Finally, tensor product is distributive over arbitrary direct sums.
Given any family of modules , we have:
Take the map which takes . Note that this is well-defined: since only finitely many are non-zero, only finitely many are non-zero. It is A-bilinear so we have an induced A-linear map
The reverse map is left as an exercise. ♦
Let us prove that the first example of the previous article is truly the tensor product of V and W. More generally suppose M (resp. N) is a finite free module over A with basis (resp. . Write and . Distributivity together with gives us:
with basis given by over . More generally:
We have , i.e. a tensor product of free modules is free.
Prove that a tensor product of two projective modules is projective.
Recall that localization commutes with arbitrary direct sums; tensor product also commutes with arbitrary direct sums. Does that mean tensor product commutes with localization? The answer is yes: the reader should be able to prove this as an easy exercise by the end of this article.
We imagine the class of A-modules as having a “semi-ring”-like structure, where
- A ↔ multiplicative identity;
- direct sum ↔ addition;
- tensor product ↔ multiplication.
⊗ is Right-Exact
First, we re-interpret the universal property of tensor product as follows.
For any A-modules M, N, P, we have a natural isomorphism of A-modules
where both sides are treated as functors in M, N and P (contravariant in M, N, covariant in P).
Indeed, the universal property of tensor product says: the LHS corresponds to the set of all bilinear maps where corresponds to . Clearly B corresponds to a linear map , via . Hence we get the desired bijection. It is easy to show that the map is A-linear. ♦
Let M be an A-module; recall that is functorial.
The functor is a covariant right-exact functor.
Additivity is left as a simple exercise. To prove right-exactness, let be an exact sequence.
To show that is exact, by proposition 4 here it suffices to show that for any A-module P,
is exact. But by the previous result, this sequence is isomorphic to
Now we know this is exact by apply left-exactness of the Hom functor twice, once for and once for . ♦
Let B be an A-algebra.
Given an A-module M, we write .
The notation is identical to our earlier one for induced modules, for a good reason.
Let be defined by . Then is the induced module.
Let us refresh the reader’s memory: we want to show that for any B-module N and A-linear , there is a unique B-linear such that .
Proof (with bug)
Consider the map which takes . This is clearly A-bilinear so it induces an A-linear which takes . It follows that .
For uniqueness of f, by condition ; since f is B-linear, . Since every element of is a finite sum of we see that f must be as defined above. ♦
The above proof has a huge bug in it. Find it and fix it.
[Hint: in the first paragraph, we did not prove that f is B-linear. But for that, we need MB to have the structure of a B-module, which we forgot to define… ]
The various constructions we have seen for modules are actually induced. Thus is a generalization of these cases we have already seen.
We have a natural isomorphism
Take the A-bilinear map , by . Show that this is well-defined (i.e. independent of choice of a and s). So we get an A-linear map as above.
For the reverse map we take via . To show this is well-defined, if then for some which gives
Both maps are clearly mutually inverse.
Let be an ideal.
We have a natural isomorphism
Hence, together with right-exactness of tensor product, we recover the earlier result (corollary 2 here) that is right-exact.
Sketch of Proof
Define an A-linear map which takes . Also define the reverse map by . ♦
Fill in the details.
3 Polynomial (Exercise C)
Given and an A-module M, describe the induced A[X]-module .
The construction is functorial.
Indeed if is A-linear, let and be the maps in the universal property. We leave it to the reader to fill out the remainder of the proof to obtain a B-linear .
It is also additive and right-exact, as we saw earlier.
Find an A-algebra B and an injective A-linear map of A-modules such that is not injective.
Module induction is transitive. Thus if C is a B-algebra and B is an A-algebra then
as an isomorphism of C-algebras.
Apply Yoneda lemma. For any C-module N we have a series of natural isomorphisms
so as C-modules. ♦
The earlier general properties of tensor product further imply the following.
Properties of Induced B-Module.
- We have .
- For any family of A-modules we have
- Given A-modules M and N, we have
In each case, the earlier properties imply we have an isomorphism of A-modules on both sides. E.g. for the last case we have:
Now to check that the maps are actually B-linear, we let act on the pure tensors on both sides and show that they match. E.g. for the last case, the isomorphism takes . Multiplying both sides by gives:
Since every element of is a finite sum of , the result follows. ♦