Say we have a UFD A and a prime ideal P. Is A/P a UFD?
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What are your thoughts on the problem? – Rory Daulton Mar 13 '16 at 22:32
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What's $U$?${}{}$ – Mar 13 '16 at 22:48
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Duplicate of http://math.stackexchange.com/questions/1023990/r-mathfrak-p-not-always-a-ufd. – user26857 Mar 13 '16 at 22:51
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Not necessarily. $\Bbb{Z}[x]$ is a UFD, but $$\Bbb{Z}[x]/(x^2+5) \cong \Bbb{Z}[\sqrt{-5}]$$ is not.
Crostul
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Not necessarily. $\mathbb C[X,Y,Z]$ is a UFD, but $$\mathbb C[X,Y,Z]/(Z^2-XY)=\mathbb C[x,y,z] $$ is not.
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Although it is certainly possible to give an elementary proof of non UFDness based on the two different factorizations $z^2=xy$ of the same element, I can't resist the temptation to say that $\mathbb C[x,y,z]$ is not a UFD because its class group is $\mathbb Z/(2)$ (Hartshorne, example 6.5.2, page 133), whereas a UFD has zero class group (Hartshorne, Proposition 6.2, page 131).
Georges Elencwajg
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1Thanks! What distinction are you making between these two rings when you write the indeterminates of one in capitals and of the other in lowercase, by the way? – Vik78 Mar 14 '16 at 00:56
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Capital letters are for the indeterminates in $\mathbb C[X,Y,Z]$, whereas lowercase letters are the classes of those indeterminates modulo the ideal $(Z^2-XY)$, namely $x=X \operatorname {mod} (Z^2-XY)$ etc. – Georges Elencwajg Mar 14 '16 at 08:40
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Alright. Also, I have an additional question: we know that integral domains arise as quotients by prime ideals, fields as quotients by maximal ideals, etc. What conditions can we place on an ideal to ensure that a quotient by it is a UFD? – Vik78 Mar 14 '16 at 15:21
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I don't think there is such a criterion. Anyway, I'm sure that I don't know any. – Georges Elencwajg Mar 14 '16 at 18:18