# Finitely generated ring with zero Krull dimension

I'm trying to prove the following:

Every finitely generated ring with Krull dimension equal to zero is finite.

I'm trying to show that the ring is a domain, hence a field, in order to use the property that claims that every finitely generated field is finite, but I'm not seeing how I can show that this ring is a domain, so I guess that this way may, may be not the best way to show this.

Thank you for any help!

• What notion of finite generation are you using that excludes infinite fields? – rschwieb Aug 12 '13 at 21:31
• What do you mean by a finitely generated ring that is finite? – user38268 Aug 12 '13 at 21:35
• @Benja: presumably it means finite as a set. – Qiaochu Yuan Aug 12 '13 at 21:43
• @Charles: the ring is not necessarily a domain, e.g. a finite product of finite fields works. – Qiaochu Yuan Aug 12 '13 at 21:43
• It seems you can use the fact that since your ring is finitely generated Z algebra that it is Noetherian. But, Noetherian and dimension zero is equivalent to Artinian. Decompose your ring into the product of local Artinian rings, and go from there. – Alex Youcis Aug 12 '13 at 21:52

To see this, note that since $R$ is a finitely generated $\mathbb{Z}$-algebra, that it is necessarily Noetherian. Thus, $R$, being Noetherian and dimension zero, it is necessarily Artinian. Thus, we have a decomposition of $R$ as $R\cong R_1\times\cdots\times R_n$ where each $R_i$ is a local Artinian ring, which is necessarily also finitely generated as a $\mathbb{Z}$-algebra.
Now, it suffices to show that each $(R_i,\mathfrak{m}_i)$ is finite. It is a common fact that a local Artinian ring with finite residue field is necessarily finite itself. But, note that $R_i/\mathfrak{m}_i$ is a field which is finitely generated as a $\mathbb{Z}$-algebra, and so necessarily finite (this follows by the Nullstellansatz).
• I didn't downvote, but the $R_i$ are not necessarily finite fields, e.g. they may have the form $F[x]/x^n$ where $F$ is a finite field. This doesn't seem too hard to fix though. – Qiaochu Yuan Aug 12 '13 at 22:06