Commutative Algebra 21

Exact Sequences

When studying homomorphisms of modules over a fixed ring, we often encounter sequences like this:

\ldots \longrightarrow M_{n+1} \stackrel {f_{n+1}} \longrightarrow M_n \stackrel {f_n}\longrightarrow M_{n-1} \stackrel {f_{n-1}} \longrightarrow \ldots

where each f_n : M_n \to M_{n-1} is a homomorphism of modules. This sequence may terminate (on either end) or it may continue indefinitely.

Note on indices: usually the index of the map takes its source; thus f_2 maps M_2 \to M_1 for instance.

Definition.

The sequence is said to be exact at M_n if

\mathrm{ker} f_n = \mathrm{im} f_{n+1}.

The whole sequence is said to be exact if it is exact at each term except the ends. A short exact sequence is an exact sequence of the form

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

Note

Having a short exact sequence above is equivalent to:

f:N\to M is injective, g:M\to P is surjective, \mathrm{ker} g = \mathrm{im} f.

Note that there is no harm in replacing N by N' = f(N) \subseteq M and f:N\to M by i : N' \hookrightarrow M as the inclusion of a submodule. Then P can be replaced with M/N'.

Why do we care about short exact sequences? In studying an A-module M, one trick is to decompose it as M \cong N \oplus P, then study the components separately. But unlike vector spaces, modules do not decompose so easily: if N\subseteq M is a submodule, we do not have M \cong N \oplus (M/N) in general. This is already clear for A = \mathbb Z, by taking M = \mathbb Z / (4) and N = \{0, 2\} \subset M.

In many cases, decomposing M into N and M/N is good enough.

Examples

1. If 0 \to N \to M \to P \to 0 is an exact sequence of vector spaces, then M is finite-dimensional if and only if N and P are, in which case \dim M = \dim N + \dim P.

2. If 0 \to N \to M \to P \to 0 is an exact sequence of abelian groups, then M is finite if and only if N and P are, in which case |M| = |N| \times |P|.

3. If 0\to N \to M \to P\to 0 is an exact sequence of A-modules, and NP are both finitely generated, then so is M. [Proof: exercise.]

We will see many more examples later.

Note that in example 3, the converse is not true since a submodule of a finitely generated submodule is not finitely generated in general.

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Decomposing an Exact Sequence

Suppose we have the following exact sequence of modules:

\ldots \stackrel{f_{n+2}} \longrightarrow M_{n+1} \stackrel {f_{n+1}} \longrightarrow M_n \stackrel {f_n}\longrightarrow M_{n-1} \stackrel {f_{n-1}} \longrightarrow \ldots

Define additional modules K_i = \mathrm{ker} f_i = \mathrm{im} f_{i+1}. Then we obtain multiple short exact sequences

0 \longrightarrow \overbrace{\mathrm{ker} f_i}^{K_i} \stackrel \subseteq \longrightarrow M_i \stackrel {f_i}\longrightarrow \overbrace{\mathrm{im} f_i}^{K_{i-1}} \longrightarrow 0.

In diagram form, this gives us:

long_exact_to_short_exact

where the diagonals give us short exact sequences.

Example

Suppose we have the following long exact sequence of finite abelian groups.

0 \longrightarrow M_4 \stackrel{f_5} \longrightarrow M_3 \stackrel {f_3} \longrightarrow M_2 \stackrel {f_2} \longrightarrow M_1 \stackrel {f_1} \longrightarrow M_0 \longrightarrow 0.

Breaking this up as above gives us short exact sequences

\left.\begin{aligned} 0 \to M_4 \to M_3 \to K_2 \to 0, \\ 0\to K_2 \to M_2 \to K_1 \to 0 \\ 0\to K_1 \to M_1 \to M_0 \to 0 \end{aligned}\right\} \implies \begin{aligned} |M_3| = |M_4| \times |K_2|, \\ |M_2| = |K_2| \times |K_1|, \\ |M_1| = |K_1| \times |M_0|.\end{aligned}

Thus |M_4| \times |M_2| \times |M_0| = |M_3| \times |M_1|. The upshot: if we know the orders of all terms but one, we also know the order of the remaining term.

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Functoriality

Through the remaining of this article, we will look at functors

F : A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}.

for fixed rings A and B. Recall that \mathrm{Hom}_A(M, N) is an A-module.

Definition.

A covariant F is said to be additive if, for any M, N \in A\text{-}\mathbf{Mod},

F : \mathrm{Hom}_A(M, N) \longrightarrow \mathrm{Hom}_B(F(M), F(N))

is a homomorphism of additive groups.

Note

We only require F(f+g) = F(f) + F(g) for f,g \in \mathrm{Hom}_A(M, N). Since the base and target rings are different, we cannot specify A-linearity.

Examples

1. Let BA and F(M) = M\oplus M. Any A-linear map \phi :M\to N induces an A-linear \phi\oplus \phi : M\oplus M \to N\oplus N, (x,y) \mapsto (\phi(x), \phi(y)).

2. Suppose f:A\to B is a homomorphism (i.e. B is an A-algebra). Each B-module M becomes an A-module via (a, m) \mapsto f(a)m. This gives a functor B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}.

3. Let \mathfrak a \subseteq A be an ideal and B = A/\mathfrak a. We saw that M/\mathfrak a M is canonically an A/\mathfrak a-module. This gives a functor A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod} since an A-linear M\to N induces a B-linear M/\mathfrak a M\to N/\mathfrak a N.

4. Let BA. For any A-module M, we have a functor \mathrm{Hom}_A(M, -). This is additive since for f : N\to N', any g_1, g_2 \in \mathrm{Hom}_A(M, N) give

f_* (g_1 + g_2) = f \circ (g_1 + g_2) = (f\circ g_1) + (f\circ g_2) = f_*(g_1) + f_*(g_2).

Exercise A

Let F be an additive functor.

  • Prove that F takes the zero module (over A) to the zero module (over B).
  • Prove that for any A-modules M and N, we have F(M\oplus N) \cong F(M) \oplus F(N).

[Hint for direct sum: let M\to M\oplus N, N\to M\oplus N, M\oplus N \to M and M\oplus N\to N be the obvious maps. Find relations among them.]

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Exact Functors

Definition.

An additive functor F is said to be exact if, for any exact sequence N\stackrel f\to M\stackrel g\to P,

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

is also exact.

Exercise B

1. Prove that we always have \mathrm{im} F(f) \subseteq \mathrm{ker} F(g) if F is additive.

2. Among the four examples above, decide which are exact.

Lemma 1.

An additive functor F is exact if and only if it takes a short exact sequence 0 \to N \stackrel f\to M \stackrel g\to P \to 0 to a short exact sequence

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

Proof

(⇒) Obvious: just take 3 terms at a time.

(⇐) Suppose N\stackrel f\to M \stackrel g\to P is a 3-term exact sequence. Let K = \mathrm{im} f = \mathrm{ker} g. Break the sequence into:

\left.\begin{aligned} 0 \to \mathrm{ker } f \stackrel \subseteq \to N \to K \to 0, \\ 0 \to K \stackrel \subseteq \to M \to \mathrm{im} g \to 0 \\ 0 \to \mathrm{im} g \to P \to P/\mathrm{im} g \to 0\end{aligned} \right\} \implies \begin{aligned} 0 \to F(\mathrm{ker } f) \to F(N) \to F(K) \to 0, \\ 0 \to F(K) \to F(M) \to F(\mathrm{im} g) \to 0 \\ 0 \to F(\mathrm{im} g) \to F(P) \to F(P/\mathrm{im} g) \to 0\end{aligned}

Since the LHS are short exact sequences, so are the RHS. Also since f: N\to M is obtained by composing N\to K \to M on the left, F(f) is obtained by composing F(N)\to F(K) \to F(M) on the right. Same goes with F(g). Piecing the short exact sequences then gives an exact F(N) \stackrel{F(f)}\to F(M) \stackrel{F(g)} \to F(P). ♦

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Implications of Exact Functor

There are many reasons why it is desirable for F to be exact.

Property 1.

If f:N\to M is injective, so is F(f) : F(N) \to F(M).

Indeed, injectivity of N\to M is equivalent to exactness of 0\to N \to M. Hence, without loss of generality, for a submodule N\subseteq M, we identify F(N) as a submodule of F(M).

Property 2.

Under this identification, if N\subseteq M is a submodule, F(M/N) \cong F(M)/F(N).

Indeed since 0 \to N \stackrel \subseteq \to M \to M/N \to 0 is a short exact sequence, so is:

0 \longrightarrow F(N) \longrightarrow F(M) \longrightarrow F(M/N) \longrightarrow 0

where F(N) is identified as a submodule of F(M). Thus F(M/N) \cong F(M) / F(N).

Property 3.

If g:M\to P is surjective, so is F(g) : F(M) \to F(P).

Indeed surjectivity of M\to P is equivalent to exactness of M\to P\to 0.

Property 4.

For f:N\to M, we have F(\mathrm{ker} f) = \mathrm{ker} F(f), as submodules of F(N).

Indeed, we have the exact sequence

0 \longrightarrow \mathrm{ker} f \longrightarrow N \stackrel f \longrightarrow M

which upon applying F gives us an exact sequence

0 \longrightarrow F(\mathrm{ker} f) \longrightarrow F(N) \stackrel {F(f)} \longrightarrow F(M)

which says that F(\mathrm{ker} f) is the kernel of F(f).

Property 5.

If N, N'\subseteq M are submodules, then F(N + N') = F(N)+ F(N') as submodules of F(M).

From N \subseteq N + N' we get F(N)\subseteq F(N+N'). Similarly F(N') \subseteq F(N+N') and thus F(N) + F(N') \subseteq F(N+N'). To show equality, since N\oplus N' \to N+N' is surjective, so is

F(N) \oplus F(N') \cong F(N\oplus N') \to F(N+N'),

where the isomorphism was proven in exercise A above. Hence F(N+N') is generated by F(N) and F(N').

Exercise C

1. For an A-linear f:M\to N, the cokernel of f is defined to be \mathrm{coker} f := N/(\mathrm{im} f). Prove that \mathrm{coker} F(f) = F(\mathrm{coker} f).

[ Hint: show that the sequence M\stackrel f\to N \to \mathrm{coker} f \to 0 is exact. ]

2. For an A-linear f:M\to N, prove that \mathrm{im} F(f) = F(\mathrm{im} f) as submodules of F(N).

3. Prove that for submodules N, N' \subseteq M, we have F(N\cap N') = F(N)\cap F(N'), as submodules of F(M).

warningIt is not true that F(\cap_i N_i) = \cap_i F(N_i) or F(\sum_i N_i) = \sum_i F(N_i) for any collection of submodules N_i \subseteq M. Counter-examples are a bit hard to find at the moment, but we will encounter one later.

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