Must a ring (commutative, with 1), in which every non-zero ideal is prime, be a field?

Solution 1:

This is false. For instance, let $R=K\times L$ where $K$ and $L$ are fields. Then the only nonzero proper ideals in $R$ are $K\times 0$ and $0\times L$, which are both prime, but $R$ is not a field.

For another example, consider $R=\mathbb{Z}/(p^2)$ for any prime $p$. The only nonzero proper ideal is $(p)$ which is prime.

Here is a classification of all the examples. Suppose $R$ is a ring in which every nonzero proper ideal is prime. For any prime $P\subseteq R$, then $R/P$ has the same property but is a domain, and so must be a field. Thus in fact every nonzero proper ideal is maximal.

If $R$ has two different nonzero proper ideals $P$ and $Q$, then we must have $P\cap Q=0$ (since the intersection is a non-maximal proper ideal). By the Chinese remainder theorem we then get an isomorphism $R\cong R/P\times R/Q$ and so $R$ is a product of two fields.

If $R$ has exactly one nonzero proper ideal $P$, then $P$ is the nilradical of $R$ (since it is the unique prime ideal) and is principal (generated by any of its nonzero elements). This implies $P^2=0$ (otherwise it would be a smaller nonzero proper ideal) and that $P\cong R/P$ as an $R$-module (otherwise $P$ would be an $R/P$-vector space of dimension greater than $1$ and so would have a nontrivial proper subspace). If the quotient map $R\to R/P$ has a section which is a ring-homomorphism, then we can identify $R$ with $K[x]/(x^2)$ where $K$ is the field $R/P$. But such a section may not exist, as shown by the example $R=\mathbb{Z}/(p^2)$ above.

Finally, if $R$ has no nonzero proper ideals, it is either a field or the zero ring.

All of these cases can be joined together into the following equivalent characterization: $R$ is a ring in which every nonzero proper ideal is prime iff $R$ is an artinian ring of length at most $2$ as a module over itself.