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International Journal of Mathematical Archive-4(5), 2013, 235-239 Available online through www.ijma.info ISSN 2229 – 5046

International Journal of Mathematical Archive- 4(5), May – 2013 235

HOMEOMORPHISMS OF THE PRIME SPECTRA

A.V. S. N. Murty1*, A. V. H. Sastry2 and A. V. G. S. Sastry3

1Professor in Mathematics, Srinivasa Institute of Engineering & Technology,

Cheyyeru (V), NH-216, Amalapuram-533222, (A.P.), India

2Head Master, Z. P. High School, S. Rayavaram, Visakhapatnam District, (A.P.), India

3Assistant Professor in Mathematics, Srinivasa Institute of Engineering & Technology, Cheyyeru (V), NH-216, Amalapuram-533222, (A.P.), India

(Received on: 12-04-13; Revised & Accepted on: 01-05-13)

ABSTRACT In this paper it is proved that closed subspaces of Spec R are homeomorphic to the spectra of the quotient rings of R. If R is a finite direct product of rings 1R , 2R ,…, nR then it is proved that Spec R is the disjoint union of open sets that

are homeomorphic to Spec iR . Keywords: Prime ideals, Prime Spec R, Quotient Rings and Homeomorphism. AMS Mathematics Subject Classifications (2000): 06E15, 16D25. 1. INTRODUCTION The notion of points in geometry and the study of prime ideals occupy the central part in the study of Commutative Algebra ([1] and [2]). Let R be a commutative ring with identity and X be the set of all prime ideals of R. For any A⊆R, let X (A)={P∈X | A⊈P}. If A = {a}, then we write simply X(a) for X({a}). It is clear that the class {X (A) | A ⊆R} forms a topology on X, for which {X (a) | a ∈R} is a base. The set of prime ideals of a commutative ring R with identity together with this topology is called the Prime Spectrum of R and is denoted by Spec R ([3], [5], [6], [8] and [9]). Also studied some properties of homeomorphisms of the Prime Spectra ([3], [6] and [8]). Note: Throughout this paper, only commutative rings with identity are considered and hence it was preferred to call these as rings for simplicity. 2. PRELIMINARIES Definition 2.1: Let R be a ring and X be the set of all prime ideals of R. For any A⊆R, let X (A) = {P∈X | A ⊈ P}. If A = {a}, then we write simply X(a) for X({a}). It is clear that the class {X (A) | A ⊆R} forms a topology on X, for which {X (a) | a ∈R} is a base. Definition 2.2: The set X of all prime ideals of a ring R together with the topology {X (A) | A ⊆R} is called the Prime Spectrum of R and is denoted by Prime Spec R or simply, Spec R. Definition 2.3: For any ring R, A⊆R and Y⊆ Spec R, (i) Hull of A is defined as H(A) = {P ∈ Spec R | A ⊆P}= Spec R – X(A) (ii) Kernel of Y is defined as ( )

P Y

K Y P∈

=

If :f R S→ is any homomorphism of rings, then, for any prime ideal P of S, 1( ) { | ( ) }f P a R f a P− = ∈ ∈ is also

a prime ideal of R. In fact, this correspondence 1( )P f P−→ becomes a continuous map of Spec S into Spec R. The following can be easily proved.

Corresponding author: A.V. S. N. Murty1* 1Professor in Mathematics, Srinivasa Institute of Engineering & Technology,

Cheyyeru (V), NH-216, Amalapuram-533222, (A.P.), India

http://www.ijma.info/�

A.V. S. N. Murty1*, A. V. H. Sastry2 and A. V. G. S. Sastry3/HOMEOMORPHISMS OF THE PRIME SPECTRA/IJMA- 4(5), May-2013.

© 2013, IJMA. All Rights Reserved 236

Theorem 2.4: Let :f R S→ be a homomorphism of rings. Define * :f Spec S Spec R→ by * 1( ) ( )f P f P−=

for all P∈Spec S. Then *f is a continuous map. Theorem 2.5: Let I be a proper ideal of a ring R. Then Spec (R/I) is homeomorphic to H (I). Definition 2.6: Let :f R S→ be a homomorphism of rings. For any ideal I of R, the ideal generated by ( )f I in S is

called the extension of I with respect to f and is denoted by eI . Also, for any ideal J of S, the inverse image 1f − (J)

is an ideal of R and is called the contraction of J with respect to f and is denoted by cJ 3. HOMEOMORPHISMS Definition 3.1: Let X and Y be topological spaces. Let f : X → Y be a bijection. If both f and 1f − are continuous then f is called a homeomorphism. Theorem 3.2: Let :f R S→ be a homomorphism of rings and * :f Spec S Spec R→ be the corresponding continuous map. Then,

* 1(1) ( ( )) ( ),ef H I H I− = for any ideal I of R *(2) ( ( )) ( ),cf H J H J= for any ideal J of S

Proof: (1) Let I be an ideal of R. For any P Spec S∈ ,

* 1 *( ( )) ( ) ( )P f H I f P H I−∈ ⇔ ∈

1( ) ( )f P H I−⇔ ∈

1( )I f P−⇔ ⊆ ( )f I P⇔< >⊆

eI P⇔ ⊆

( )eP H I⇔ ∈ Therefore * 1( ( )) ( )ef H I H I− = (2) Let J be an ideal of S. First we observe that 1 1( ( )) ( ( ))r f J f r J− −= ; a∈ 1( ( ))r f J− 1( )na f J−⇔ ∈ , for some n>0

1( ) ( ) ( ) ( ( ))nf a J f a r J a f r J−⇔ ∈ ⇔ ∈ ⇔ ∈

Now, * *( ( )) ( ( ( ( ))))P f H J P H K f H J∈ ⇔ ∈

*( ( ( ))K f H J P⇔ ⊆

* ( ( ))

*

( )

1

1

1

1 1 1

1

( )

( )

( )

( ( ))( ) (sin ( ( )) ( ( )))

( ( ))

Q f H J

A H J

J A Spec S

J A Spec S

Q P

f A P

f A P

f A P

f r J Pf J P ce f r J r f JP H f J

∈

∈

−

⊆ ∈

−

⊆ ∈

−

− − −

−

⇔ ⊆

⇔ ⊆

⇔ ⊆

⇔ ⊆

⇔ ⊆

⇔ ⊆ =

⇔ ∈



Therefore *( ( )) ( )Cf H J H J= . Theorem 3.3: Let :f R S→ be a homomorphism of rings and * :f Spec S Spec R→ be the corresponding continuous map. Then (1) If f is an epimorphism, then *f is a homeomorphism of Spec S onto the closed subset H (Ker f). Moreover, Spec

R and Spec(R/N(R)) are homeomorphic. (2) *( )f Spec S is dense in Spec R if and only if ( )Ker f N R⊆ ; in particular, if f is a monomorphism, then

*( )f Spec S is dense in Spec R. Proof: (1) Suppose :f R S→ is an epimorphism. Then by the fundamental theorem of homomorphism, ( / )R Ker f S≅ and hence, by the theorem 2.5, Spec S ≅ Spec ( / )R Ker f ≅ H(Ker f). Also Spec (R/N(R)) ≅ H(N(R))=Spec R. Thus Spec R and Spec (R/N(R)) are homeomorphic. (2) We have *( )f Spec S is dense in Spec R if and only if *( ( ))K f Spec S = N(R)

Now, *( ( ))K f Spec S = *( )Q Spec S

f Q∈

= 1( )Q Spec S

f Q−

∈

= 1(Q Spec S

f Q−

∈

)

= 1( {0})f r−

= 1( (0))r f − = ( )r Ker f

Also, Ker f ⊆ N(R) if and only if r(Ker f) = N(R). Therefore, *( )f Spec S is dense in Spec R if and only if Ker f⊆N(R). If f is a monomorphism, then Ker f = {0}⊆N(R) and hence *( )f Spec S is dense in Spec R. Theorem 3.4: Let R1 and R2 be rings and R= R1× R2. Then Spec R is the disjoint union of open sets X1 and X2 such that Xi is homeomorphic with Spec iR . Proof: Put I1= R1×{0} and I2= {0}×R2. Then I1 and I2 are ideals of R= R1×R2. Let X = Spec R. Put X1 = X (I1) and X2 = X (I2). Then 1 2 1 2 1 2( ) ( ) ( ) ( )X X X I X I X I I X R X= = + = =

and 1 2 1 2 1 2( ) ( ) ( ) ({0})X X X I X I X I I X φ= = = = . Therefore 1X and 2X are open sets in Spec R which is the disjoint union of 1X and 2X . Now, we shall prove that iX is homeomorphic to Spec iR . For this, consider the projection map 1 1 2 1:f R R R× →

defined by 1 1 2 1( , )f a a a= . Then 1f is an epimorphism. By the theorem 3.3, 1 1 2 1 1( ) ( ) ( )Spec R H Ker f H I X I X≅ = = = .



Thus 1X is homeomorphic with Spec 1R . Similarly, 2X is homeomorphic with Spec 2R . The following is the immediate consequence of the above.

Corollary 3.5: Let 1 2, ,..., nR R R be rings and1

n

ii

R R=

=∏ be the direct product. Then Spec R is the disjoint union of

open sets iX such that iX is homeomorphic with Spec iR . Corollary 3.6: If 1 2, ,..., nK K K are fields, then Spec ( 1K × 2K ×…× nK ) is the n-element discrete space. It is proved that * :f Spec S Spec R→ is a continuous map if :f R S→ is a homomorphism of rings. This

*f may not be an open map. In the following it come across a situation where *f is a continuous bijection which is not a homeomorphism Theorem 3.7: Let R be an integral domain with exactly one non-zero prime ideal P and let K be the field of fractions of R. Let S = (R/P)×K. Define :f R S→ by f(a)=(a+P, a/1) for all a∈R. Then *f is a bijection but not a homeomorphism. Proof: Let X = Spec R. Since R is an integral domain, {0} is a prime ideal of R. Therefore, {0} and P are the only prime ideals of R and hence X = {{0}, P}. By hypothesis, we get that P is a maximal ideal of R and hence R/P is a field. Therefore, {0} × K and (R/P) × {0} are the only prime ideals of (R/P) × K. Let Y = Spec S. Then Y = {P1, P2}, where P1= {0}×K and P2=(R/P) ×{0}. Define * :f Spec S Spec R→ by * 1( ) ( )f P f P−= , for all P Spec S∈ . We shall prove that *f is a bijection but not homeomorphism. To prove that *f is a bijection. Now,

* 11 1 1( ) ( ) { | ( ) }f P f P a R f a P−= = ∈ ∈

| , {0}1aa R a P K = ∈ + ∈ ×

{ | 0 }a R a P P= ∈ + = + { | }a R a P= ∈ ∈ P= and

* 12 2 2( ) ( ) { | ( ) }f P f P a R f a P−= = ∈ ∈

| , ( / ) {0}1aa R a P R P = ∈ + ∈ ×

0|

1 1aa R = ∈ =

{0}=

Thus *f is a bijection. To prove that *f is not homeomorphism. Let Y (0, 1), Y (1, 0) be basic open sets in Spec S respectively. Then, clearly Y(0,1) = { 2P } and Y(1,0) = { 1P }.



Now, *

2({ })f P = {{0}}= X(a) if 0 a P≠ ∈ and *1({ })f P = {P}≠ X(a) for all a R∈ .

If a P∈ , then X(a) = {P, {0}}. If 0 a P≠ ∈ , then X(a) = {{0}}. If a = 0, then X(0) = φ . Therefore, *

1({ })f P is not open in Spec R, though 1{ }P is open in Spec S and hence *f is not continuous. Thus *f is not homeomorphism.

REFERENCES [1] Hochster, M., Prime ideal structure in commutative rings, Trans. Amer. Math. Soc. 142 (1969), 43-60. [2] Hochster, M., Existence of topologies for commutative rings with identity, Duke Math. J. 38 (1971), 551-554. [3] Lu, C. P., The Zariski topology the Prime Spectrum of a Module, Houston J. Math, 25 (3), 1999, 417-425. [4] Atiyah, M. F., and Macdonald., I. G., Introduction to Commutative Algebra, Univ. of Oxford, Addison-Wesley Publishing Co., 1969. [5] Jacobson, N., Basic Algebra, Voi.II, Hindustan Publishing Corporation, India, 1984. [6] James, R. Munkers., Topology a First course, Prentice-Hall of India Pvt. Ltd., 1975. [7] Joachim, L., Lectures on Rings and Modules, McGill University, Blaisdell Publishing Company, A Division of Ginn and Company, 1966. [8] Swamy, U. M., Commutative Algebra, Course Book for M.Sc., programme in Mathematics, Dr. B. R. Ambedkar Open University, Hyderabad, India. [9] Murty, A.V.S.N., Sastry, A.V.G.S., Prime and Maximal Spectra, International Journal of Mathematical Archive-4(3), 2013, 1-5.

Source of support: Nil, Conflict of interest: None Declared

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