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A field has a finite maximal ideal. Of course the converse is not true. So is there a characterization for those rings having a finite maximal ideal ??

EDIT: The characterization below is very nice. I would like to know What happens in non-commutative case ? I mean, Rings with a finite maximal left ideal. Of course, it can be shown easily that $_RR$ has finite composition length.

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I don't understand why this question is on hold ! –  user38215 Aug 11 '13 at 14:28
    
I think the question is too elementary. I recommend asking on stack exchange –  David White Aug 11 '13 at 15:08
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@DavidWhite So, I would be very glad if you could give me a hint for non-commutative case ! –  user38215 Aug 11 '13 at 18:35

2 Answers 2

Let $R$ be a commutative ring with identity and let $M$ be a finite maximal ideal of $R$. If $R$ is local then it is not hard to show that either $R$ is finite or $R$ is a field.

For the general case, since $M$ and $R/M$ are Artinian, $R$ is also an Artinain ring. So we can find local rings $R_1, ..., R_n$ such that $R\cong R_1\times ...\times R_n$ Thus the image of $M$ is in the form $R_1\times ...\times m_i\times ...\times R_n$ with $m_i$ maximal in $R_i$. This mean that the rings $R_i$ are all finite and $R_i$ is a field (as I mentioned above). So the characterization (in commutative case) is as follows: Fields or direct product of a field and a finite ring.

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EDIT : The first version of my answer was unnecessarily complicated. The following avoids lifting of idempotents and structure theory of artinian rings and answers the original question:

So let $R$ be a ring with $1$ and finite maximal ideal $M$. (Note that if $R$ contains a finite maximal left ideal, then it contains a finite maximal ideal.) Let
$$ I = \{ r\in R \mid rM = 0 \}. $$ I claim that if $R/M$ is infinite, then $R = I \oplus M \cong (R/M)\times (R/I)$. Thus the answer in the noncommutative case is: finite ring or direct product of simple ring and finite ring.

Proof of $R=I+M$: Note that $I$ is the kernel of the map $$ R \to \bigoplus_{m\in M} (Rm)\subseteq M^M,\quad r\mapsto (rm)_{m\in M}. $$ As $M$ is finite, $R/I$ is finite. Since we assume $R/M$ infinite, it follows that $I+M> M$ and so $I+M=R$ by maximality.

Proof of $I\cap M=0$: Assume $I\cap M\neq 0$. Since $M$ is finite, $(I\cap M)_R$ contains a simple right $R$-module $S_R$. As $IM=0$, we may view $S$ as a module over $R/M$. Let $D= \text{End}(S_R)$. Since $|S|$ is finite, it follows $S\cong D^n$ for some $n$ and $D$ is finite. By the density theorem of Jacobson-Chevalley then $R/M$ is isomorphic to an $n\times n$-matrix ring over $D$ and thus finite, contradiction. Thus $I\cap M=0$ as claimed.

In the special case where $R$ has a finite maximal left ideal, an occuring infinite simple ring factor is necessarily a division ring.

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