Hartshorne I.2.6 - questions about Boercherds' solution

Hartshorne Exercise I.2.6: $$Y$$ is a projective variety with homogeneous coordinate ring $$S(Y)$$, then $$\dim (Y) = \dim S(Y) + 1$$.

I think I'm not understanding a simple thing. It relates to the part of the hint "show $$A(Y_i)$$ can be identified with the subring of degree zero in the localization $$S(Y)_{x_i}$$".

I've read two solutions now (including Borcherds') which basically say the following:

1. $$S(Y)_{x_i}$$ is the affine coordinate ring of the part of the cone in $$\mathbb{A}^{n+1}$$ where $$x_i \neq 0$$. (Fine, that has nothing to do with the grading of $$S(Y)_{x_i}$$.)

2. Therefore its homogeneous part of degree zero, is "the coordinate ring of the cone with $$x_i = 0$$". (What? Wouldn't that be a proper closed subvariety?)

I want to understand why people are drawing the conclusion 2 from 1. I am surely missing something going on.

• By the way, $\dim$ will produce $\dim$, which will display a bit more nicely than what you had written before. Commented Jul 19 at 4:01

I think you've been tricked by a typo. The homogeneous part of degree zero of $$S(Y)_{x_i}$$ is isomorphic to $$S(Y)_{x_i}/(x_i-1)\cong S(Y)/(x_i-1)$$, which is the coordinate ring of the cone with $$x_i=1$$ (not zero!).

Now let's use this to connect the various dimensions. Assume that $$Y_i= Y \cap D_+(x_i)$$ is nonempty. We know that $$S(Y)_{(x_i)}$$, the homogeneous localization of $$S(Y)$$ at $$x_i$$, is the affine coordinate algebra of $$Y_i$$, so $$\dim Y_i = \dim S(Y)_{(x_i)}$$. Via the above work giving $$S(Y)_{(x_i)}\cong S(Y)/(x_i-1)$$, we have $$\dim S(Y)_{(x_i)} = \dim V(x_i-1)\cap C(Y)$$, so as intersecting with a hyperplane drops dimension by one, we have $$\dim Y_i = \dim (C(Y)\cap D(x_i)) -1$$.

Since $$Y_i$$ is nonempty, $$C(Y)$$ contains a line, so $$C(Y)$$ is of dimension at least one. Since $$\bigcup_i (C(Y)\cap D(x_i))$$ is $$C(Y)$$ without the origin and $$C(Y)$$ is of positive dimension, we have that $$\dim C(Y) = \dim C(Y)\setminus0$$, so from the properties of dimension and open covers, we know $$\dim C(Y) = \dim C(Y)\setminus 0 = \max_i(\dim C(Y)\cap D(x_i)).$$ As $$\dim C(Y)\cap D(x_i) = \dim Y_i + 1$$, we know $$\dim C(Y) = 1 + \max_i(\dim Y_i) = \dim Y$$, again by the properties of dimension on open covers. But then as $$\dim C(Y) = \dim S(Y)$$, we have $$\dim S(Y)=\dim Y+1$$.

• Great! This is maybe less algebraic and more geometric than the hinted solution. Also technically maybe part of the problem has to do with hartshorne's $\alpha$ and $\beta$ maps? You say "$S(Y)_{x_i}$ is the affine coordinate algebra of $Y_i$", but your $Y_i$ is technically still in $\mathbb{P}^n$. Hartshorne's is its image under $\phi$. So maybe part of the point is something like that the dehomogenization $\alpha$ is precisely what transfers functions under $\phi$? Commented Jul 22 at 5:28
• @eggselent Sure, the solution I wrote when I went through these exercises as a student follows the hint a little more closely, and I'm definitely breezing through the identification of $Y_i\subset\Bbb P^n$ and $Y_i\subset\Bbb A^n$. But since your post focused on the cones, I aimed my answer a bit more in that direction. Were you looking for something else? Commented Jul 22 at 5:31
• No. Thanks. Why do comments have to be so many characters. Oh I see why now lol, it's written in the box. Sorry and thanks for your help! Commented Jul 22 at 5:33