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

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.

• 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.) – Eric Wofsey Aug 8 '19 at 22:09
• @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. – Rob Arthan Aug 8 '19 at 22:34
• I have never seen any author that allows $1=0$ in a field. – Eric Wofsey Aug 8 '19 at 22:37
• 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.) – Rob Arthan Aug 8 '19 at 23:15
• ... 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). – 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.