## Rings with finitely generated additive group [closed]

The question is the following: let A be a ring (commutative, with 1) whose additive group is a finitely generated abelian group. If $P\subset A$ is a maximal ideal, then $A/P$ is a finite field? To show this, I've argued as follows, but I'm not sure that this is the "right" way: A has a structure of finitely generated $\mathbb{Z}$-algebra. If $p\mathbb{Z} = P \cap \mathbb{Z}$, then, as a Corollary of a generalized form of Hilbert's Nullstellensatz, $A/P$ is a finite field extension of $\mathbb{Z}_p$, hence a finite field.

Is this a correct proof? thanks again!

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I'm quite sure that $A/P$ is a field of positive characteristic $p$. If not, it would contain a copy of $\mathbb{Q}$ but this - I think - contradicts the hypothesis of being itself finitely generated as abelian group. – Jack Mild Dec 9 2010 at 21:53
-1. not a research level question – a-fortiori Dec 9 2010 at 21:54
Close this homework!!! – Bugs Bunny Dec 9 2010 at 22:56
Having said this, $P\cap Z$ makes no sense but otherwise the proof is OK, although too heavy... – Bugs Bunny Dec 9 2010 at 22:57
Matsumura uses the notation $P\cap\mathbf Z$ for preimages even for non-injective homomorphisms. – a-fortiori Dec 9 2010 at 23:19