Let $A=\mathbb Q[x_1,...,x_n]$ and let $M$ be an $A$-module. Show that $l_A(M)$ is finite if and only if $\dim_{\mathbb Q}(M)$ is finite.

Question

Let $A=\mathbb Q[x_1,...,x_n]$ and let $M$ be an $A$-module.

$l_A(M)$ stands for the length of $M$, regarded as an $A$-module.

Show that $l_A(M)$ is finite if and only if $\dim_{\mathbb Q}(M)$ is finite.

First suppose that $\dim_{\mathbb Q}(M)$ is finite. Then say ${v_1,...,v_n}$ generates $M$, so $M$ is a finitely generated $\mathbb Q$-module. Since $\dim_{\mathbb Q}(M)$ = $l_{\mathbb Q}(M)$, M is noetherian and artinian as $\mathbb Q$-module. I hope I can show $M$ is noetherian and artinian as $A$ module from it. I think I can show it is noetherian, but I was stuck on how to show it is artinian.

I think maybe I can prove it by contrapositive. Suppose $l_A(M)$ is infinite, then $l_{\mathbb Q}(M)$ cannot be finite. I wonder whether there is any connection between simple module in $A$-module and simple module in $\mathbb Q$-module.

Answer 1

If $\dim_{\mathbb{Q}}(M)=n$, then any proper ascending/descending chain of $A$-submodules must terminate in $n$ steps (since a proper $A$-submodule $L\subsetneq N$ should satisfy $\text{dim}_{\mathbb{Q}}(L)<\text{dim}_{\mathbb{Q}}(N)$). Hence $M$ is both Noetherian and Artinian as an $A$-module.

The other implication is the harder part- this requires knowing that every simple $A$-module $N$ has $\text{dim}_{\mathbb{Q}}(N) < \infty$. Assuming this, show that a finite length $A$-module $M$ necessarily has $\text{dim}_{\mathbb{Q}}(M)<\infty$ by looking at a composition series and using additivity of $\mathbb{Q}$-dimension over short exact sequences.

Finally, observe/argue that a simple $A$-module is of the form $A/\mathfrak{m}$ for a maximal ideal $\mathfrak{m}$. Now, $A/\mathfrak{m}$ is a field which is also finitely generated as an algebra over $\mathbb{Q}$, hence is a finite extension of $\mathbb{Q}$ (see, for example, Proposition 7.9 in Atiyah-Macdonald) and in turn has finite dimension over $\mathbb{Q}$.