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.
Best Answer
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.