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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 ?

azimut
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Idonknow
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  • Yes, that looks fine (note that only one of the directions is of interest as the other is trivial). – Tobias Kildetoft Apr 03 '13 at 12:50
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    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. – Karl Kroningfeld Apr 03 '13 at 12:50
  • @user1 no, since $R$ is assumed finite, that is correct. – Tobias Kildetoft Apr 03 '13 at 12:52
  • @user1: I thought $R$ is a field $\Leftrightarrow$ $R$ is a fintie integral domain ? Can you explain why it is invalid? – Idonknow Apr 03 '13 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" – DonAntonio Apr 03 '13 at 12:54
  • Apparently there is disagreement over this small point. Looking back, I would classify it as pedentry since it does not apply to the situation anyway. – Karl Kroningfeld Apr 03 '13 at 12:57

4 Answers4

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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.

Community
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lhf
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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".)

rschwieb
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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.

Thrash
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$R/p$ 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.

Rajdeep
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user134158
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