# Distributivity

Finally, tensor product is distributive over arbitrary direct sums.

Proposition 1.Given any family of modules , we have:

**Proof**

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. ♦

## Example

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:

Corollary 1.We have , i.e. a tensor product of free modules is free.

**Exercise A**

Prove that a tensor product of two projective modules is projective.

**Note**

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.

Summary.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.

Proposition 2.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).

**Proof**

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.

Proposition 3.The functor is a covariant right-exact functor.

**Proof**

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 . ♦

# Induced *B*-Modules

Let *B* be an *A*-algebra.

Definition.Given an A-module M, we write .

The notation is identical to our earlier one for induced modules, for a good reason.

Proposition 4.Let be defined by . Then is the induced module.

**Note**

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. ♦

**Important Exercise**

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 *M ^{B}* to have the structure of a

*B*-module, which we forgot to define… ]

# Special Cases

The various constructions we have seen for modules are actually induced. Thus is a generalization of these cases we have already seen.

### 1. Localization

Proposition 5.We have a natural isomorphism

**Proof**

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.

### 2 Quotient

Let be an ideal.

Proposition 6.We have a natural isomorphism

.

**Note**

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 . ♦

**Exercise B**

Fill in the details.

### 3 Polynomial (Exercise C)

Given and an *A*-module *M*, describe the induced *A*[*X*]-module .

# Properties

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.

**Exercise D**

Find an *A*-algebra *B* and an injective *A*-linear map of *A*-modules such that is not injective.

Proposition 7.Module induction is transitive. Thus if C is a B-algebra and B is an A-algebra then

as an isomorphism of C-algebras.

**Proof**

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

**Proof**

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. ♦

There is a typo at the end of the proof of “Module induction is transitive”

Thanks! Fixed.