Commutative Algebra (IV): Finitely Presented Modules

29 Jan 2019

• commutative algebra

In this post, we take a closer look at some finiteness properties for modules, namely finite generation and finite presentation.

Finite modules

Proposition: Let $M$ be an $A$-module. Then $M$ is finite (ie. finitely generated) if and only if there is a surjective $A$-module map $\varphi:A^q\to M$ for some $q\geq0$.

Proof: ($\Rightarrow$) If $M$ is generated by $\{\omega_1,\ldots,\omega_q\}$, consider the free $A$-module $A^q$ with generators $\{e_1,\ldots,e_q\}$. By the universal property of $A^q$, there is an $A$-module map $\varphi:A^q\to M$ with $\varphi(e_i)=\omega_i$ for all $i$. Now $\{\omega_1,\ldots,\omega_q\}\subseteq\im\varphi$ implies $\im\varphi=M$, ie. $\varphi$ is surjective.

($\Leftarrow$): If $A^q$ has generators $\{e_1,\ldots,e_q\}$, then for all $m\in M$ there exists $a_i\in A$ with

$$m=\varphi\left(\sum_ia_ie_i\right)=\sum_ia_i\varphi(e_i),$$

so $\{\varphi(e_1),\ldots,\varphi(e_q)\}$ generates $M$ as an $A$-module. Hence $M$ is finite. $\qed$

Corollary: An $A$-module is finite if and only if it is isomorphic to a quotient of $A^q$ for some $q\geq0$. $\qed$

Corollary: Let $M$ be a finite $A$-module, and let $f:M\to N$ be a surjective $A$-module map. Then $N$ is also finite. $\qed$

Corollary: Quotients of finite $A$-modules are also finite. $\qed$

In the other direction, we have:

Proposition: Let $M$ be an $A$-module, and let $N\subseteq M$ be a submodule. If $N$ and $M/N$ are both finite, then so is $M$.

Proof: Suppose that $\{n_1,\ldots,n_k\}$ generates $N$, and $\{m_1+N,\ldots,m_l+N\}$ generates $M/N$. Then for all $m\in M$, there exists $a’_ i\in A$ with

$$m+N=\sum_{i'}a'_ {i'}(m_{i'}+N),$$

so $m-\sum_{i’}a’_ {i’}m_{i’}\in N$, and thus there are $a_i\in N$ with

$$m-\sum_{i'}a'_ {i'}m_{i'}=\sum_ia_in_i.$$

Hence $\{n_1,\ldots,n_k,m_1,\ldots,m_l\}$ generate $M$, so $M$ is finite. $\qed$

The above argument can be reformulated in a more categorical way, as follows:

Proposition: Let

$$0\to M_1\to M_2\to M_3\to0$$

be a short exact sequence of $A$-modules. If $M_1$ and $M_3$ are finite, then so is $M_2$.

Proof: Let $f:A^q\to M_1$ and $g:A^r\to M_3$ be surjective $A$-module maps. Then we have the following commutative diagram:

$$\def\ra#1{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \def\da#1{\bigg\downarrow\raisebox{.5ex}{\mathrlap{$\scriptstyle{#1}$}}} \begin{matrix} 0&\ra{}&A^q&\ra{}&A^q\oplus A^r&\ra{}&A^r&\ra{}&0\\ &&\da{f}&{}_ \tca\smash\searrow&\da{h\tcc}&\smash\swarrow_ \tcb&\da{g}\\ 0&\ra{}&M_1&\ra{}&M_2&\ra{}&M_3&\ra{}&0 \end{matrix}$$

• $\tcb$: by projectivity of the free module $A^r$;
• $\tcc$: by universal property of $A^q\oplus A^r$.

Now by the snake lemma, we have an exact sequence

$$\coker f\to\coker h\to\coker g,$$

and $f,g$ surjective implies $\coker f=\coker g=0$, so $\coker h=0$ implies $h:A^{q+r}\to M_2$ is surjective; hence $M_2$ is finite. $\qed$

In general, it is not true that submodules of finite $A$-modules are finite. For example, take $A=\bb Z[X_1,X_2,\ldots]$, the polynomial ring over countably many variables, and take $M=A=A^1$, which is clearly finite over $A$. Now consider the submodule $N=(X_1,X_2,\ldots)$ of polynomials with constant term $0$. For any $f_1,\ldots,f_k\in N$, let $n\geq1$ be the largest index $m$ for which $X_m$ appears in some $f_k$; this is well-defined, since each $f_k$ only has finitely many terms with nonzero coefficients. Then

$$(f_1,\ldots,f_k)\subseteq(X_1,\ldots,X_n),$$

so $X_{n+1}\not\in(f_1,\ldots,f_k)$. Thus $N$ is not finitely generated over $A$.

To say more about the finiteness of quotient modules and submodules, we need to introduce a finiteness condition on modules which is stronger than finite generation; this is the goal of the next section.

Finitely presented modules

An $A$-module $M$ is of finite presentation, or finitely presented, if there is a surjective $A$-module map $\varphi:A^q\to M$ (ie. $M$ is finite), and $\ker\varphi$ is finite. Informally, $\ker\varphi$ gives the relations among the generators for $M$, so a finitely presented module is a finite generated module where the relations are also finitely generated.

Equivalently, $M$ is finitely presented if there is an exact sequence of the form

$$A^p\xrightarrow\psi A^q\xrightarrow\varphi M\to0$$

for some $p,q\geq0$. Indeed, we can just take any surjective map $A^p\xrightarrow\psi\ker\varphi$.

Proposition: Let

$$0\to M_1\to M_2\to M_3\to0$$

be a short exact sequence of $A$-modules.

1. If $M_1$ and $M_3$ are finitely presented, then so is $M_2$;
2. If $M_1$ is finite and $M_2$ is finitely presented, then $M_3$ is finitely presented;
3. If $M_2$ is finite and $M_3$ is finitely presented, then $M_1$ is finite.

Proof: (1) Let $f:A^q\to M_1$ and $g:A^r\to M_3$ be surjective $A$-module maps. As before, we have the following commutative diagram:

$$\def\ra#1{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \def\da#1{\bigg\downarrow\raisebox{.5ex}{\mathrlap{$\scriptstyle{#1}$}}} \begin{matrix} 0&\ra{}&A^q&\ra{}&A^q\oplus A^r&\ra{}&A^r&\ra{}&0\\ &&\da{f}&&\da{h}&&\da{g}\\ 0&\ra{}&M_1&\ra{}&M_2&\ra{}&M_3&\ra{}&0 \end{matrix}$$

We have already shown that $h$ is surjective. Again, by the snake lemma, we have the following exact sequence:

$$\ker f\xrightarrow{\varphi}\ker h\to\ker g\to0$$

Since $\ker f$ and $\ker g$ are finite, we have

$$\im\varphi\cong\ker f/\ker\varphi\quad\text{and}\quad\ker h/\im\varphi\cong\ker g$$

are both finite, so $\ker h$ is also finite. Thus $h:A^{q+r}\to M_2$ is surjective and has finitely generated kernel, so $M_2$ is finitely presented.

(2) Let $f:A^q\to M_1$ and $g:A^r\to M_3$ be surjective $A$-module maps. Then we have the following commutative diagram:

$$\def\ra#1{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \def\da#1{\bigg\downarrow\raisebox{.5ex}{\mathrlap{$\scriptstyle{#1}$}}} \begin{matrix} 0&\ra{}&A^q&\ra{}&A^q\oplus A^r&\ra{}&A^r&\ra{}&0\\ &&\da{f}&{}_ \tca\smash\searrow&\da{h\tcb}&\smash\swarrow_ g&\da{g'\tcc}\\ 0&\ra{}&M_1&\ra{}&M_2&\ra{}&M_3&\ra{}&0 \end{matrix}$$

• $\tcb$: by universal property of $A^q\oplus A^r$.

Note that $g$ surjective and $M_2\to M_3$ surjective implies $g’$ surjective.

Now by the snake lemma, we have the following exact sequence:

$$\ker f\to\ker h\xrightarrow{\psi}\ker g'\to0$$

Since $\ker h$ is finite and $\psi$ is surjective, we have $\ker g’$ is also finite. Thus $g’:A^r\to M_3$ is surjective and has finitely generated kernel, so $M_3$ is finitely presented.

(3) Take an exact sequence

$$A^p\xrightarrow\psi A^q\xrightarrow\varphi M_3\to0,$$

so we have the following commutative diagram:

$$\def\ra#1{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \def\da#1{\bigg\downarrow\raisebox{.5ex}{\mathrlap{$\scriptstyle{#1}$}}} \begin{matrix} 0&\ra{}&A^q&\ra{}&A^r&\ra{}&M_3&\ra{}&0\\ &&\da{f_1\tcd}&\smash\searrow^\tcc&\da{f_2\tcb}&\smash\searrow^\tca&\parallel\\ 0&\ra{}&M_1&\ra{}&M_2&\ra{}&M_3&\ra{}&0 \end{matrix}$$

• $\tcb$: by projectivity of the free module $A^r$;
• $\tcd$: since $A^q\to M_2\to M_3$ is the zero map.

Now by the snake lemma, we have the following exact sequence:

$$0\to\coker f_1\to\coker f_2\to0$$

Hence $M_1/\im f_1\cong M_2/\im f_2$, which is finite since $M_2$ is finite. Also, $\im f_1$ is finite since $A^q$ is finite. Hence $M_1$ is finite, and we are done. $\qed$