# Projective Spectrum of $K[X,Y]$

Let's assume that $K$ is algebraically closed.
I'm having some difficulties figuring out what $\text{proj}\;K[X,Y]$ is, where $K[X,Y]$ is interpreted as a graded ring.

Any hints? So far I have only figured out that $(X,Y)$ obviously cannot be in $\text{proj}\;K[X,Y]$ since it contains all the elements of $K[X,Y]$ without summands coming from the ground field. Also, $(aX + Y)$ is prime and thus should be in the projective spectrum if I am not mistaken. What about the ideals generated by higher powers of $X$ and $Y$?

However, I fail to see how the fact that $K$ is algebraically closed comes to play.

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## 2 Answers

The only homogeneous prime ideals generated by one element are of the form $(bX-aY)$ for $(a,b) \neq (0,0)$ (these are maximal) and $(0)$ (which is not maximal). All other principal ideals will not be prime because $K$ is algebraically closed, so you will be able to factor the polynomial. For instance, with $K = \mathbb{C}$, you have $(X^2+Y^2) = (X+iY)(X-iY)$.

All other prime ideals, that is to say those that are not generated by at most one element, will contain the irrelevant ideal $(X,Y)$.

Hence $\mathrm{Proj}(K[X,Y])$ is what you'd expect - it contains a point $(a:b) = (ka:kb)$ (for $0 \neq k \in K$) for each maximal ideal $(bX-aY) = (kaX-kaY)$, and so the closed points form a projective line over $K$, and then you also have the ideal $(0)$ which is the generic point of the whole projective line. It's a very similar answer to the case of $\mathrm{Spec}(K[X])$, where you have a point for each $(X-a)$ and the generic point $(0)$. The difference is that with Proj, we also get a point at infinity, and so we get the projective line over $K$ instead of the affine line over $K$.

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Thanks, that was a nice explanation. However, one thing remains for me - is there some general theorem on why polynomials of higher degree can be factored, for an arbitrary algebraically closed field? In one variable this is of course clear, but I don't see how this can be generalized. – Wendelin Oct 15 '11 at 15:00
@Wendelin, note that the polynomials in question are homogeneous, say of degree d, so factoring out Y^d gives a univariate polynomial in (X/Y), which can be factored, then multiplying the Y^d back in gives factors of the form (X-cY) – Brent Baccala Jun 6 '13 at 0:51

Another way to help understand the Proj construction better is to understand the homogenization / dehomogenization process that relates to the standard cover by affine opens. For $\text{Proj } K[x,y] \cong \mathbb{P}^1$, there are two open sets $U_0, U_1 \cong \mathbb{A}^1$. It is worthwhile to understand which homogeneous prime ideals in $\text{Proj } K[x,y]$ correspond to prime ideals in $U_0$ and which correspond to prime ideals in $U_1$.

This is along the lines of P.A.I.'s last paragraph, with the extra information that there are natural morphisms $\phi_0, \phi_1 : \text{Spec } K[t] \rightarrow \text{Proj } K[x,y]$ that map isomorphically onto $U_0, U_1$.

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Not sure! I think there must have been a response from a user whose screenname was P.A.I. but it appears that reply has since been deleted. – Michael Joyce Jun 7 '13 at 20:12