Commutative Algebra 26

Left-Exact Functors

We saw (in theorem 1 here) that the localization functor M \mapsto S^{-1}M is exact, which gave us a whole slew of nice properties, including preservation of submodules, quotient modules, finite intersection/sum, etc. However, exactness is often too much to ask for.

Throughout this article, A and B are fixed rings and F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod} is a covariant additive functor.

Definition.

We say F is left-exact if it takes a short exact sequence of A-modules

0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0

to an exact sequence of B-modules

0 \longrightarrow F(N) \stackrel {F(f)}\longrightarrow F(M) \stackrel {F(g)} \longrightarrow F(P).

Immediately we can weaken the condition.

Lemma 1.

F is left-exact if and only if it takes an exact sequence of A-modules

0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P

to an exact sequence of B-modules

0 \longrightarrow F(N) \stackrel {F(f)}\longrightarrow F(M) \stackrel {F(g)} \longrightarrow F(P).

Proof

(⇐) is obvious. For (⇒) given an exact 0 \to N \stackrel f\to M \stackrel g\to P, we extend it to an exact 0 \to N \to M \to P \to \mathrm{coker } g \to 0. [Recall that \mathrm{coker } g = P/\mathrm{im } g is the cokernel of g.] We can split this exact sequence into short exact ones:

\begin{aligned} 0 \to N \stackrel f\to M \to \mathrm{im } g  \to 0 &\implies 0 \to F(N) \stackrel{F(f)}\to F(M) \to F(\mathrm{im } g)\\ 0 \to \mathrm{im } g \to P \to \mathrm{coker } g \to 0 &\implies 0 \to F(\mathrm{im } g) \to F(P) \to F(\mathrm{coker } g).\end{aligned}

which gives an exact sequence 0\to F(N) \to F(M) \to F(P) as desired. ♦

Note

Left-exact functors are not as nice as exact ones but we still get some useful results out of them. For example, if f : N\to M is injective, so is F(f) : F(N) \to F(M) so for a submodule N\subseteq M, we can consider F(N) as a submodule of F(M).

Also, for any f:N\to M, since 0\to \mathrm{ker } f \to N \to M is exact, applying F gives an exact 0 \to F(\mathrm{ker}  f) \to F(N) \stackrel{F(f)}\to F(M) and thus F(\mathrm{ker} f) = \mathrm{ker } F(f).

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Hom Functors Are Left-Exact

The main result we wish to show is

Proposition 1.

For any A-module M, the functor \mathrm{Hom}_A(M, -) is a left-exact functor.

Proof

Take the exact sequence 0 \to N_1 \stackrel f\to N_2 \stackrel g\to N_3. We need to show that

0 \longrightarrow \mathrm{Hom}_A(M, N_1) \stackrel {f_*}\longrightarrow \mathrm{Hom}_A(M, N_2) \stackrel{g_*}\longrightarrow \mathrm{Hom}_A(M, N_3)

is exact. It is easy to show f_* is injective (easy exercise). Next, we have

g_*\circ f_* = (g\circ f)_* = 0 \implies \mathrm{im } f_* \subseteq \mathrm{ker } g_*.

Conversely suppose h:M\to N_2 satisfies g_*(h) = g\circ h = 0. Then \mathrm{im } h \subseteq \mathrm{ker } g = \mathrm{im } f. Since f is injective it follows that for each m\in M, we have h(m) = f(x) for a unique x\in N_1. This map h' : M \to N_1, m\mapsto x is clearly A-linear so h = f\circ h' = f_*(h'). ♦

Here is one application of this result.

Corollary 1.

Let a\in A; for each A-module, let M[a] = \{m \in M : am = 0\}. Then

0 \longrightarrow N \stackrel f \longrightarrow M \stackrel g \longrightarrow P \text{ exact } \implies 0 \longrightarrow N[a] \stackrel f\longrightarrow M[a] \stackrel g \longrightarrow P[a] \text{ exact.}

Proof

Indeed the functor M\mapsto M[a] is naturally isomorphic to \mathrm{Hom}_A(A/(a), -). Now apply the above. ♦

Exercise A

Find a surjective M\to P for which M[a] \to P[a] is not surjective. This shows that the functor \mathrm{Hom}_A(M, -) is not exact in general.

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Converse Statement

Next we have the following converse.

Proposition 2.

Suppose f:N_1 \to N_2 and g:N_2\to N_3 are A-linear maps such that for any A-module M,

0\longrightarrow \mathrm{Hom}_A(M, N_1) \stackrel {f_*} \longrightarrow \mathrm{Hom}_A(M, N_2) \stackrel {g_*} \longrightarrow \mathrm{Hom}_A(M, N_3)

is exact. Then 0 \to N_1 \stackrel f \to N_2 \stackrel g \to N_3 is exact.

Proof

First we show f is injective. Let M =\mathrm{ker } f so that 0 \to M \to N_1 \stackrel {f} \to N_2 is exact. By left-exactness of Hom we have an exact:

0 \to {\mathrm{Hom}_A(M, M)} \to \mathrm{Hom}_A(M, N_1) \stackrel {f_*} \to \mathrm{Hom}_A(M, N_2).

Since f_* is injective, \mathrm{Hom}_A(M, M) = 0 and in particular 1_M = 0\implies M=0.

Next we show g\circ f = 0. Setting M = N_1 gives:

0 = g_* f_* (\mathrm{Hom}_A(N_1, N_1)) = (g\circ f)_*(\mathrm{Hom}_A(N_1, N_1))

so g\circ f = (g\circ f)_* (1_{N_1}) = 0. This gives \mathrm{im } f \subseteq \mathrm{ker } g.

Finally we set M = \mathrm{ker } g. The following sequence:

0\longrightarrow \mathrm{Hom}_A(M, N_1) \stackrel {f_*} \longrightarrow \mathrm{Hom}_A(M, N_2) \stackrel {g_*} \longrightarrow \mathrm{Hom}_A(M, N_3)

is exact. The inclusion i : M\hookrightarrow N_2 gives g_*(i) = g\circ i = 0 and so i = f_*(j) = f\circ j for some j : M \to N_1. Then M = \mathrm{im } i \subseteq \mathrm{im } f. ♦

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Another Left-Exactness

Now suppose F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod} is a contravariant additive functor.

Definition.

We say F is left-exact if it takes a short exact sequence of A-modules

0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0

to an exact sequence of B-modules

0 \longrightarrow F(P) \stackrel {F(g)}\longrightarrow F(M) \stackrel {F(f)} \longrightarrow F(N).

Again we have:

Lemma 2.

F is left-exact if and only if it takes an exact sequence of A-modules

N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0

to an exact sequence of B-modules

0 \longrightarrow F(P) \stackrel {F(g)}\longrightarrow F(M) \stackrel {F(f)} \longrightarrow F(N).

Proof

Exercise. ♦

Now we show, as before:

Proposition 3.

For any A-module M, the functor \mathrm{Hom}_A(-, M) is a left-exact functor.

Proof

Take the exact sequence N_1 \stackrel f\to N_2 \stackrel g\to N_3 \to 0. We need to show that

0 \longrightarrow \mathrm{Hom}_A(N_3, M) \stackrel {g^*}\longrightarrow \mathrm{Hom}_A(N_2, M) \stackrel{f^*}\longrightarrow \mathrm{Hom}_A(M, N_1)

is exact for any A-module M. Injectivity of g^* is obvious. Also f^* \circ g^* = (g\circ f)^* = 0 so \mathrm{im } g^* \subseteq \mathrm{ker } f^*.

It remains to show \mathrm{ker } f^* \subseteq \mathrm{im } g^*. Pick h: N_2 \to M such that f^*(h) = h\circ f = 0 so that \mathrm{ker } g = \mathrm{im } f \subseteq \mathrm{ker } h. Hence h factors through N_2 / \mathrm{ker } g \to M. Since N_2 / \mathrm{ker } g \cong N_3, we have h = h'\circ g = g^*(h'') for some h' : N_3 \to M. ♦

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Another Converse Statement

Finally, the reader should expect the following.

Proposition 4.

Suppose f:N_1 \to N_2 and g:N_2\to N_3 are A-linear maps such that for any A-module M,

0\longrightarrow \mathrm{Hom}_A(N_3, M) \stackrel {g^*} \longrightarrow \mathrm{Hom}_A(N_2, M) \stackrel {f^*} \longrightarrow \mathrm{Hom}_A(N_1, M)

is exact. Then N_1 \stackrel f \to N_2 \stackrel g \to N_3 \to 0 is exact.

Proof

We leave it as an exercise to show: g is surjective, g\circ f = 0.

It remains to show \mathrm{ker } g \subseteq \mathrm{im } f; let M = \mathrm{coker } f and take the exact sequence

0\longrightarrow \mathrm{Hom}_A(N_3, M) \stackrel {g^*} \longrightarrow \mathrm{Hom}_A(N_2, M) \stackrel {f^*} \longrightarrow \mathrm{Hom}_A(N_1, M)

If \pi : N_2 \to M is the canonical map, f^*(\pi) = \pi\circ f = 0 so we have \pi = g^*(j) = j\circ g for some j : N_3 \to M. Thus \mathrm{ker } g \subseteq \mathrm{ker } \pi = \mathrm{im } f. ♦

As an application, let us prove the following.

Corollary 2.

Let \mathfrak a \subseteq A be an ideal. If N \to M \to P \to 0 is an exact sequence of A-modules, then we get an exact sequence of (A/\mathfrak a)-modules:

N/\mathfrak a N \longrightarrow M/\mathfrak a M \longrightarrow P/\mathfrak a P \longrightarrow 0.

Proof

Let B = A/\mathfrak a. It suffices to show that for any B-module Q, the sequence

0 \longrightarrow \mathrm{Hom}_B(P/\mathfrak a P, Q) \longrightarrow \mathrm{Hom}_B(M/\mathfrak a M, Q) \longrightarrow \mathrm{Hom}_B(N/\mathfrak a N, Q)

is exact. But since M\mapsto M/\mathfrak a M gives the B-module induced from M, the above sequence is naturally isomorphic to:

0 \longrightarrow \mathrm{Hom}_A(P, Q) \longrightarrow \mathrm{Hom}_A (M, Q) \longrightarrow \mathrm{Hom}_A(N, Q)

which we know is exact because the Hom functor is left-exact. ♦

This leads to the following definition.

Definition.

Suppose F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod} is a covariant additive functor. We say F is right-exact if it takes a short exact sequence of A-modules

0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0

to an exact sequence of B-modules

F(N) \stackrel {F(f)}\longrightarrow F(M) \stackrel {F(g)} \longrightarrow F(P) \longrightarrow 0.

Exercise B

Prove that F is right-exact if and only if it takes an exact sequence of A-modules N \to M \to P \to 0 to an exact sequence of B-modules F(N) \to F(M) \to F(P) \to 0.

Thus, we have shown that the functor

A\text{-}\mathbf{Mod} \longrightarrow (A/\mathfrak a)\text{-}\mathbf{Mod}, \quad M \mapsto M/\mathfrak aM

is right-exact. This will be generalized in future chapters.

Exercise C

  • Prove that the functor M\mapsto M/\mathfrak a M is not exact in general.
  • Use a direct proof to show M\mapsto M/\mathfrak a M is right-exact.

In summary, we have covered the following concepts:

exact_functor_types

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