1
$\begingroup$

Let $R$ be a commutative ring with unity such that for every prime ideal $P$ of $R$ , $R/P$ is a finite ring , so a finite field ; also suppose $R$ is artinian ring ; then can we say that $R$ is finite ?

$\endgroup$

closed as off-topic by user26857, Arnaud D., Shaun, Xander Henderson, dantopa Sep 17 '17 at 16:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Arnaud D., Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

Being Artinian, you can factor it into a product of local Artinian rings, and they all have to have finite residue fields. Clearly this allows us to reduce the problem to local Artinian rings, since the product will be finite iff the summands are finite.

Let $R$ be local, Artinian, with maximal ideal $M$, such that $|R/M|=n<\infty$. Actually $R/M$ is the only isotype of simple $R$ module that exists! Since $R$ is Artinian, it has a finite composition series

$$ \{0\}\subseteq I_1\subseteq\ldots\subseteq I_k\subseteq R $$

Each factor of which is isomorphic to $R/M$.

Basic counting says that $|R|=[R:I_k][I_k:I_{k-1}]\cdots[I_1:\{0\}]$. There are $k+1$ factors here, each one with finite size $|R/M|$.

So we see then that $|R|=|R/M|^{k+1}<\infty$

For Noetherian rings, the answer is still true, because such a ring is actually still Artinian: if all primes are maximal, the nilradical equals the Jacobson radical and it is nilpotent. Furthermore $R/J(R)$ is semisimple Artinian, and at this point Hopkins-Levitzki says $R$ was Artinian all along.

If you drop Noetherianity, then it is obviously false. You can just take $R=\prod _{i=1}^\infty F_2$.

$\endgroup$
  • $\begingroup$ Why is $R/J(R)$ artinian ? $\endgroup$ – user Aug 25 '17 at 14:37
  • $\begingroup$ @users Because (when all prime ideals are maximal) $R/J(R)$ is von Neumann regular, and a Noetherian von Neumann regular ring is Artinian. $\endgroup$ – rschwieb Aug 25 '17 at 15:15
  • $\begingroup$ Ah yes true . Thanks . Do you think that same conclusion that $R$ is finite would hold if we assumed $R$ is Noetherian and every minimal prime ideal of $R$ has finite index ? $\endgroup$ – user Aug 25 '17 at 15:26
  • $\begingroup$ @users finite index in the sense that $|R/P|$ is finite? That still implies every prime has finite index. $\endgroup$ – rschwieb Aug 25 '17 at 15:29
  • $\begingroup$ Yes in the sense $R/P$ has finite index ... I see .. because in Noetherian ring , every prime ideal contains a minimal prime ideal .. right ? $\endgroup$ – user Aug 25 '17 at 15:38

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