Finite generating sets and finitely-generated modules

Question

(All my rings are commutative with $1$.)

Reworded a little, a question in a previous Commutative Algebra exam goes like this:

Let

• $A$ denote a ring,
• $X$ denote an $A$-module
• $F$ denote a finite generating subset of $X$, and
• $\Lambda$ denote a (not-necessarily finite) generating subset of $X$.

Show that $\Lambda$ includes a finite generating subset of $X$.

The obvious approach is to try to inductively choosing elements of $\Lambda$ in a way that is guided by $F$ somehow, until we've built a finite subset that generates all of $\Lambda$ and hence of all of $X$. I really have no idea how to do this though.

Ideas, anyone?

Answers phrased in the language of finitary closure operators and/or algebraic posets are especially welcome!

Answer 1

Each element of $F$ can be written as a finite linear-combination of elements of $\Lambda$. Since there are finitely many elements in $F$, this gives rise to a finite subset $\Omega \subset \Lambda$ with $A\langle \Omega \rangle \supset A \langle F \rangle = X$.

Answer 2

Okay, so I've taken the answer MooS gave and generalized it a lot, by formulating everything in the language of closure systems. The essence of the proof is the same, however. Also, its a little wordy because I wanted to be 100% rigorous and precise. But fundamentally, the argument is the same.

For the remainder of this answer, let $X$ denote a closure system.

Definition 0. A subset $A \subseteq X$ is algebraic iff $A = \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F).$

Proposition 0. For all $A \subseteq X$, the following are equivalent:

• $A$ is algebraic

• $\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(A) \rightarrow \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}x \in \mathrm{cl}(F)\right)$

Proof. The following are equivalent.

• $A = \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F)$

• $A \subseteq \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F)$

• $\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(A) \rightarrow x \in \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F)\right)$

• $\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(A) \rightarrow \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}x \in \mathrm{cl}(F)\right)$

Proposition 1. Let $\Lambda$ denote an algebraic generating subset of $X$.

Then for all finitely-generated $G \subseteq X$, there exists finite $\Lambda_{\mathrm{fin}} \subseteq \Lambda$ such that $\mathrm{cl}(\Lambda_{\mathrm{fin}}) \supseteq G$.

Proof.

Since $\Lambda$ is algebraic, hence by Proposition 0, we have

$$\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(\Lambda) \rightarrow \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(\Lambda)}x \in \mathrm{cl}(F)\right)$$

Therefore, since $\Lambda$ is a generating set, we have: $$\mathop{\forall}_{x \in X} \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(\Lambda)}x \in \mathrm{cl}(F)$$

Hence letting $G_{\mathrm{fin}}$ denote a finite generating set of $G$, as a special case, we get: $$\mathop{\forall}_{g \in G_{\mathrm{fin}}} \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(\Lambda)}g \in \mathrm{cl}(F)$$

Using the principle of finite choice, this can be rewritten:

$$\mathop{\exists}_{F : G \rightarrow \mathcal{P}_{\mathrm{fin}}(\Lambda)}\mathop{\forall}_{g \in G_{\mathrm{fin}}} g \in \mathrm{cl}(F(g))$$

So let $F : G_{\mathrm{fin}} \rightarrow \mathcal{P}_{\mathrm{fin}}(\Lambda)$ denote any such function. Define:

$$\Lambda_{\mathrm{fin}} = \bigcup_{g \in G_{\mathrm{fin}}} F(g)$$

Since $G_{\mathrm{fin}}$ is finite, hence $\Lambda_{\mathrm{fin}}$ is finite.

We need to show that $\mathrm{cl}(\Lambda_{\mathrm{fin}}) \supseteq G$.

It suffices to show that $\mathrm{cl}(\Lambda_{\mathrm{fin}}) \supseteq G_{\mathrm{fin}}$.

We know that

$$\mathop{\forall}_{g \in G_{\mathrm{fin}}} g \in \mathrm{cl}(F(g))$$

In other words:

$$G_{\mathrm{fin}} \subseteq \bigcup_{g \in G} \mathrm{cl}(F(g)).$$

So

$$G_{\mathrm{fin}} \subseteq \mathrm{cl}\left(\bigcup_{g \in G} F(g)\right).$$

In other words:

$$G_{\mathrm{fin}} \subseteq \mathrm{cl}(\mathrm{\Lambda}_{\mathrm{fin}}).$$

as required.

Corollary. Let $\Lambda$ denote an algebraic generating subset of $X$.

If $X$ is finitely-generated, then $\Lambda$ includes a finite generating set of $X$.