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I am trying to show that $k[x,y,z]/(xz-y^2)\not\cong k[x,y]$. The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $x|xz$ which implies $x|y^2$ but $x|y$. So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irred implies prime in UFDs.

Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists a $c$ in $k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irred there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.

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up vote 7 down vote accepted

It's easier just to observe that $xz=y^2$ in your ring gives a decomposition into irreducibles that is not unique ($x$, $y$, and $z$ are irreducible because they are of degree one; your quotient is homogeneous). Of course, this also shows $x$ is not prime since it divides $y^2=y y$ but divides neither $y$ nor $y$.

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The algebra $k[x,y,z]/(xz-y^2)$ is a graded domain, with (the classes of) $x$, $y$ and $z$ all in degree $1$.

In a graded domain, the product of two non-homogeneous elements is non-homogeneous. So if $x$ is a product of two elements, these must be homogeneous. Since $x$ has degree $1$, it can only be a product of an element of degree $1$ and an element of degree $0$. Since the only elements of degree zero are units, we are done.

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For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$. –  fpqc Apr 3 '13 at 22:31
    
@Steve, $xy$ is also linear in $x$ :-) –  Mariano Suárez-Alvarez Apr 3 '13 at 23:08
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My point is that it does not follow «just» from the fact that that it is linear in $x$, because there exist other polynomials which are linear in $x$ and which are reducible. You need to observe that $x$ is linear in $x$ and use that to start some argument or another. –  Mariano Suárez-Alvarez Apr 3 '13 at 23:18
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«The obvious way» is rarely as obvious to everyone. Surely you do not need to be told this. In any case, I don't see what your point is and, by now, I have stopped being interested. –  Mariano Suárez-Alvarez Apr 3 '13 at 23:27
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