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:
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$S(Y)_{x_i}$ 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 $S(Y)_{x_i}$.)
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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.
Best Answer
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 $\dim S(Y)=\dim Y+1$.