My question is in the title:

Given a commutative ring $R$ with identity $1\neq 0$ and an ideal $M$ of $R$. For what condition of $M$, the quotient ring $R/M$ is finite?

My question comes from the theorem: if $M$ is a maximal ideal, then $M$ is a prime ideal. In general, the converse is not true. However, we know that if $M$ is a prime ideal then $R/M$ is an integral domain, and if $R/M$ is also finite, then it is a field, and the converse holds.

Thank you so much for any help.

  • 4
    $\begingroup$ There isn't really any condition that is simpler than saying "$R/M$ is finite" in general. Is there some particular class of rings $R$ you have in mind? $\endgroup$ Dec 3, 2015 at 16:40
  • $\begingroup$ I did not think of any particular ring. For example, take the ring $\mathbb{Z}[x]$ and the ideal $(p,x)$ generated by $p$ and $x$, where $p$ is a prime number. Then the quotient ring $\mathbb{Z}[x]/(p,x)$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$, which is finite. $\endgroup$ Dec 3, 2015 at 16:46
  • $\begingroup$ You mean besides the very evident one of $R=\cup_{a\in A} M$ where $A$ is a finite set of elements? $\endgroup$ Dec 3, 2015 at 17:39
  • $\begingroup$ @ZelosMalum Meant to write $R=\bigcup_{a\in A} a+M$ for a finite set $A$, I think? You could repost and then I'll blow this comment away... $\endgroup$
    – rschwieb
    Dec 3, 2015 at 17:51
  • $\begingroup$ That was exactly what I was intending $\endgroup$ Dec 3, 2015 at 17:51

1 Answer 1


Speaking from my own experiences, I'm not aware of a nontrivial characterization of when a quotient ring is finite.

Actually, you could generalize your question by asking "When is $R/P$ Artinian?" since an Artinian domain is also a field. There is not, as far as I'm aware, any useful characterization of when $R/I$ is Artinian.

Rings for which $R/J(R)$ is Artinian are called semilocal rings. There is a slightly useful characterization of commutative semilocal rings: they're exactly the commutative rings with finitely many maximal ideals.

This doesn't generalize your original post, but if $R/P$ is von Neumann regular, $R/P$ is also a field.

So "when is $R/I$ von Neumann regular?" is another question, but again there does not seem to be any useful characterization. Rings for which $R/J(R)$ is von Neumann regular are called semiregular rings.

Since you seem to be interested in cases where prime ideals turn out to be maximal, I'll link you to a characterization of commutative rings whose prime ideals are all maximal.


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