# Does there exist a non-PIR in which every countably generated prime ideal is principal?

Is there a commutative ring $R$ such that all the countably generated primes are principal, but $R$ is not a principal ideal ring?

I know that if all the prime ideals are principal, then all the ideals are principal (see here ; this answer doesn't need $R$ to ba a domain at all).

On the other hand, any commutative ring such that countably generated ideals are principal, is a PIR (see here).

Notice that since the ring $R$ of algebraic integers is a non-Noetherian Bezout domain, all the finitely-generated (prime) ideals are principal, but $R$ is not a PIR.

I tried some examples of non-PIR rings without too many prime ideals. The extreme case is only one prime ideal, for instance Artin and local, but the unique prime ideal wasn't principal in the examples I found. As pointed out by Alex Youcis in the comment below, this can't work since any Artin ring is noetherian. I wanted to search for local zero-dimensional rings, but I didn't know how.

• I don't know an example off hand, but let me point out that I've very confused about your sequence of statements. Namely, if $A$ is Artin local then, as you pointed out, it has only one prime, and so it's a terrible place to look for this property since $A$ is Noetherian, and so your condition forces the maximal ideal to be principal and thus by your first statement a PIR. – Alex Youcis Nov 26 '16 at 10:18
• @AlexYoucis : you are right. Actually I wanted to search for local zero-dimensional rings, but I didn't know how, so I search for local Artin rings. – Watson Nov 26 '16 at 10:20

Let $k$ be a field and $X$ be an uncountable set, and let $R$ be the ring of functions $X\to k$ that are constant off of a finite set. If $P\subset R$ is a prime ideal, there are two cases. The first case is that $P$ contains a function with cofinite support, in which case by primality it must contain a function whose support is the complement of a singleton $\{x\}$. It then follows that $P$ must be equal to the ideal of functions which vanish at $x$, which is principal (generated by the characteristic function of $X\setminus\{x\}$).

The second case is that $P$ contains no functions of cofinite support. By primality, $P$ must then contain every function of finite support (since if $f$ has finite support, $fg=0$ where $g$ is the characteristic function of the complement of the support of $f$, and $g\not\in P$). Thus $P$ is the set of all functions with finite support, which is indeed a prime ideal. This prime ideal is not countably generated, since $X$ is uncountable.

Another similar example is the ring of all functions $X\to k$, where $X$ is any infinite set. The proof that no nonprincipal prime is countably generated is more complicated in that case, and is equivalent to the statement that no nonprincipal ultrafilter on a set is countably generated.

For a totally different kind of example, let $\{x_i\}$ be an uncountable set of variables and let $R=k[x_i]/(x_i^2)$. Then the unique prime ideal of $R$ is the ideal generated by all the $x_i$, since they are all nilpotent and the ideal they generate is already maximal (since the quotient is $k$). This ideal is not countably generated since any countable set of generators would only involve countably many of the variables.

Here's one last example, which unlike the previous examples is a domain. First, let me start with a couple lemmas.

Lemma 1: Let $A$ be a domain and $a\in A$ be a nonzero element. Then the ring $B=A[x,y]/(xy-a)$ is a domain.

Proof: Note that $B[1/x]=A[x,1/x][y]/(xy-a)=A[x,1/x][z]/(z-a/x)=A[x,1/x]$ by letting $z=y/x$. So $B[1/x]$ is a domain, and to show $B$ is a domain it suffices to show $x$ is not a zero divisor in $B$. If $x$ were a zero divisor, then there would be polynomials $f,g\in A[x,y]$ with $xf=g(xy-a)$ and $f$ not divisible by $xy-a$. But $x$ is prime in $A[x,y]$, so if $xf=g(xy-a)$ then $x$ divides $g$, so $f=(g/x)(xy-a)$ is divisible by $xy-a$.

Lemma 2: Let $A$ be a domain and let $a\in A$ be a nonzero element. Then the ring $C=A[x,y,s,t]/(xy-a,sx+ty-1)$ is a domain and no nonunit element of $A$ is a unit in $C$.

Proof: By Lemma 1, $B=A[x,y]/(xy-a)$ is a domain. Now $C[1/x]=B[1/x][s,t]/(sx+ty-1)=B[1/x][s,u]/(s+uy-1/x)=B[1/x][u]$ (by letting $u=t/x$). So $C[1/x]$ is a domain, and to show $C$ is a domain it suffices to show $x$ is not a zero divisor in $C$. If $x$ were a zero divisor in $C$, there would be polynomials $f,g\in B[s,t]$ such that $xf=g(sx+ty-1)$ but $f$ is not divisible by $sx-ty-1$. But if $x$ divides $g(sx+ty-1)$, then by induction $x$ must divide each of the homogeneous parts of $g$, and thus $x$ divides $g$ (here we use the fact that the constant term of $sx+ty-1$ is a unit). So $f=(g/x)(sx+ty-1)$ is divisible by $sx+ty-1$.

Thus $C$ is a domain. To prove that no nonunit in $A$ is a unit in $C$, note that there is a homomorphism of $A$-algebras from $C$ to $A$ which sends $x$ to $a$, $y$ to $1$, $s$ to $0$, and $t$ to $1$. Any unit in $C$ maps to a unit in $A$ under this homomorphism, and so any unit of $C$ which came from an element of $A$ must be a unit in $A$.

OK, now we can finally construct the example. Let $U$ be an uncountable set. Given a domain $A$, let $F(A)$ be the ring obtained by adjoining elements $x_{a,u},y_{a,u},s_{a,u},t_{a,u}$ such that $x_{a,u}y_{a,u}=a$ and $s_{a,u}x_{a,u}+t_{a,u}y_{a,u}=1$ for each nonzero element $a\in A$ and each $u\in U$. By iterating Lemma 2, $F(A)$ is a domain and no nonunit in $A$ is a unit in $F(A)$.

Now let $A_0$ be a domain which is not a field and define $A_1=F(A_0)$, $A_2=F(A_1)$, and so on, and let $A_\omega$ be the direct limit of the rings $A_n$. Then $A_\omega$ is a domain, and is not a field, since $A_0$ was not a field and any non-unit in $A_0$ is still a non-unit in each $A_n$. But I claim that no nonzero prime ideal in $A_\omega$ is countably generated.

Indeed, suppose $P\subset A_\omega$ is prime and $a\in P$ is a nonzero element. Then $a\in A_n$ for some $n$, and for each $u\in U$ in $A_{n+1}$ there are elements $x_{a,u},y_{a,u},s_{a,u},t_{a,u}$ such that $x_{a,u}y_{a,u}=a$ and $s_{a,u}x_{a,u}+t_{a,u}y_{a,u}=1$. Since $P$ is prime, it must contain exactly one of $x_{a,u}$ and $y_{a,u}$ for each $u$ (if it contained both, then $s_{a,u}x_{a,u}+t_{a,u}y_{a,u}=1$ would be in $P$). Now if $P$ were countably generated, its generators would involve only countably many of the elements of $U$, and we could find an automorphism of $A_\omega$ that fixes each of the generators of $P$ but swaps $x_{a,u}$ and $y_{a,u}$ (and also swaps $s_{a,u}$ and $t_{a,u}$) for some $u$ that is not involved in any of the generators of $P$. This automorphism would fix $P$, and so $P$ contains $x_{a,u}$ iff it contains $y_{a,u}$. This is a contradiction, since $P$ must contain exactly one of them.

Thus no nonzero prime in $A_\omega$ is countably generated. Since $A_\omega$ is a domain which is not a field, it does contain nonzero primes, so it is not a principal ideal ring.

• Not the OP but I have a followup question. Do you know an example of a domain with this property? I think the examples you gave were all non-domains – Jay Nov 27 '16 at 10:40
• Good question! I've added an example that is a domain. – Eric Wofsey Nov 27 '16 at 21:51
• Thanks, that's a nice example. It looks pretty complicated at first but on second glance it seems like each step is straightforward and maybe necessary – Jay Nov 29 '16 at 0:14
• Incidentally, I suspect that the $s$ and $t$ variables are not actually necessary for the example. But without them, I don't know how to prove no prime can be countably generated, since then you only know that $P$ must contain at least one of $x_{a,u}$ and $y_{a,u}$ and I don't quite see how to prove that a countable set of generators couldn't somehow generate both $x_{a,u}$ and $y_{a,u}$ for all $u$. – Eric Wofsey Nov 29 '16 at 0:23