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.

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:

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.

Functoriality

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

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.]

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 $F(N) \stackrel{F(f)}\to F(M) \stackrel{F(g)} \to F(P)$. ♦

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 $F(M)$.

It 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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