5
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ What notion of finite generation are you using that excludes infinite fields? $\endgroup$ – rschwieb Aug 12 '13 at 21:31
  • $\begingroup$ What do you mean by a finitely generated ring that is finite? $\endgroup$ – user38268 Aug 12 '13 at 21:35
  • 1
    $\begingroup$ @Benja: presumably it means finite as a set. $\endgroup$ – Qiaochu Yuan Aug 12 '13 at 21:43
  • 1
    $\begingroup$ @Charles: the ring is not necessarily a domain, e.g. a finite product of finite fields works. $\endgroup$ – Qiaochu Yuan Aug 12 '13 at 21:43
  • 3
    $\begingroup$ 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. $\endgroup$ – Alex Youcis Aug 12 '13 at 21:52
8
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Why the down vote? Did I make a mistake somewhere? $\endgroup$ – Alex Youcis Aug 12 '13 at 22:06
  • 1
    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Aug 12 '13 at 22:06
  • 1
    $\begingroup$ @QiaochuYuan I somehow made a stupid mistake! Fixed $\endgroup$ – Alex Youcis Aug 12 '13 at 22:07
  • $\begingroup$ @AlexYoucis Thank you! $\endgroup$ – User43029 Aug 12 '13 at 23:03
  • 2
    $\begingroup$ @Charles What is the definition of a finitely generated ring? $\endgroup$ – user38268 Aug 12 '13 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.