Let be a ring, let be a finite family of (left, unital) -modules, and let be a submodule for each . Prove that .

We did this previously. D&F, why repeat an exercise?

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let be a ring, let be a finite family of (left, unital) -modules, and let be a submodule for each . Prove that .

We did this previously. D&F, why repeat an exercise?

Let be an infinite dimensional vector space over a field , say with basis . Prove that the dual space has strictly larger dimension than does .

We claim that . To prove this, for each let be a copy of . Now define by . By the universal property of direct products, there exists a unique -linear transformation such that for all . We claim that is an isomorphism. To see surjectivity, let . Now define by letting and extending linearly; certianly . To see injectivity, suppose . Then , so that , and thus for all . Thus for all . Since is a basis of , we have . Thus is an isomorphism, and we have .

By this previous exercise, has strictly larger dimension than does .

Let be a field. Prove that a vector space over having basis (regardless of the cardinality of ) is isomorphic as a vector space to . Prove that is also an -vector space which has strictly larger dimension than that of .

(So a free module on any set is isomorphic to a direct sum. We’ve never gotten around to proving this in the best possible generality, though, so we’ll just prove it for vector spaces here.)

Note that, by the universal property of free modules, the natural injection which sends to the tuple which has 1 in the th component and 0 elsewhere induces a vector space homomorphism . This mapping is clearly surjective, and also clearly injective. So .

Certainly is an -vector space which contains , so that . Suppose these two dimensions are in fact equal. Identify with the usual basis of . By this previous exercise, there is a basis of which contains , and as argued above, . By our hypothesis, in fact and have the same cardinality, and so there exists a bijection . Now induces a vector space isomorphism .

However, note that , while . Since , we have a contradiction. Thus the dimension of is strictly smaller than that of .

Let be a ring with 1. Let be a left unital -module, and let be a nonempty family of left unital -modules. Prove that, as abelian groups, and . Prove also that if is commutative, these pairs are -module isomorphic.

First we show that as abelian groups. Recall that for each , we have the canonical injection . Define for each the map by ; certainly each is well defined. By the universal property of direct products of abelian groups, there exists a unique group homomorphism such that , where denotes the th natural projection from a direct product. We claim that this is a group isomorphism.

Suppose , so . Then for each , so that for each . Thus for all . That is, applied to the th component of any element in is zero, for all . So , and thus . So is injective.

Now suppose . Define by ; this map is well defined since only finitely many terms of are nonzero. Moreover, it is clear that is an -module homomorphism, so in fact . Note that for all , . Thus for all , and so . Thus is surjective.

So is an isomorphism of abelian groups. Suppose now that is commutative, so that both and are naturally left -modules. For all and all module homomorphisms , we have . Thus is an isomorphism of left -modules.

Next we show that as abelian groups. For each , define by , where denotes the th canonical projection from a direct product. By the universal property of direct products, we have a unique group homomorphism such that for all . We claim that this is an isomorphism.

Suppose . So , and thus for all . Thus for all , so for all . That is, every coordinate of any element in the image of is 0. So . Thus , and so is injective.

Now let . Define by . Certainly is a module homomorphism. Note that, for all , . So for all , and we have . So is surjective, and thus an isomorphism of groups.

Finally, suppose is commutative. If and is a module homomorphism, then . Thus in this case is an isomorphism of -modules.

Let be an -algebra via the inclusion map and consider . Let , , , and . Note that every simple tensor (hence every element) of can be written in the form for some real numbers .

- Given two elements and in , compute their product as a linear combination of the .
- Let and . Show that , , , and . Deduce that as rings, .
- Prove that given by is -bilinear, where overbar denotes complex conjugation.
- Show that the -module homomorphism induced by is also -linear, taking care to note how acts on and . Show that is bijective.

We can give the following multiplication table for the (with the left factor on the top row and the right factor on the left column):

Evidently, then, .

Evidently, we have , , and . By this previous exercise, we have a ring isomorphism .

Now is clearly bilinear (since real numbers are fixed by complex conjugation). Letting be the induced -module homomorphism , we in fact have , so that is a -module homomorphism. Evidently, and . In particular, is surjective, since and generate as a -module. Now . Thus is a ring homomorphism.

Finally, suppose is in . Evidently, . Thus , and so , and thus . Likewise, , , and are all zero. Thus , and so . So is injective.

Thus, as rings, we have .

Let be a commutative ring and let be a free -module of finite rank ; say is a basis for .

- Let be any nonzero -module. (Since is commutative, is naturally an -bimodule.) Show that . In particular, every element of can be written uniquely in the form .
- Show that if , where the are merely assumed to be -linearly independent in , it need not be the case that the are all zero. (That is, it is crucial that the generate as an -module.)

Define a mapping by . (This is well defined since is a basis for .) Moreover, , , and , so that is bilinear. By the universal property of tensor products, induces a unique group homomorphism (indeed, -module homomorphism) such that .

Now define by . Clearly is an -module homomorphism.

Moreover, note that . Similarly, . Thus and are mutual inverses, and we have .

In particular, if and only if for all .

Now for the counterexample, let , let be the free -module of rank 1, and let be a -module in the usual way. Consider the simple tensor in ; note that , but that .

Let and let . Consider to be a left -module in the usual way (so is the direct product of free modules), and consider as a submodule. Suppose is free as a -module with basis .

- Show that is countable.
- Show that there is a countable subset such that is contained in . Prove also that is countable.
- Show that is free as a -module. Deduce that if is nonzero, then there exist at most finitely many such that for some .
- Let . Prove that is uncountable. Deduce that there exists some such that .
- Show that for every , there exists such that . Deduce that is not free as a module.

- For each , we have a natural inclusion given by if and 0 otherwise. Certainly each element of is in the image of some . Thus . In particular, is a countable union of countable sets, and hence is countable.
- For each , there is a finite subset such that . Let ; as the countable union of finite sets, is countable. Certainly . Certainly is also countable.
- We wish to show that is free. To that end, note that we have the natural inclusion given by . By the universal property of free modules, there is a unique -module homomorphism such that for all . We claim that is an isomorphism. To that end, let . Say . Now , so that . Thus , and we have . So is injective. Now let ; since is free on , we have for some and . Certainly then , so that is surjective. Thus is an -module isomorphism, and is free as a -module.
- We will now prove some lemmas.
Lemma 1: If is divisible by infinitely many integers, then . Proof: Certainly 0 is divisible by infinitely many integers, since for all . Suppose now that is nonzero and that for some infinite set , for all . There must exist some with , a contradiction.

Lemma 2: Let be a free -module with free basis . If is nonzero, then there are only finitely many such that for some . Proof: Suppose to the contrary that a nonzero exists such that, for infinitely many , there exists such that . Say . Since , some is nonzero. Looking at the coefficient in the (unique) expansion of , we have for infinitely many integers . By the lemma, , a contradiction. .

We certainly have a bijection . A diagonalization argument shows that is uncountable as follows. Suppose is a bijection. Construct as follows: if and otherwise. Since differs from every element in in some component, is not in , a contradiction. Thus is uncountable. Since is countable, there must exist such that .

- Let be positive. Define by if and if , where if and if . Now for all , so that . Thus . This is a contradiction of part (4) above. Thus is not free as a -module.

Let be a ring and let be a family of left -modules indexed by a (nonempty) set . Prove that is a left -module via the action (Called the *direct product* of the modules ). Moreover, prove that the subset is a submodule of (Called the *direct sum* of the modules ). Moreover, show that and are not isomorphic as -modules.

We saw in this previous exercise that is an abelian group under pointwise addition and multiplication. Thus it suffices to show that the three left-module axioms are satisfied. To that end, let and let . We have , , and . Thus is a left -module. Suppose further that has a 1 and that each is unital; then , for all , so that is unital.

Now we will use the submodule criterion to show that is a submodule of . To that end, note first that is nonempty since . (The set of indices such that is nonzero is empty, which is certainly finite). Now let and let . Say such that for and for . Consider . Note that if , then . Since is finite, . By the submodule criterion, is an -submodule.

Now consider and as -modules in the natural way. We intend to show that these two modules are not isomorphic by showing that one is torsion while the other is not. To that end, let . There is a natural number such that, for all , . Now let (where this product is taken over the natural numbers). Certainly then ; that is, every element of is torsion, so that is torsion. However, note that is nonzero for all integers since (for instance) the component of is 1. So is not torsion. Thus and are not isomorphic as -modules.

Let be a ring with 1 and let , , and be unital left -modules. Prove that and . (Where means “isomorphic as -modules”.)

Given -module homomorphisms and , define by . Note that, for all , , and , we have

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So is an -module homomorphism. Consider the map . We claim that is an -module homomorphism. To see this, let and be -module homomorphisms and let . Now

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So is an -module homomorphism. Now suppose . In particular, for all , so that . Similarly, . Thus , and in fact . Hence is injective. Finally, let be an -module homomorphism, and define and by and . Note that , so that is an -module homomorphism. Similarly, is an -module homomorphism. Since , we have . Hence is surjective. Thus, we have .

Now let and be the left and right coordinate projections; certainly these are -module homomorphisms. Define by . If and , then

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Thus is an -module homomorphism. Now suppose . In particular, if , then . So , hence , and thus is injective. Finally, suppose . Define by . Certainly is an -module homomorphism. Moreover, we have . Thus , and so is surjective. Hence .

Let be a ring with 1, let , and let . Recall that is a left -module via the action . Let be a family of ideals of . Prove that the following subsets of are submodules.

We use the submodule criterion for both examples.

- Note that for each . Then ; hence is nonempty. Now suppose , and let . Because each is an ideal, for each . Then . Thus is a submodule of .
- Certainly , so that . Now suppose ; then . Let . Then , and hence . Thus is a submodule of .