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Does there exist a non-unital commutative ring such that all its proper ideals are prime?

Note also that that equipping the abelian group $\mathbb Z/p\mathbb Z$ with trivial multiplication $xy=0$ for all $x,y$ does not work. Though the resulting ring does not contain any nontrivial proper ideals, it does contain the trivial ideal $(0)$ which is not prime since the ring in question is not integral.

This question is inspired by an exercise which asked one to prove that all nontrivial rings with no non-prime proper ideals are fields. I suspect though that the text of the exercise simply forgot to specify the rings having units or was operating under some quiet "all rings are unital" assumption, as then it is rather simple to prove.

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Such a ring would be a domain, of course. – egreg Jan 8 '14 at 23:41
And, after it's proven to have an identity, it's actually a field. – rschwieb Jan 9 '14 at 18:01
up vote 5 down vote accepted

Let $A$ be such a ring. If $ab = 0$ in $A$, then $a = 0$ or $b = 0$. Thus if $a c = bc$, we find $a = b$.

Now suppose $A \neq 0$, and let $a \neq 0$ be an element. The ideal $I = \{n a^2 + b a^2 \, | \, n \in \mathbb Z, b \in A\}$ is prime (actually $I$ may not be proper, but then certainly $a \in I$; thanks to user115654 for suggesting that this be made explicit), so since $a^2 \in I$, we find $a \in I$. Thus $a = na^2 + b a^2$ for some $n \in \mathbb Z, b \in A$.

Then $ac = na^2c + bc a^2$ for all $c \in A$. Cancelling $a$, we find that $c = (na + b a)c.$ Thus $na + ba $ is a unity in $A$, and so we find that $A$ is necessarily unital.

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It might be worth mentioning that even if the ideal $I$ is not proper (so not prime), that $a \in I$ anyways. By the way, I think this a great answer, and upvoted – zcn Jan 9 '14 at 0:29
Now, if $I$ and $J$ are proper ideals of $A$, $IJ\subseteq I\cap J$, which is prime. Therefore either $I\subseteq I\cap J$ or $J\subseteq I$ and so $A$ is a valuation domain. – egreg Jan 9 '14 at 0:29
@user115654: Dear user, Thanks for pointing this out; I'll make an edit to this effect. Cheers, – Matt E Jan 9 '14 at 0:30
@user115654: Dear user, Actually, I forgot that I had thought about this: the whole ring $A$ is automatically prime (modulo the fact that prime ideals are normally defined to be proper), since if $ab \in A$ then $a$ or $b$ in $A$ for trivial reasons; both $a$ and $b$ are in $A$! Of course, this is just your observation again. Best wishes, – Matt E Jan 9 '14 at 0:33
And, having an identity and all proper ideals prime, the ring is a field. – rschwieb Jan 9 '14 at 18:00

If the ring doesn't have a $1$, what would a unit in such a ring be? A unit is defined to be an element that has a multiplicative inverse, i.e., an element such that the product of the two elements is $1$. But if $1$ doesn't exist, how can any units?

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Thank you very much for quickly pointing out the utter stupidity of the previous version fo the question. I have absolutely no idea what was going through my head as I wrote it. What I meant was whether such a ring can even exist, as I have now corrected in the question as well. – user82454 Jan 8 '14 at 23:42
No problem. It happens to everyone. I felt a bit awkward about keeping my answer after you edited the question. – Nishant Jan 9 '14 at 0:20

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