10,1,2Recall that is indeed a group; certainly the restriction of to is a mapping

. Now for all , we have , and if , then . So acts on by the restriction of our module operator .

10,7We use the submodule criterion.Certainly for some . Thus , and hence .Now suppose and . Now and for some and . Suppose without loss of generality that ; then . Since is a submodule, . Thus .Hence is a submodule of .

10,1,10We use the submodule criterion.Certainly ; then .Let , let , and let . Now since is a right ideal. Then

. Thus .Thus is a submodule of .

10,1,111. We claim that . Let be in the annihilator of . In particular,

. Thus , , and , so that divides . Thus . Let . Now let . Note that

. So annihilates . Thus .

2. We claim that . Let annihilate . In particular, . Then mod 24,

mod 15, and mod 50. Thus mod 12, mod 15, and mod 25. So . Suppose

, and let . Now ; thus annihilates . So

.

10,11,12

1. Let , and let . Certainly then . Thus .

2. Now let , let , and let . We claim that . To see this, suppose annihilates . Then in particular,

. Then mod 6, so that mod 2. Now let and note that . So . Now we claim that

. To see this, suppose . Then in particular, , and we have mod 2 and mod 6. Then mod 3 and we have no restrictions on ; thus

. Suppose now that and let . Then , as desired. Note that .

3. Let and let . Certainly then , so that in fact .

4. Now let , let , and let . We claim that . To see this, suppose annihilates . Then in

particular, . Then mod 6, and we have mod 3. hence . Suppose now that and let

. Then as desired. Thus . Next we claim that . To see this,

suppose annihilates . In particular, . So mod 6, and thus mod 2. Suppose now that .

Since , we have . Note then that .

10,1,13We claim that for each , . To see this, let and let . Now by definition, is a finite sum of -fold products of elements in . Note that every -fold product of elements in is annihilated my , so that in fact . Thus

.

So is the union of a chain of submodules. By 10,1,7 previous exercise, it is a submodule of .

10,1,14Let . We use the submodule criterion to show that is a submodule of . Note first that , so that is not empty. Now suppose and

. Then . Thus is a submodule of .

Now for the example, note that . Letting and

, we have . Thus is not a submodule of .

10,1,16We claim that the -submodules of are precisely the sets of the form

for . (Note that and .) We prove this in two parts: first, that each is an -submodule, and second, that every -submodule is equal to some .Let . Note that , so that is nonempty. Now let and

. Note that since, for , the th component of is , and the th coordinate of is 0. Moreover, for all . Now say

; then . This is a finite sum of elements whose th coordinate is 0 for all , and thus is in . So by the submodule criterion, is an -submodule.Now suppose is an -submodule. Let be maximal such that there exists an element with . If no such exists, then . Thus we may assume that exists and thus . Certainly ; we claim that these two sets are in fact equal. To show this, we first prove by induction that for all , where eij=1 if and 0 otherwise. (The “standard basis vectors”, whatever that means.) We begin with some lemmas.Lemma 1: Let and . Let . Then the -th component of is

for and is 0 for . Proof: We proceed by induction on . For the base case, suppose . Let ; then for . Now if , then

. By definition, then, the th component of is for and is 0 for . For the inductive step, suppose that for some , the conclusion holds for all

. Let . We now proceed by induction on . For the base case, let . By definition, then, the th component of is for , is for

, and is 0 for . For the inductive step, suppose that for some , the conclusion holds for all . The th component of is $latex for and 0 for . By the induction hypothesis, the th component of is for and is 0 otherwise. By induction, the lemma holds. Recall that has the property that . We claim that . To see this, note that by Lemma 1, the th component of is 0 if and is

for .

Now we claim that if, for some , we have for all , then .

Specifically, we claim that . To see this, note that the th component of is for and 0 otherwise. Then each

cancels out the th component for , leaving 1 in the th component and 0 elsewhere.By induction, then, for all . If , then . Thus

.

10,1,23We begin with a lemma.Lemma: Let be a ring and let be a family of ring homomorphisms. Define by . Then is a ring homomorphism. Moreover, if some

is injective, then is injective. Moreover, if each is unital, then is unital. Proof: For all and , we have

and . So and , and hence is

a ring homomorphism. Now suppose some is injective. If , then in particular . So , and is injective. Now suppose each is unital;

then , and thus . Using the lemma, since and are both injective unital ring homomorphisms (in fact isomorphisms) , the mappings are all injective unital ring homomorphisms

. Moreover, because is commutative, .Thus each yields a -algebra structure on , whose induced module action is given by

. Specifically, we have (letting bars denote complex conjugation) , , , and .We can see that these -algebra structures are different:

●

●

●

●

We let denote the -algebra structure induced by . Suppose now that . If the identity mapping is in fact a -algebra homomorphism, then we have $latex . However, we showed above that this is not the case for ; hence the identity mapping is not a -algebra homomorphism .

Note that both and are ring isomorphisms. Given , define . We claim that is a -algebra isomorphism.

Certainly respects addition, and moreover is a bijection. Thus it suffices to show that it respects scalar multiplication. To that end, let and let . Then

, as desired. Thus and are isomorphic as

-modules.