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

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 ?

• Yes, that looks fine (note that only one of the directions is of interest as the other is trivial). Commented Apr 3, 2013 at 12:50
• 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. Commented Apr 3, 2013 at 12:50
• @user1 no, since $R$ is assumed finite, that is correct. Commented Apr 3, 2013 at 12:52
• @user1: I thought $R$ is a field $\Leftrightarrow$ $R$ is a fintie integral domain ? Can you explain why it is invalid? Commented Apr 3, 2013 at 12:52
• @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" Commented Apr 3, 2013 at 12:54

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.

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".)

If a finite commutative ring $$R$$ is without unity, then only one direction of your statement is true, that is

$$I$$ prime ideal $$\iff$$ $$R/I$$ finite integral ring $$\iff$$ $$R/I$$ finite field $$\Longrightarrow$$ $$I$$ maximal ideal

Why can't be the last implication reverted? Example: $$R=2\mathbb Z / 4 \mathbb Z = \{0,2\}$$ and $$I = \{0\}$$. Then $$I$$ is maximal but not prime because $$2\cdot 2 = 0 \in I$$, but $$2 \notin I$$. In other words: $$R/I = R$$ is not a field.

$$R/p$$ 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.