.11 Undone

6.12 Undone

7 NOETHERIAN RINGS 67

7 Noetherian Rings

7.1

8 ARTIN RINGS 68

8 Artin Rings

9 DISCRETE VALUATION RINGS AND DEDEKIND DOMAINS 69

9 Discrete Valuation Rings and Dedekind Domains

10 COMPLETIONS 70

10 Completions

11 DIMENSION THEORY 71

11 Dimension Theory

12 CONVENTIONS 72

12 Conventions

1. Don’t use “∈”.

2. Don’t use “$”.

3. Direct proofs when possible.

4. Give constructive and non-constructive proof when possible.

5. Colon for definitions.

6. Reader should not need to look anything up.

13 REMARKS 73

13 Remarks

Remark 1. I would like to prove an analogue of Exercise 1.2 (iii) by the following induction: If f is a zero-divisor,then by Exercise 1.2 (iii), there is an element a of Ar−1 such that af = 0. Then claim that the lead coefficient ofa as a polynomial in xr−1 kills f as well. However, the argument in Exercise 1.2 (iii) made use of the hypothesisthat g was of minimum degree, and we have no obvious such control on the degree of the lead coefficient of a withrespect to a single variable. If you find an example showing that this method of proof will not work, or if you finda way of making it work, please send me an email.

Remark 2. The following proof is for the case that A is reduced. I would like to extend this approach to thenon-reduced case, but I am skeptical that this is possible. In particular, a regular function on a reduced schemeis uniquely determined by its values. This is no longer true for non-reduced schemes. I suspect that this featureis at the heart of the matter.

Remark 3. The following proof uses machinery from Chapter 3. The merit of the proof is that it has a geometricinterpretation: Z/mZ corresponds to a module supported on the finite closed subset V (m) of SpecZ. If m,n arecoprime, then V (m) ∩ V (n) is empty. Therefore the tensor product module has empty support.

Remark 4. There are two other proofs of this fact that are somewhat straightforward, albeit tedious. You can provethat M ⊗AN satisfies the universal property of the direct sum or you can prove that

⊕i∈I (Mi ⊗A N) satisfies the

universal property of the tensor product. Since universal objects are unique up to unique isomorphism, in eithercase you have the desired result.

Remark 5. More generally, suppose M is an A-module with endomorphism φ such that φ2 = φ. Then

M = (kerφ)⊕ (imφ) .

This can be seen by applying φ and idM −φ to the decomposition m = (m−φ(m))+φ(m). Such an endomorphismis called a linear projection.

Suppose that 0 → Kf−→ M

g−→ N → 0 is a short exact sequence of A-modules. If there is σ : N → M suchthat g σ = idN , then σ g is a projection on M . If there is τ : M → K such that τ f = idK , then f τ is aprojection on M . In either case, we obtain an identification of M with K ⊕N .

Remark 6. Here we give an alternative construction of the direct limit of the directed system (Mi, µij : Mi →Mj).Let X =

∐i∈IMi be the disjoint union of the sets (Mi)i∈I . Define the relation ∼ on X by (x, i) ∼ (y, j) if there

is k ≥ i, j such that µik(x) = µjk(y).

We claim that ∼ is an equivalence relation on X. Reflexivity and symmetry follow immediately from thoseproperties of equality. Suppose that (x, i) ∼ (y, j) and (y, j) ∼ (z, k). Then there are m,n ∈ I such that m ≥ i, j;n ≥ j, k; µim(x) = µjm(y); and µjn(y) = µkn(z). Let p ∈ I such that p ≥ m,n. Then we have

µip(x) = µmp(µim(x)) = µmp(µjm(y)) = µjp(y) = µnp(µjn(y)) = µnp(µkn(z)) = µkp(z).

That is, (x, i) ∼ (z, k).

Let M = X/ ∼. We will put the structure of an A-module on M . Let [(x, i)], [(y, j)] ∈M . Define

[(x, i)] + [(y, j)] := [(µik(x) + µjk(y), k)],

where k ∈ I is such that k ≥ i, j. Let us show that addition is well-defined. Suppose that l ≥ i, j. Then there ism ≥ k, l such that

µkm(µik(x) + µjk(y)) = µim(x) + µjm(y) = µlm(µil(x) + µjl(y)).

So the equivalence class [(µik(x) + µjk(y), k)] does not depend on the choice of k.

Suppose (u, p) ∼ (x, i) and (v, q) ∼ (y, j). Pick k ≥ i, j, p, q such that µpk(u) = µik(x) and µqk(v) = µjk(y).Then

µik(x) + µjk(y) = µpk(u) + µqk(v).

Therefore, addition does not depend on the choice of class representatives.

13 REMARKS 74

Let a ∈ A and definea · [(x, i)] := [(ax, i)].

If (x, i) ∼ (y, j), let k ≥ i, j. Then

µik(ax) = aµik(x) = aµjk(y) = µjk(ay).

So scalar multiplication is well-defined.

The axioms for M to be an A-module follow from the maps µij being A-linear and from each Mi being anA-module. The can be tediously verified.

Each map µi : Mi → M is defined as the inclusion Mi → X followed by the quotient X → M . It is easy tocheck that these maps are A-linear. Let i ≤ j and let x ∈Mi. Then µj(µij(x)) = [(µij(x), j)] and µi(x) = [(x, i)].Take k = j in the definition of ∼ to see that [(x, i)] = [(µij(x), j)].

Remark 7. I have a suspicion that there is a problem with the given proof.

Remark 8. A useful consequence of Exercise 2.16 is that the construction of the direct limit given in the text ofExercise 2.14 and that in Remark 6 are canonically isomorphic.

Remark 9. Let A be a ring and let (Mi)i∈I , (µij)i,j∈I be a directed system of A-modules. Let i0 ∈ I and let Ji0 bethe set of indices j ∈ I such that j ≥ i0. Then lim−→j∈Ji0

Mj is canonically isomorphic to lim−→i∈IMi as an A-module.

Below is a proof.

First, we need to define the structure maps µ′i : Mi → lim−→j∈Ji0Mi to see lim−→j∈Ji0

Mi as a direct limit over I.

Let (µi)i∈Ji0 be the structure maps of lim−→j∈Ji0Mi from Exercise 2.14. If i ∈ I, then there is j ∈ Ji0 such that i ≤ j.

Define µ′i := µj µij . The identity stated in the text at the end of Exercise 2.14 shows that µ′i is well-defined.

We will use the characterization of lim−→Mi given in Exercise 2.16. Let N be an A-module and let (αi : Mi →N)i∈I be a family of A-linear maps such that αi = αj µij for all i ≤ j in I. Let µ′i(x) ∈ lim−→j∈Ji0

Mi. Define

α(µ′i(x)) := αi(x).

Let us show that α is well-defined. Let µ′i(x) = µ′j(y) and let k ≥ i, j. Then

µ′k(µik(x)) = µ′i(x) = µ′j(y) = µ′k(µjk(y)).

So µ′k(µik(x)− µjk(y)) = 0. By Exercise 2.15 there is l ≥ k such that µil(x) = µjl(y). Therefore,

αi(x) = αl(µil(x)) = αl(µjl(y)) = αj(y),

and α is well-defined. On the other hand, the definition of α was forced by the condition α µ′i = αi. So α isunique.

Remark 10. Here is a characterization of the Tor functor due to J. P. May (we have adapted it to commutativerings):

For every n ∈ Z there is a functor

TorAn : A−Mod×A−Mod→ A−Mod

such that for every A-module M and short exact sequence of A-modules

0→ N ′ → N → N ′′ → 0

There are connecting homomorphisms

∂ : TorAn (N ′′,M)→ TorAn−1(N ′,M) and ∂ : TorAn (M,N ′′)→ TorAn−1(M,N ′),

natural in M and in the choice of short exact sequence that satisfy the following properties:

1. TorAn (M,N) = 0 for all n < 0.

13 REMARKS 75

2. TorA0 (M,N) is naturally isomorphic to M ⊗A N .

3. TorAn (M,N) = 0 for n > 0 if either M or N is projective.

4. The following sequences are exact:

· · · → TorAn+1(N ′′,M)→ TorAn (N ′,M)→ TorAn (N,M)→ TorAn (N ′′,M)→ TorAn−1(N ′,M)→ · · ·· · · → TorAn+1(M,N ′′)→ TorAn (M,N ′)→ TorAn (M,N)→ TorAn (M,N ′′)→ TorAn−1(M,N ′)→ · · ·

Furthermore, for each fixed M , the functors TorAn (M, ·) and TorAn (·,M) with their associated connecting homo-morphisms are uniquely determined up to isomorphism by properties 1-4.

Here are some further properties from Weibel’s An Introduction to Homological Algebra:

• (Weibel 3.2.8) Let M,N be A-modules and let F∗ →M be a flat resolution of M . Then

TorA∗ (M,N) ∼= H∗(F∗ ⊗A N) and TorA∗ (N,M) ∼= H∗(N ⊗A F∗).

Note that free ⇒ projective ⇒ flat.

•

Remark 11. Let a be an ideal of A and let M be an A-module. Tensor the short exact sequence

0→ a→ A→ A/a→ 0

with M to get the following commutative diagram with exact rows:

TorA1 (A,M) TorA1 (A/a,M) a⊗AM A⊗AM (A/a)⊗AM 0

0 0 aM M M/aM 0

Taking the kernel complex yields the exact sequence

0 = TorA1 (A,M)→ TorA1 (A/a,M)→ ker (a⊗AM → aM)→ 0.

As a consequence, we have the following chain of implications:

“M is flat”

⇒ “a⊗AM = aM for all ideals a of A”

⇒ “a⊗AM = aM for all finitely generated ideals of A”

⇒ “M is flat.”

Remark 12. A consequence of Exercise 2.27: Every finitely generated ideal of an absolutely flat ring is projective.

Remark 13. Here I will develop an alternative construction of the localization of a ring A with respect to amultiplicative subsemigroup S.

Construction of As: Let s ∈ A. Define As := A[x]/(sx − 1) and let f : A → As be the composition of thecanoncial maps A→ A[x] and A[x]→ As. In the ring As, we will denote the class of x as s−1.

Representation of Elements: Every element of As has a representative of the form as−n for some n ≥ 0and a ∈ A/

Given any polynomial anxn + an−1x

n−1 + . . .+ a0 in A[x] we have the following congruence mod (sx− 1):

anxn + an−1x

n−1 + . . .+ a0 ≡ anxn + an−1sxn + . . . a0s

nxn = (an + an−1s+ . . . a0sn)xn.

13 REMARKS 76

Universal Property of As: If B is an A-algebra, then # homA−alg(As, B) ≤ 1, with equality if and only ifs is invertible in B.

Let g ∈ homA−alg(As, B). Then1 = g(1) = g(ss−1) = sg(s−1)

implies that s is invertible in B. Furthermore, let f : A→ As, h : A→ B be the structure maps. Then

g(f(a)f(s)−1) = h(a)h(s)−1

shows that g is uniquely determined by the structure map of B. Therefore, # homA−alg(As, B) ≤ 1.

Conversely, suppose B is an A algebra such that s is invertible in B. By the Universal Property of PolynomialRings, there is a unique A-algebra homomorphism A[x]→ B sending x to s−1. Clearly, sx− 1 is in the kernel. Sothis A-algebra homomorphism lifts uniquely to an A-algebra homomorphism As → B such that s−1 7→ s−1.

Characterization of ker(A→ As): a ∈ ker(A→ As) if and only if s ∈√

AnnA(a).

Suppose a ∈ A maps to 0 in As. Then there is

q(x) = bmxm + bm−1x

m−1 + . . . b0 ∈ A[x]

such thata = (sx− 1)q(x) in A[x]. (4)

Looking at the degree m+ 1 term on the right hand side of (4) shows that sbm = 0. Therefore,

sa = (sx− 1)(sbm−1xm−1 + . . .+ sb0).

Repeating this same reasoning a total of m times yields

sma = (sx− 1)smb0 = sm+1b0x− smb0.

By comparing degrees, we getsma = −smb0 and sm+1b0 = 0.

Therefore,sm+1a = −sm+1b0 = 0.

Conversely, suppose that a ∈ A and sma = 0 for some m ≥ 1. Then

a ≡ smaxm mod (sx− 1) = 0.

Construction of S−1A: We will construct S−1A as a direct limit of A-algebras.

Let S ⊆ A be a multiplicatively closed set containing 1. Let S ′ be the category with objects Ass∈S andA-algebra homomorphisms as morphisms. By the “Universal Property 1” each hom-set has cardinality at most 1.Let S be a subcategory of S ′ obtained by selecting a single representative of each isomorphism class of S ′. Then Sis a partial order. If s ∈ S, then As is uniquely isomorphic to a unique object As′ of S and we may identify theseA-algebras without further comment. By abuse of notation, we will refer to an object As of S as s. In particular,s1 ≤ s2 if and only if there is an A-algebra homomorphism σs1s2 : As1 → As2 .

Suppose that s1, s2 ∈ S. In A(s1s2) we have the identities s1 ·(s2(s1s2)−1

)= 1 and s2 ·

(s1(s2s1)−1

)= 1.

Therefore, s1, s2 ≤ s1s2. So S is a directed set. Properties (1) and (2) of Exercise 2.14 follow immediately fromthe cardinality of hom-sets. Therefore, Ass∈S , σs1s2 : As1 → As2s1≤s2 is a directed system of A-algebras. Wedefine

S−1A := lim−→s∈S

As.

If we picked a different subcategory S of S ′, then our previous remarks would show that we obtain an isomorphicpartially ordered set and all objects are uniquely isomorphic as A-algebras. Therefore, S−1A has been constructedup to unique isomorphism.

13 REMARKS 77

Universal Property of S−1A: (Proposition 3.1) If B is an A-algebra, then # homA−alg(S−1A,B) ≤ 1, with

equality if and only if every element of S is invertible in B.

By the Universal Property of As and the identification,

homA−alg(S−1A,B) = lim←−

s∈ShomA−alg(As, B),

we can see that the set homA−alg(S−1A,B) has at most one element and is non-empty precisely when every element

of S is invertible in B.

A Third Construction: Let A be a ring with multiplicatively closed subset S. Let B := A [xss∈S ]be the polynomial ring over A with an indeterminate for each element of S. Let I be the ideal of B generatedby the family sxs − 1s∈S , and define S−1A := B/I. It is straightforward to show that the ring B/I satisfiesthe universal property of S−1A by applying the Universal Property of Polynomial Rings and using the FirstIsomorphism Theorem.

Documents

Solutions to the Problems in Introduction to Commutative ...jtaylor/notes/atiyah... · Solutions to the Problems in Introduction to Commutative Algebra by M. F. Atiyah and I. G. MacDonald