# UFD containing element with finite quotient

Since this question could not be answered at math.stackexchange, I would like to try my luck here now:

Does anyone know an example of a unique factorization domain $R$ that is

(i) not a Dedekind domain (or equivalently, not a principal ideal domain) and

(ii) contains some irreducible element $r \in R$ such that the quotient $R/rR$ is finite?

I am grateful for any suggestions.

• Let me just point out why this is not so obvious (I don't have an answer to your question): one way to get (ii) would be to take an algebraic variety over a finite field such that the hypersurface $r=0$ is a single point. But being factorial limits how bad the singularity can be, and a point being a hypersurface is "bad". – Gro-Tsen May 31 '18 at 11:05
• I think there is no such example where $R$ is an algebra over a (necessarily finite) field. Namely, it is known that the hypersurface $V(r) \subset \operatorname{Spec} R$ has dimension $\dim R -1$, hence $R$ is the coordinate ring of an affine curve, and $R$ being a UFD implies that the curve is non-singular, hence $R$ is a Dedekind domain. – François Brunault May 31 '18 at 15:28
• @FrançoisBrunault: Good point, although I guess you need to ask that $R$ is noetherian, right? – Filippo Alberto Edoardo May 31 '18 at 18:34
• @FilippoAlbertoEdoardo You're absolutely right. I meant to assume $\operatorname{Spec} R$ is an affine (irreducible) variety over $k$, which means in particular that $R$ is of finite type over $k$. Thanks for pointing this out. – François Brunault May 31 '18 at 18:42
• Remark: if $r$ is irreducible, then $rR=(r)$ is prime (since $R$ is a UFD). Then $R/(r)$ will be a finite integral domain which means that it is a field. So $(r)$ will in fact be maximal. – Jonathan Dunay May 31 '18 at 19:58

Let $S,k,{\frak m}_1, {\frak m}_2$ be as in that answer, and choose the field $k$ to be a finite field. There I say that $S$ is normal and that ${\frak m}_1$ has height 1, but in fact $S$ is regular (in the sense that its local rings are regular local rings) and ${\frak m}_1$ is principal (as you can see in either Nagata's book or the Stacks tag mentioned in my answer linked above). A regular local ring is a UFD, and a semilocal domain all of whose localizations are UFDs is itself a UFD; hence $S$ is a UFD. Since $S$ is 2-dimensional, it is not a Dedekind domain. Since ${\frak m}_1$ is prime, its generator is irreducible. Since $k$ is finite and $S/{\frak m}_1 \cong k$, the ring $S$ satisfies all the desired conditions, and is moreover Noetherian.

This is an attempt to give a more specific and concrete version of Neil Epstein's answer. The initial version was incorrect, and the current version is incomplete.

Let $k$ be a finite field, and fix a sequence of polynomials $u_i(x)\in k[x]$. For $1\leq n<\infty$ put $$R_n = k[x,y_0,\dotsc,y_n]/(y_i = x(y_{i+1}+u_i(x))+1),$$ then let $R_\infty$ be the colimit of the rings $R_n$. Because the relations give $y_i$ in terms of $y_{i+1}$ we just have $R_n=k[x,y_n]$ for $n<\infty$, and this is a UFD. We have $y_i=1\pmod{x}$ for all $i$ so $R_\infty/x=k$. On the other hand, we have $y_i=y_0-i\pmod{x-1}$ for all $i$, so $R_\infty/(x-1)=k[y_0]$. From this it follows that the ideal $(x-1,y_0)$ cannot be principal. Thus, if we can prove that $R_\infty$ is a UFD, then we are done.

One can check that if $f$ is irreducible in $R_n$ and does not lie in $R_n.(x,y_n-1)$ then $f$ remains irreducible in $R_{n+1}$.

Initially I had hoped to take $u_i(x)=0$ for all $i$. However, in this case we find that the elements $p_n=(1-x)y_n-1\in R_n$ satisfy $p_n=x\,p_{n+1}$ in $R_{n+1}$, and it follows that $p_0$ cannot be factored as a product of irreducibles in $R_\infty$.

I still think (by comparison with the details of the example mentioned by Neil Epstein) that it should be possible to produce an example by choosing the polynomials $u_k(x)$ appropriately, possibly as $u_k(x)=x^{m_k}$ for some rapidly increasing sequence $m_k$, perhaps $m_k=k!$. The point is that a certain power series defined in terms of the numbers $m_k$ should be transcendental over $k(x)$. However, I have not understood all the details yet.

• Very nice argument! I guess that in the definition of $R_n$ you want $1\leq n <\infty$ rather than $\leq \infty$: but do you have a reference for the claim that an injective limit (colimit) of UFD's with isomorphic units is still a unit? Is it a direct application of Nagata which I don't see? – Filippo Alberto Edoardo Jun 1 '18 at 20:12
• In the UPDATE answer, I don't think the ring in question is a UFD. The problem is that $y$, while not a unit, has no factorization into irreducibles. We have $(y) \subsetneq (y/x) \subsetneq (y/x^2) \subsetneq \cdots$. I haven't determined whether the PREVIOUS ANSWER ring satisfies the OP's desired conditions, though. – Neil Epstein Jun 4 '18 at 13:59
• @NeilEpstein You are right, of course. I will roll back to the previous answer, and think some more. – Neil Strickland Jun 4 '18 at 14:21
• @FilippoAlbertoEdoardo I can answer narrowly your question about whether an injective colimit of UFDs with isomorphic units is still a UFD. The answer is no. For example let $R_n = k[x^{1/2^n}]$, with the maps being the obvious inclusion within the algebraic closure of the field $k(x)$. Then in the colimit, the element $x$ is not a unit, but is infinitely reducible. The problem is the failure of ACCP (the Acending Chain Condition for Principal ideals), a condition necessary for being a UFD. – Neil Epstein Jun 4 '18 at 17:06
• Well (reacting to the latest attempt), Nagata's construction has your $u_k(x) \in k$, and the requirement is that $\sum_{k=0}^\infty u_k x^k$ is transcendental over $k(x)$. I suspect that choosing $u_k$ to be $0$ when $k \neq n!$ and $1$ when $k=n!$ for some $n$ will satisfy these conditions. – Neil Epstein Jun 5 '18 at 3:14

Let $D:=\mathbb{Z}[X]$, and let $S:=D\setminus(2D\cup(3,X)D)$. Let
Take a DVR $V$, and let $R=V[X]$ be the polynomial ring on $V$. Then, $R$ is a unique factorization domain (since $V$ is a UFD) but not a principal ideal domain (since $R$ has dimension 2).
Let $\pi$ be a uniformizer of $V$, and let $r:=\pi X-1$. Then, $r$ is irreducible, and $R/rR$ is isomorphic to $V[1/\pi]$, which is equal to the quotient field of $V$. In particular, you can choose it to have any (infinite) cardinality.
The same works if you take any PID with a finite number of maximal ideals instead of $V$, and then instead of $\pi$ take an element in every maximal ideal.
• Since $R/rR$ is infinite, this does not answer the question. – Laurent Moret-Bailly May 31 '18 at 8:06