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It is well-known that if we pass from a UFD to a new ring where we have factored one of the primes, it does not need to stay a UFD. The classic example is passing from $\mathbb{Z}$ to $\mathbb{Z}[\sqrt{-5}]$, where $-5$ has a new (repeated) factor.

My current research student, Caleb Dastrup, recently pointed out to me that we do not lose the UFD property if we adjoin new factors freely. In other words, if $R$ is a UDF and if $p\in R$ is prime, then the new ring $S:=R[s,s'\, :\, ss'=p]$ is a UFD. Additionally, all the primes of $R$ not associate to $p$ are still prime, and $s,s'$ are both prime, and the units of $S$ are the same as in $R$.

We've developed two different proof strategies. First, one can write elements of $S$ in a reduced form, where they are $R$-linear combinations of powers of $s$ and $s'$ (separately, never jointly). Then some ad hoc computations verify all those claims. Second, one can think of $S$ as the subring $R[s,p/s]$ of $R[s,s^{-1}]$, which is the valuation subring where we give both $s$ and $p$ valuation $1$. Again, some ad hoc work suffices to finish proving the theorem.

I'm wondering if this result already appears in the literature, or follows from some known results.

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    $\begingroup$ I suppose a third approach, which might generalize, is to try to compute the (divisor) class group of $R[s,s':ss' = p] = R[X,Y]/(XY-p)$ in terms of the class group of $R$, using techniques of Fossum's book, The Class Group of a Krull domain. If $p$ is prime in $R$, then probably the class groups are the same. If $p$ is the square of a prime in $R$, then I'd guess the class group is cyclic of order $2$ if $R$ is a UFD. That's the only type of generalization I can think of to generalize your result. $\endgroup$
    – Jesse Elliott
    Commented Apr 17 at 1:34
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    $\begingroup$ I recall that Fossum's book has examples computing the class group of $k[X_1,X_2,\ldots,X_n]/(F)$ for various quadratic forms $F$, but I don't own a copy of the book and can't check if it has an example that might include yours. $\endgroup$
    – Jesse Elliott
    Commented Apr 17 at 1:45
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    $\begingroup$ Not sure if the following is less ad hoc than what you already have: (1) The element $s$ of $S = R[s, p/s]$ is prime because $S/(s) \simeq R/(p)[s']$. (2) A non-zero element of $S$ has only finitely many divisors up to multiplication by a unit of $R$; for this use the $p$-valuation of the coefficient of least degree to reduce to a similar statement in the UFD $R[s]$. Thus $S$ has ACCP. By Nagata's criterion, the ring $S$ is a UFD; indeed its localization at $s$ is the UFD $R[s, s^{-1}]$. $\endgroup$
    – Luc Guyot
    Commented Apr 17 at 22:45
  • $\begingroup$ @JesseElliott Thanks for that reference. Lemma 11.1 is quite close to what we want. $\endgroup$
    – Pace Nielsen
    Commented Apr 18 at 13:30
  • $\begingroup$ @LucGuyot I was not familiar with Nagata's criterion. It is very convenient in this setting, and streamlines things. Thanks! $\endgroup$
    – Pace Nielsen
    Commented Apr 18 at 13:59

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