# Finitely-generated algebra over Z

Let A be an artin ring which is also a finitely generated algebra over Z.

Show that $|A|<\infty$.

If A would have been a field then I know how to prove it. I know that A is a product of local rings, so I could restrict the question to Local artin rings that are finitely generated algebra over Z. But how does this help?

Thanks, Yatir

• Your algebra is a finite-length module over a polynomial ring of the form $\mathbb Z[x_1,\dots,x_n]$, and the simple modules of such a ring are of finite cardinal. – Mariano Suárez-Álvarez Jul 5 '10 at 9:15
• @yatir: It seems the hardest case is when $A$ is a field... – fherzig Jul 5 '10 at 9:39
• @Mariano: "the simple modules of $\mathbb Z[x_1,\dots,x_n]$ are of finite cardinal" is (pretty much) equivalent to the question asked. (The case when $A$ is a field.) I'm not saying this isn't a standard fact... – fherzig Jul 5 '10 at 9:50

Take $A$ local (you already reduced to it), with $m$ the max. ideal. I claim that $A/m$ is a finite field. Suppose first that it has char. 0. Then we get injections $\mathbb Z \to \mathbb Q \to A/m$. By Zariski's lemma, $\mathbb Q \to A/m$ is finite, since it is of finite type. Now (unfortunately I don't have it on me), Atiyah-Macdonald have a beautiful lemma which says that if $A \subset B \subset C$ are (comm.) rings, $A$ noetherian, $A \subset C$ of finite type, $B \subset C$ finite, then $A \subset B$ is of finite type.

In our case, $\mathbb Z \to \mathbb Q$ is of finite type, contradiction. Thus $\mathbb Z/p \to A/m$ is of finite type, hence finite for some prime number $p$. So $A/m$ is a finite field. Also $m^n = 0$ for some $n$ since $A$ is artin local. Finally, $m^i/m^{i+1}$ is a f.d. $A/m$-vector space (since $A$ is noetherian), so it is finite as well. And $|A| = \sum |m^i/m^{i+1}|$.

• I remember now that Emerton gave a different proof that $A/m$ is a finite field in his notes on Jacobson rings posted here. – fherzig Jul 5 '10 at 9:30
• Just one question, How did you derive the last equation? – yatir Jul 5 '10 at 9:47
• Use the short exact sequences $0 \to m^i/m^{i+1} \to A/m^{i+1} \to A/m^i \to 0$ and induct. – fherzig Jul 5 '10 at 9:54
• Shortcut: since {(0)} is not constructible in Spec Z, Chevalley's theorem also rules out char. 0. – user2035 Jul 6 '10 at 7:56
• For reference, the lemma you're quoting in AM is known as the "Artin Tate Lemma". It can be used to prove Zariski's lemma quite nicely, too. – Pedro Tamaroff Nov 1 '14 at 17:50

As you already mentioned, it is enough to show that every local artinian ring $A$, which is of finite type over $Z$, is finite. Let $m$ be the maximal ideal of $A$. By a standard filtration argument, we may assume $m^2=0$. Now $A/m$ is a finite field (since it is of finite type over some $Z/p$, apply Noether Normalization). Also, $m$ is an artinian, thus finite-dimensional $A/m$-vector space, and thus finite. Hence, also $A$ is finite.

• How did you see that $A/m$ has positive characteristic? – fherzig Jul 5 '10 at 9:32
• There are several ways to prove it. You mentioned one. Of course, our proofs are identical. – Martin Brandenburg Jul 5 '10 at 10:23