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Let $K := (uw - v^2, w^3 - u^5)$. Show that $V(K)$ consists of two irreducible components, one of which is $V(uw - v^2, w^3 - u^5, u^3-vw) = V(J)$.

I don't know how to start this. I see that $V(K)$ is symmetric under the interchange of $v \to -v$ but that $V(J)$ isn't. Does this mean that the other component should be symmetric under this exchange?

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This is copy and pasted homework for the Algebraic Geometry course at the University of Edinburgh, coincidentally due in tomorrow. Mary, I suggest you do your own work or at least tag this post as homework. – user26482 Mar 7 '12 at 20:17
Not the most productive thing to write, A.non. – Stephen Mar 7 '12 at 20:25

1 Answer 1

First, we need to find out where $K$ is not prime. More precisely, we need $ab\in K$ but neither $a\in K$ nor $b\in K$. The definition of $J$ helps a bit - notice that

$$(u^3-vw)(u^3+vw) = u^6 - v^2w^2 = uw^3 - w^2v^2 - uw^3 + u^6 = w^2(uw-v^2) - u(w^3-u^5) \in K$$

So we're looking for a prime ideal $I$ with $I\cap J=K$ since $V(I\cap J)=V(I)\cup V(J)$. In this case, that'd be $I=(uw-v^2,w^3-u^5,u^3+vw)$.

What is left to do is check that both $I$ and $J$ are actually prime.

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