Ring whose proper ideals are finite is field.

Question

How can I prove that if I have a commutative ring $R$ with unity that is domain which all its proper ideals are finite then it is a field?

The only thing that I'm able to see is that $R$ is artinian and noetherian, but no more. Can anyone help me?

Answer 1

An Artinian domain (with unity) is always a field - note that the ideals $ (a) \supset (a^2) \supset (a^3) \supset \ldots $ form a descending chain for nonzero $ a $, and being Artinian implies that we have $ (a^k) = (a^{k+1}) $ for some $ k $, so that $ x a^{k+1} = a^k $ for some $ x \in R $, thus $ xa^{k+1} - a^k = a^k (ax - 1) = 0 $. Now, since $ R $ is a domain we conclude that $ ax = 1 $, and $ x $ is the desired inverse.

Answer 2

Let $R$ be an integral domain whose proper ideals are finite. Assume that $R$ is not a field, Then $R$ has a nonzero maximal ideal $M\lhd R$, and we may choose $a\in M-\{0\}$. The ideal $M$ contains all powers $a, a^2, a^3,\ldots$, so by finiteness we must have $a^m=a^{m+k}$ for some positive $m$ and $k$. Choose positive $\ell$ so that $m+\ell$ is a multiple of $k$. Then for $e=a^{m+\ell}$ we have $e^2 = e\in M$.

Now $R$ is a domain and $a\neq 0$, so $e = a^{m+\ell}\neq 0$. Since $0 = e^2-e = e(e-1)$, we derive that $e=1$, again using that $R$ is a domain. But now we have $1 = e\in M$, a contradiction.