# Distributivity

Finally, tensor product is distributive over arbitrary direct sums.

Proposition 1.

Given any family of modules $(N_i)_{i\in I}$, we have:

$M \otimes_A \left( \oplus_{i\in I} N_i\right) \cong \oplus_{i\in I} (M\otimes_A N_i).$

Proof

Take the map $M \times (\oplus_i N_i) \to \text{ RHS }$ which takes $(m, (n_i)_{i\in I}) \mapsto (m \otimes n_i)_{i \in I}$. Note that this is well-defined: since only finitely many $n_i$ are non-zero, only finitely many $m\otimes n_i$ are non-zero. It is A-bilinear so we have an induced A-linear map

$f : M\otimes_A (\oplus_i N_i) \longrightarrow \oplus_{i\in I} (M\otimes_A N_i), \quad m \otimes (n_i)_{i\in I} \mapsto (m\otimes n_i)_{i\in I}.$

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 $e_1, \ldots, e_n$ (resp. $f_1, \ldots, f_m)$. Write $M \cong A^{\oplus n}$ and $N \cong A^{\oplus m}$. Distributivity together with $A\otimes_A A \cong A$ gives us:

$M\otimes_A N \cong \overbrace{(A \oplus\ldots \oplus A)}^{n \text{ terms}} \otimes \overbrace{(A \oplus\ldots \oplus A)}^{m \text{ terms}} \cong A^{\oplus mn}$

with basis given by $e_i \otimes f_j$ over $1 \le i \le n, 1 \le j \le m$. More generally:

Corollary 1.

We have $\left(A^{\oplus I}\right) \otimes \left(A^{\oplus J}\right) \cong \left(A^{\oplus I \times J}\right)$, 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

\begin{aligned}\mathrm{Hom}_A(M\otimes_A N, P) &\cong \mathrm{Hom}_A(M, \mathrm{Hom}_A (N, P)), \\ \phi &\mapsto (m \mapsto (n \mapsto \phi(m\otimes n)),\end{aligned}

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 $B : M\times N\to P$ where $\phi$ corresponds to $B(m, n) = \phi(m\otimes n)$. Clearly B corresponds to a linear map $M \to \mathrm{Hom}_A(N, P)$, via $m\mapsto (n \mapsto B(m, n))$. 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 $M\otimes -$ is functorial.

Proposition 3.

The functor $M\otimes_A -$ is a covariant right-exact functor.

Proof

Additivity is left as a simple exercise. To prove right-exactness, let $N_1 \stackrel f\to N_2 \stackrel g \to N_3 \to 0$ be an exact sequence.

To show that $M\otimes_A N_1 \stackrel {1_M \otimes f} \longrightarrow M\otimes_A N_2 \stackrel {1_M \otimes g} \longrightarrow M\otimes_A N_3 \longrightarrow 0$ is exact, by proposition 4 here it suffices to show that for any A-module P,

$0 \to \mathrm{Hom}_A(M \otimes_A N_3, P) \stackrel {(1 \otimes g)^*}\longrightarrow \mathrm{Hom}_A(M \otimes_A N_2, P) \stackrel {(1\otimes f)^*}\longrightarrow \mathrm{Hom}_A(M \otimes_A N_1, P)$

is exact. But by the previous result, this sequence is isomorphic to

$0\longrightarrow \mathrm{Hom}_A (M, \mathrm{Hom}_A(N_3, P)) \longrightarrow \mathrm{Hom}_A (M, \mathrm{Hom}_A(N_2, P)) \longrightarrow \mathrm{Hom}_A (M, \mathrm{Hom}_A(N_1, P)).$

Now we know this is exact by apply left-exactness of the Hom functor twice, once for $\mathrm{Hom}_A(-, P)$ and once for $\mathrm{Hom}_A(M, -)$. ♦

# Induced B-Modules

Let B be an A-algebra.

Definition.

Given an A-module M, we write $M^B := B\otimes_A M$.

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

Proposition 4.

Let $\phi : M\to M^B = B\otimes_A M$ be defined by $m\mapsto 1_B\otimes m$. Then $(M^B, \phi)$ 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 $\psi : M\to N$, there is a unique B-linear $f:M^B \to N$ such that $f\circ \phi = \psi$.

Proof (with bug)

Consider the map $B \times M \to N$ which takes $(b,m) \mapsto b\cdot \psi(m)$. This is clearly A-bilinear so it induces an A-linear $f : B\otimes_A M \to N$ which takes $b\otimes m \mapsto b\cdot\psi(m)$. It follows that $f(\phi(m)) = f(1\otimes m) = \psi(m)$.

For uniqueness of f, by condition $f(1\otimes m) = f(\phi(m)) = \psi(m)$; since f is B-linear, $f(b\otimes m) = b\cdot f(1\otimes m) = b\cdot \psi(m)$. Since every element of $B\otimes_A M$ is a finite sum of $b\otimes m$ 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 MB 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 $M\mapsto M^B$ is a generalization of these cases we have already seen.

### 1. Localization

Proposition 5.

We have a natural isomorphism

$S^{-1}A \otimes_A M \cong S^{-1}M, \quad \frac a s \otimes m \mapsto \frac{am}s.$

Proof

Take the A-bilinear map $S^{-1}A \times M \to S^{-1}M$, by $(\frac a s, m) \mapsto \frac{am}s$. Show that this is well-defined (i.e. independent of choice of a and s). So we get an A-linear map $S^{-1}A \otimes_A M \to S^{-1}M$ as above.

For the reverse map we take $S^{-1}M \to S^{-1}A \otimes_A M$ via $\frac m s \mapsto \frac 1 s \otimes_A m$. To show this is well-defined, if $\frac m s = \frac {m'}{s'}$ then $ts'm = tsm'$ for some $t\in S$ which gives

$\frac 1 s \otimes_A m = \frac {s't}{ss't} \otimes_A m = \frac 1 {ss't} \otimes_A s'tm = \frac 1 {ss't} \otimes_A stm' = \frac {st}{ss't} \otimes_A m' = \frac 1 {s'} \otimes_A m'$.

Both maps are clearly mutually inverse.

### 2 Quotient

Let $\mathfrak a\subseteq A$ be an ideal.

Proposition 6.

We have a natural isomorphism

$(A/\mathfrak a) \otimes_A M \cong M/\mathfrak a M$.

Note

Hence, together with right-exactness of tensor product, we recover the earlier result (corollary 2 here) that $M\mapsto M/\mathfrak a M$ is right-exact.

Sketch of Proof

Define an A-linear map $f:(A/\mathfrak a) \otimes_A M \to M/\mathfrak a M$ which takes $\overline a \otimes m \mapsto \overline {am}$. Also define the reverse map $g : M/\mathfrak a M\to (A/\mathfrak a)\otimes_A M$ by $\overline m\mapsto \overline 1 \otimes m$. ♦

Exercise B

Fill in the details.

### 3 Polynomial (Exercise C)

Given $B = A[X]$ and an A-module M, describe the induced A[X]-module $B\otimes_A M$.

# Properties

The construction $M\mapsto M^B$ is functorial.

Indeed if $f:M\to N$ is A-linear, let $\phi_M : M\to M^B$ and $\phi_N : N \to N^B$ 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 $f^B : M^B \to N^B$.

It is also additive and right-exact, as we saw earlier.

Exercise D

Find an A-algebra B and an injective A-linear map $f:M\to N$ of A-modules such that $f^B : M^B \to N^B$ is not injective.

Proposition 7.

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

$C \otimes_B (B\otimes_A M)\cong C \otimes_A M, \quad c \otimes_B (b\otimes_A m) \mapsto bc \otimes_A m,$

as an isomorphism of C-algebras.

Proof

Apply Yoneda lemma. For any C-module N we have a series of natural isomorphisms

\begin{aligned} \mathrm{Hom}_C ( C\otimes_B (B\otimes_A M), N) &\cong \mathrm{Hom}_B(B\otimes_A M, N) \\ &\cong \mathrm{Hom}_A(M, N) \\ &\cong \mathrm{Hom}_C( C\otimes_A M, N)\end{aligned}

so $C\otimes_A M \cong C\otimes_B (B\otimes_A M)$ as C-modules. ♦

The earlier general properties of tensor product further imply the following.

Properties of Induced B-Module.

• We have $A^B \cong B$.
• For any family of A-modules $(M_i)_{i\in I}$ we have

$\left( \oplus_{i\in I} M_i \right)^B \cong \oplus_{i\in I} (M_i)^B.$

• Given A-modules M and N, we have

$M^B \otimes_B N^B \cong (M \otimes_A N)^B.$

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:

$M^B \otimes_B N^B \cong (B\otimes_A M) \overbrace{\otimes_B (B}^{\text{cancel}}\otimes_A N)\cong (B\otimes_A M) \otimes_A N \cong B\otimes_A (M\otimes_A N).$

Now to check that the maps are actually B-linear, we let $b\in B$ act on the pure tensors on both sides and show that they match. E.g. for the last case, the isomorphism takes $(b\otimes_A m) \otimes_B (b' \otimes_A n) \mapsto bb' \otimes_A m \otimes_A n$. Multiplying both sides by $b''\in B$ gives:

Since every element of $M^B \otimes_B N^B$ is a finite sum of $(b\otimes_A m) \otimes_B (b' \otimes_A n)$, the result follows. ♦

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### 2 Responses to Commutative Algebra 29

1. Vanya says:

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