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This is a follow-up to Projective Spectrum of $K[X,Y]$ .

I see why the given ideals are prime or even maximal, however, I have yet to prove that they in fact make up the entire spectrum of $K[X,Y]$.

Why is it that any polynomial other than the generators mentioned in the previous thread can be factored into degree one polynomials in X and Y?

I know that fixing, say, Y as an element of K will yield a polynomial in $K[X]$ which splits, but this decomposition shouldn't generally be an element of $K[X,Y]$.

Thanks in advance!

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I didn't register back then, and I didn't again today. It'd be great if a mod could merge both guest accounts, but I'll register anyways once I have a new mail account set up. –  Wendelin Oct 23 '11 at 16:12
@Wendelin: I've merged your accounts. –  Zev Chonoles Oct 23 '11 at 16:19
It would be nice if you mad this post self-contained... –  Mariano Suárez-Alvarez Oct 23 '11 at 23:38

1 Answer 1

up vote 1 down vote accepted

Suppose $K$ is algebraically closed. If $P(X,Y)$ is homogeneous of degree $d$, then $P(X,Y)=Y^dP(X/Y,1)$. Let $Q(T)=P(T, 1)$ viewed as an element of $K[T]$, of degree $r$. Then $r\le d$. As $K$ is algebraically closed, $Q(T)=c\prod_{1\le i\le r}(T-\alpha_i)$ with $c, \alpha_i\in K$. Thus $$P(X,Y)=Y^{d-r}\prod_{i\le r}(X-\alpha_i Y).$$

If $K$ is not necessarily algebraically closed, the same method shows that $P(X,Y)$ is a product of homogeneous irreducible polynomials.

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Thanks! I got it. –  Wendelin Oct 24 '11 at 17:28

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