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Let $R = \Bbbk[x_1, \dots, x_n]/(p)$, where $\Bbbk$ is an algebraically closed field of characteristic zero and where $p \in \Bbbk[x_1, \dots, x_n]$ is prime, so that $R$ is a domain. In general, $R$ is not a UFD: one such example is $\Bbbk[w,x,y,z]/(wx-yz)$, where $wx=yz$ has two factorizations into irreducibles. Obviously, one can invert every element of $R$ to get a field, and hence a UFD, but I'm interested in only inverting a subset of elements of $R$:

Let $p \in \Bbbk[x_1, \dots, x_n]$ be prime. Is $\Bbbk[x_1^{\pm 1}, \dots, x_n^{\pm 1}]/(p)$ a UFD? If this isn't true in general, can any simple hypotheses be placed on $p$ to guarantee that it is a UFD?

lokodiz
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1 Answers1

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No, $k[x_1^{\pm 1}, \dots, x_n^{\pm 1}]/(p)$ is not a UFD in general.
For example if $p(x,y)=(y-1)^2-(x-1)^3\in k[x^{\pm 1},y^{\pm 1}]$, then the quotient ring $R=k[x^{\pm 1},y^{\pm 1}]/(p)$ is not a UFD, as witnessed by the two factorizations of $(\bar y-1)^2=(\bar x-1)^3\in R$ ( There are details to check, like irreducibility of $\bar y-1$ and $\bar x-1$).

user26857
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Georges Elencwajg
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  • Let me see if I've understood well this example: one can start with $k[t^2,t^3]$ which is isomorphic to $k[x,y]/((y-1)^2-(x-1)^3)$. But if one inverts $t^2$ and $t^3$ we get $k[t,t^{-1}]$ which is a UFD. Then remark that $k[t^2,t^3]=k[t^2+1,t^3+1]$ and invert $t^2+1$ and $t^3+1$. This way $t^2$ and $t^3$ remain irreducible and $(t^2)^3=(t^3)^2$ still provides a non-unique decomposition. – user26857 Jun 13 '17 at 09:23
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    Dear @ user26857: yes, you are of course right (as usual). – Georges Elencwajg Jun 13 '17 at 10:32