Commutative Algebra 29


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).


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. ♦


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.


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.

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).


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.


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.


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.


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.


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.


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.



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.


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.


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”

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