An early exercise in Irving Kaplansky's commutative rings asks:

Let R be a ring. Suppose that every ideal in R (other than R) is prime. Prove that R is a field.

This is easy if we assume the zero ideal is prime. But is this assumption necessary?

If every non-zero ideal is prime, then for any non-unit $x \in R$ and with $x^{n+1} \ne 0$ we must have $\langle x \rangle \subseteq \langle x^{n+1} \rangle$, which requires the existence of an element $y$ satisfying: $$ x(1-x^ny) = 0 $$ The collection of these and similar relations on the elements seems rather restrictive, but I would appreciate a simple and incisive argument to show that the condition that all non-zero ideals are prime can only be met by rings with trivial spectrum, or, if my guess is incorrect and this is untrue, a counter-example.

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    $\begingroup$ Note that Kaplansky's original statement is not quite right, since you also have to require that $R$ is nonzero. (If $R$ is the zero ring, then there are no ideals other than $R$ so the condition is vacuously true.) $\endgroup$ Aug 8 '19 at 22:09
  • $\begingroup$ @EricWofsey: I think your point is that fields are usually required to have $1 \neq 0$, hence (somewhat unusually by comparison with groups and rings and modules and ...) the trivial structure with just one element isn't a field. If Kaplansky allows $1 = 0$ in a field as some authors do, then his statement is right. $\endgroup$
    – Rob Arthan
    Aug 8 '19 at 22:34
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    $\begingroup$ I have never seen any author that allows $1=0$ in a field. $\endgroup$ Aug 8 '19 at 22:37
  • $\begingroup$ I agree that it is usual to require $1 \neq 0$, but Weber's original definition allowed it: he says explicitly that $0$ is distinct from $1$ except in the uninteresting case when the field has only one element. See gdz.sub.uni-goettingen.de/id/PPN235181684_0043?tify page 527. (Please forgive me for going back to the 19th century, but I've become a fan of going back to original sources in mathematics recently.) $\endgroup$
    – Rob Arthan
    Aug 8 '19 at 23:15
  • $\begingroup$ ... and in modern definitions the requirement that $1 \neq 0$ is often sneaked in surreptitiously by saying that that then non-zero elements of the field form a group under multiplication (which implies there is at leat one non-zero element, because the traditional definition of a group requires a group to have a non-empty universe). $\endgroup$
    – Rob Arthan
    Aug 8 '19 at 23:32

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.


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