11
$\begingroup$

Let $R$ be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.

My attempt: Let $I$ be an ideal of $R$. Then we have $I$ is maximal $\Leftrightarrow$ $R/I$ is a finite field $\Leftrightarrow$ $R/I$ is a finite integral domain $\Leftrightarrow$ $I$ is a prime ideal.

Is my proof valid ?

$\endgroup$
  • $\begingroup$ Yes, that looks fine (note that only one of the directions is of interest as the other is trivial). $\endgroup$ – Tobias Kildetoft Apr 3 '13 at 12:50
  • 1
    $\begingroup$ The step $R/I$ is field $\iff R/I$ is a finite integral domain is technically invalid. You need something like $R/I$ is a finite field for the biconditional to hold. However, as Tobias said the proof is fine when proceed in only the nontrivial direction. $\endgroup$ – Karl Kronenfeld Apr 3 '13 at 12:50
  • $\begingroup$ @user1 no, since $R$ is assumed finite, that is correct. $\endgroup$ – Tobias Kildetoft Apr 3 '13 at 12:52
  • $\begingroup$ @user1: I thought $R$ is a field $\Leftrightarrow$ $R$ is a fintie integral domain ? Can you explain why it is invalid? $\endgroup$ – Idonknow Apr 3 '13 at 12:52
  • $\begingroup$ @Idonknow, I think user1 is correct: the direction $\,\implies\,$ , as you wrote it, is wrong. The other one is correct, so you only have to add "finite" afdter the word "field" $\endgroup$ – DonAntonio Apr 3 '13 at 12:54
8
$\begingroup$

Yes, your proof is valid, but note that the second implication relies on $R$ being finite. It'd be clearer if written as

$R/I$ is a finite field $\Leftrightarrow$ $R/I$ is a finite integral domain

The whole thing would be even cleaner if written as

Since $R$ is finite, we have the following equivalences:

$I$ is maximal $\Leftrightarrow$ $R/I$ is a field $\Leftrightarrow$ $R/I$ is an integral domain $\Leftrightarrow$ $I$ is a prime ideal

though only the second relies on $R$ being finite.

$\endgroup$
2
$\begingroup$

The heart of the proof is good, and I wanted to comment that you could easily prove a "cousin" of that theorem for noncommutative Artinian rings!

I am, of course, using the noncommutative definition of prime ideals which generalizes the commutative one.

Proposition: A prime ideal in a right Artinian ring $R$ is maximal iff it is prime.

Proof ($\Rightarrow$) If $M$ is a maximal ideal, $R/M$ is a simple ring, which is certainly a prime ring. It follows that $M$ is a prime ideal.

Proof ($\Leftarrow$) (This is where the theme of your proof can be applied again!) Suppose $P$ is a prime ideal. Then $R/P$ is a prime ring. Since $R$ is right Artinian, so is $R/P$. But the Artin-Wedderburn theorem says that such a ring is simple, hence $P$ is maximal. $\Box$

The connection is that Wedderburn's little theorem is like the Artin-Wedderburn theorem: one says that a finite domain is a field, one says that a right Artinian prime ring is a simple ring. ("Right Artinian" is a weaker form of "finite" and "prime" is a weaker form of "domain".)

$\endgroup$
0
$\begingroup$

R/p is semiprime and right artinian, so it is semisimple. Since R/p is in fact prime, it can have only one simple component. Therefore, R/p is simple, so p is a maximal ideal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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