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 
 A: 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 $|A| = \sum |m^i/m^{i+1}|$.
A: 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.
