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Let $K$ be an algebraically closed field, why $K[x,y]$ and $K[x,y,z]/(xz-y^2)$ are not isomorphic as rings?

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  • $\begingroup$ Hint: Is $K[x,y,z]/(xz-y^2)$ a UFD? $\endgroup$ – Pierre-Yves Gaillard Mar 12 '18 at 14:14
  • $\begingroup$ Is the assumption that $K$ is an algebraically closed field necessary? Couldn't $K$ be any UFD? $\endgroup$ – Pierre-Yves Gaillard Mar 12 '18 at 14:37
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    $\begingroup$ @Pierre-YvesGaillard In $K[x,y,z]/(xz-y^2)$ i have $yy=y^2=xz$ but $x$ and $y$; and $z$ and $y$ are not associated; so $K[x,y,z]/(xz-y^2)$ is not UFD, right? And yes, if $K$ is simply an UFD, it's the same. $\endgroup$ – Minato Mar 12 '18 at 14:45
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    $\begingroup$ Yes! If we wanted to be completely rigorous, we'd have to check a couple of things, like: the images of $x,y$ and $z$ are irreducible and, as you said, they belong to different association classes. $\endgroup$ – Pierre-Yves Gaillard Mar 12 '18 at 14:57
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$K[x,y]$ is a regular ring (all of its localizations are regular). The local ring of $K[x,y,z]/(xz-y^2)$ at the origin $(0,0,0)$ is not regular: The maximal ideal $\mathfrak m$ there cannot be generated by two elements, not even mod $\mathfrak m^2$.

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  • $\begingroup$ I'm not familiar with the notion of regular ring. Do you think it is possible a more "elementary" argument? Thank-you. $\endgroup$ – Minato Mar 12 '18 at 14:09
  • $\begingroup$ @m_t_ I did my best not to make it look local, since I put that parenthetical remark in about all its local rings. $\endgroup$ – John Brevik Mar 12 '18 at 14:22
  • $\begingroup$ @Minato Proving non-isomorphism is not so easy without using something! Well, suppose that they were, and let $\mathfrak n$ be the maximal ideal of $K[x,y]$ corresponding to $\mathfrak m$ under this isomorphism. Then $\mathfrak n = (x-a,y-b)$ for some $a,b\in K$ by the Nullstellensatz. So $\mathfrak n$ can be generated by two elements. As I observed in my last comment, $\mathfrak m$ cannot. This gives a contradiction. $\endgroup$ – John Brevik Mar 12 '18 at 14:25
  • $\begingroup$ @JohnBrevik thank-you! In yuor notation, $m$ is $(x,y,z)/(xz-y^2)$ right? And i can not generete $m$ with less than 3 elements, instead every maximal ideal in $K[x,y]$ is 2-generated. Right? But since you use Nullstellensatz, this argumenti is not good if $K$ is not algebraically closed, right? $\endgroup$ – Minato Mar 12 '18 at 14:52
  • $\begingroup$ That's correct on both counts. The regularity argument should be OK in all cases, though. $\endgroup$ – John Brevik Mar 12 '18 at 15:04

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