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I have two questions in order to find general methods to find prime ideal: the first one is from Ravi Vakil:

1) Suppose $I=(wz-xy,wy-x^2,xz-y^2) \subset k[x,y,z,w]$. How can I prove that $k[x,y,z,w]/I$ is a domain?

2) Suppose $J=(x^3+y^3+z^3)$ and consider $R=\mathbb{C}[x,y,z]/J$. How can I find prime ideals of $R$? If $f=x^2$, how can I find prime ideals of localization of $R$ in $f$?

ArthurStuart
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  • Started looking at this, but got to go, so putting the 1st random thoughts that didn't produce an answer in the comment here. By multiplying the 2nd and 3rd equations for $I$ we have $x^2y^2=xywz$. This ideal is the product of ideals $x$, $y$, and $xy-wz$. The latter one is the same as the 1st equation for $I$. It looks like $spec(I)$ is a component of $spec(I')=(wy-x^2,xz-y^2)$, which is of course not irreducible. Therefore variety you are looking at should have dimension 2, not 1. Alas, that's about as far as I've got so far before having to leave; hope it helps. – Michael Sep 21 '13 at 00:29
  • Consider the map $T: k[x,y,z,w]/I \to k[s,t]$ defined so that $w \mapsto s^3, x \mapsto ts^2, y \mapsto t^2s, z \mapsto t^3$. We observe that $I$ is in the kernel of such a map so that it is indeed well defined. Moreover, I claim that such a map is an injection. Thus, the original ring is an integral domain and so $I$ is prime.
    • – Alexander Sep 21 '13 at 04:08
  • @YACP I'm sorry... a domain – ArthurStuart Sep 21 '13 at 08:22
  • @Alexander What is the geometric interpretation of your map? – ArthurStuart Sep 21 '13 at 08:37
  • The ideal $I$ is the homogeneous ideal of the twisted cubic curve as YACP pointed out. It is the rational normal curve for $\mathbb{P}^3$. – Alexander Sep 21 '13 at 13:25
  • By “domain” you mean “integral domain” ? – Ewan Delanoy Sep 29 '13 at 02:38
  • Yes.. I mean integral domain – ArthurStuart Sep 29 '13 at 06:58