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Let $R$ be a commutative ring with $1$. Suppose that every nonzero proper ideal of $R$ is maximal. Prove that there are at most two such ideals.

Help me some hints. I have no idea to start.

Starship
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user111636
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3 Answers3

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Suppose that $R$ has more than one maximal ideal, say $M_1$ and $M_2$. Then $M_1\cap M_2$ is a proper ideal and it can not be maximal (since it is contained in both $M_1$ and $M_2$) therefore we must have $M_1\cap M_2 = 0$. Now by the chinese rimainder theorem we have $$R \cong \frac{R}{0} = \frac{R}{M_1\cap M_2} \cong \frac{R}{M_1}\times\frac{R}{M_2}$$ the right hand side is a product of two fields so it has exactly two maximal ideal. Thus $R$ can not have more than two maximal ideals.

Robert M
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  • Can you explain why a product of two fields has exactly two maximal ideal? – user111636 Jan 05 '14 at 14:24
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    @user111636 If $R$ and $S$ are commutative rings with identity the the ideals of $R\times S$ are of the form $I\times J$ where $I$ is an ideal of $R$ and $J$ is an ideal of $S$. Now recall that a field has only two ideals: zero and itself. Therefore if $F_1$ and $F_2$ are fields then the ring $F_1\times F_2$ has exactly four ideals: $0\times 0, F_1\times 0, 0\times F_2, F_1\times F_2$ and among them $F_1\times 0$ and $0\times F_2$ are maximal since they are not contained in any proper ideal. – Robert M Jan 05 '14 at 20:02
  • Why $F_1 \times F_2$ is not maximal? – user111636 Jan 05 '14 at 23:48
  • @user111636 No, by maximal we mean maximal in the set of all proper ideals, i.e., a maximal ideal must be proper by its definition. – Robert M Jan 06 '14 at 07:13
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Supoose you have two nonzero ideals. The are both maximal and minimal, so their sum is the whole ring and their intersection is zero: it follows that the ring is the direct sum of the two ideals. Since they are both proper, neither of them contains $1$, so they are both rings with unit. As rings, the don't have any nonzero proper ideals, so they are in fact fields.

Our ring is therefore the direct product of two fields. By inspection, we can easily find all its ideals.

(we also get all possible examples, btw)

Mariano Suárez-Álvarez
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Let $I$ and $J$ be two distinct ideals of $R$. We know that neither is a subset of the other, so $I \subset I + J$. Since $I$ is maximal, $I + J = R$.

Assume there is a third non-zero proper ideal of $R$, call it $K$. Consider $(I + J)K$. On one hand, this must be $K$, because $I + J = R$, and $1 \in R$.

But it is also equal to $IK + JK$. If all non-zero proper ideals are maximal, what do you know about $IK$ and $JK$?

Henry Swanson
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