Exercise I-2.6 of Hartshorne: If $Y$ is a projective variety with homogeneous coordinate ring $S(Y)$, show that $\dim S(Y)=\dim Y+1.$
My approach:
I am able to verify his hints. i.e. Let $\phi:U_i\to \mathbb A^n$ be the canonical homeomorphism as defined for his proposition I-2.2. Let $Y_i:=\phi_i(Y\cap U_i)$, which is an affine variety. Then I can prove the identifications $A(Y_i)\cong (S(Y)_{x_i})_0$ and $S(Y)_{x_i} \cong A(Y_i)[x_i,x_i^{-1}]$.
Now here comes the transcendence degree part: the dimensions. I hope I can have
$$\dim S(Y)=tr.d(K(S(Y)))\\=tr.d(K(S(Y))_{x_i})
=tr.d(K(S(Y))_{x_i})_0)+1 \\=
tr.d(K(A(Y_i)_{x_i})+1
=\dim(Y\cap U_i)+1=\dim(Y)+1$$
I have difficulty in showing
$$tr.d(K(S(Y))_{x_i})=tr.d(K((S(Y)_{x_i})_0)+1.(*)$$
I expect that that $x_i$ is transcendental over $K((S(Y)_{x_i})_0)$ but don't know how to rigorously prove it. Note elements in $K((S(Y)_{x_i})_0)$ take the form $f(x_0/x_i,\dots,x_n/x_i)/g(x_0/x_i,\dots,x_n/x_i)$ ($f,g$ may not have the same degree) which already involves $x_i$. Maybe this is not the right way to prove $(*)$ or we don't need to prove $(*)$?
Alternate approaches are also appreciated!
Best Answer
You are correct that $x_i$ is transcendental over the degree-zero elements of the localization (assuming, of course, that your projective variety $Y$ is not contained in $V(x_i)$, which is an assumption from earlier on in the proof, needed to guarantee $Y_i$ is nonempty). Suppose that $x_i$ was algebraic over $K(S(Y))_0$: that is, there's some nonzero monic polynomial $p(T)$ of degree $d>0$ with coefficients from $K(S(Y))_0$ so that plugging in $x_i$ for $T$, we get that $p(x_i)=0$. This is a contradiction: $p(x_i)$ has degree-$d$ portion $x_i^d$, which is nonzero.