# Non-unital commutative ring without non-prime ideals?

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.

• 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

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.
• 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, 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