# When a quotient of a UFD is also a UFD?

Let $R$ be a UFD and let $a\in R$ be nonzero element. Under what conditions will $R/aR$ be a UFD?

A more specific question:

Suppose $R$ is a regular local ring and let $I$ be a height two ideal which is radical. Can we find an element $a\in I$ such that $R/aR$ is a UFD?

• This seems like a complicated question. Even in the special case that $R = \mathbb{Z}[x]$ and the quotient $R/a$ happens to be a Dedekind domain the answer still depends on its class number... – Qiaochu Yuan May 27 '13 at 6:32
• yesm the answer is not clear, but i want to know if there are some special cases under which the hypothesis holds. Is there anything known for my 2nd question? – messi May 27 '13 at 8:33

This is indeed a complicated question, that has also been much studied. Let me just more or less quote directly from Eisenbud's Commutative Algebra book (all found in Exercise 20.17):

The Noether-Lefschetz theorem: if $R = \mathbb{C}[x_1, \ldots x_4]$ is the polynomial ring in $4$ variables over $\mathbb{C}$, then for almost every homogeneous form $f$ of degree $\ge 4$, $R/(f)$ is factorial.

On the other hand, in dimension $3$, there is a theorem of Andreotti-Salmon:

Let $(R,P)$ be a $3$-dimensional regular local ring, and $0 \ne f \in P$. Then $R/(f)$ is factorial iff $f$ cannot be written as the determinant of an $n \times n$ matrix with entries in $P$, for $n > 1$.

• Dear user, There is something strange in your statement of Brieskorn--Lipman, since $k[[x,y,z]]/(z) = k[[x,y]]$, a regular (and hence factorial) ring. Or am I misunderstanding? Cheers, – Matt E Feb 7 '14 at 1:27
• @MattE Following the reference given in Eisenbud gives this Lipman paper which does contain the mentioned ring (pages 4-5), but I can not find the claimed statement. However, I only took a quick look through. – RghtHndSd Feb 7 '14 at 2:05
• My best guess is that the statement in Eisenbud is based off misunderstanding the statement at the end of the first paragraph on page 5. If one confuses "for which (II) holds" with "for which $R$ is a UFD" and omit the word "essentially", you get the statement in Eisenbud. – RghtHndSd Feb 7 '14 at 2:19
• My previous comment is wrong. Eisenbud cites the wrong Lipman paper. The more nuanced statement appears in this Lipman paper Theorem 25.1: If $R$ is a two-dimensional local ring with maximal ideal $m$ such that $R/m$ is algebraically closed not of characteristic 2,3, or 5, and $R$ is not regular, then $\hat{R}$ is factorial is equivalent to the existence of a basis $\{x,y,z\}$ of $m^*$ such that $(z^*)^2+(y^*)^3+(x^*)^5 = 0$. Here $m^*$ is the Henselization. – RghtHndSd Feb 7 '14 at 2:48
• @MattE: Your example certainly seems correct to me! That'll teach me to check what I say, even if quoting directly from a source. To rghthndsd: Thank you for hunting down the reference(s): the correct statement indeed seems much subtler than initially stated. – zcn Feb 7 '14 at 5:49