# 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.

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.

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I +1 vote for each answer. I can't decide to accept any answer because all of them are great. :) – user111636 Jan 5 '14 at 0:51

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.

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Can you explain why a product of two fields has exactly two maximal ideal? – user111636 Jan 5 '14 at 14:24
@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 5 '14 at 20:02
Why $F_1 \times F_2$ is not maximal? – user111636 Jan 5 '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 6 '14 at 7:13
Thank you very much – user111636 Jan 7 '14 at 1:29

Look at the intersection, the sum, and the product of the two ideals. What can you say?

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Can you explain more precise? – user111636 Jan 4 '14 at 3:59
@user111636 why don't you describe your thoughts? – Igor Rivin Jan 4 '14 at 4:02
@ Igor Rivin: with all do respect, not to put too fine a point on it, what prima facie reason do we have to believe that the OP has not thought about this? I might add that I think asking someone to describe their thoughts is an excellent idea, and I upvote your comment concerning same! – Robert Lewis Jan 4 '14 at 4:09
@RobertLewis my prima facie evidence is that the OP has explicitly stated that he did not know how to start, and my suggestions are (a) obvious places to start and (b) immediately yield a lot of information. – Igor Rivin Jan 4 '14 at 4:15
+1 for giving only a (good) hint (and being first) – Bill Dubuque Jan 4 '14 at 22:30

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)

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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 $JK$?

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Do you have an example with exactly two maximal ideals? – Igor Rivin Jan 4 '14 at 4:19
Not yet, but I can try for one. – Henry Swanson Jan 4 '14 at 4:20
I could not think of anything immediately, so I wonder if this result is actually sharp... – Igor Rivin Jan 4 '14 at 4:21
I think $\mathbb{Z}_6$ works? The ideals are $\{0, 2, 4\}$ and $\{0, 3\}$. (well, I guess $6$ can be replaced with $pq$, where $p,q$ prime) – Henry Swanson Jan 4 '14 at 4:24
Yes, I think you are right! Nice to have a complete answer :) – Igor Rivin Jan 4 '14 at 4:29