Theorem. Let $Y$ be a projective variety with homogeneous coordinate ring $S(Y)$, then $$\dim S(Y)=\dim Y+1$$
I have broken down the proof into steps. The various steps I have shown them and there are no problems, except in the final step.
Consider the multiplicative subset $$T_i=\{x_i^r+I(Y)_r\;:\; r\ge 0\}\subseteq S(Y)$$ consisting of homogeneous elements. The homogeneous localization $$S(Y)_{x_i}:=T_i^{-1} S(Y)$$ is defined by $$S(Y)_{x_i}=\bigoplus_{n\in\mathbb{Z}}\big(S(Y)_{x_i}\big)_n,$$ where
$$\big(S(Y)_{x_i}\big)_n:=\{a/s\;:\;a\in S(Y)_{n+r}, s\in T_i\cap S(Y)_r\;\text{for same}\; r\ge 0\}$$
Step 1: Consider the homeomorphism $\varphi_i\colon U_i\to \mathbb{A}^n$ given by $$\varphi([a_0:\dots :a_n])=(a_0,\dots,a_{i-1}, a_{i+1},\dots a_n).$$ The set $Y_i:=\varphi_i(Y\cap U_i)\subseteq \mathbb{A}^n$ is an affine variety. Now consider the coordinate ring $A(Y_i)=A/I(Y_i)$, where $A=k[y_1,\dots, y_n]$. I proved that:
$$
\boxed{A(Y_i)\cong \big(S(Y)_{x_i}\big)_0}
$$
Step 2. Suppose that $x_i\notin I(Y)$, then $$
\boxed{\big(S(Y)_{x_i}\big)_0[x_i,x_i^{-1}]\cong S(Y)_{x_i}}
$$
Step 3. If $x_i\notin I(Y)$, then $x_i$ is transcendental over $K((S(Y)_{x_i})_0)$, here $K(R)$ denote the quotient field of the integral domain $R$.
Concluding step: From $\color{red}{\text{step 2}}$ we have
$$K((S(Y)_{x_i})_0)(x_i)\color{red}{\cong}K(S(Y)_{x_i})\color{green}{\cong}K(S(Y))$$ For the congruence in green click here
Question 1. I did not understand the first term of the previous expression. $$K\big(\big(S(Y)_{x_i}\big)_0[x_i,x_i^{-1}]\big)\cong K((S(Y)_{x_i})_0)(x_i)$$ why?
Now we see that
\begin{equation}
\begin{split}
\dim S(Y) =&\;\text{tr.}\deg_kK(S(Y))\\
=&\;\text{tr.}\deg_kK((S(Y)_{x_i})_0)(x_i)\\
\color{blue}{=}&\; 1+\text{tr.}\deg K((S(Y)_{x_i})_0)
\end{split}
\end{equation}
Question 2 Why the blue equality holds? I know that this holds from step 3, but I don't understand why it is like this.
Thanks!
Best Answer
Question 1: since $(S(Y)_{x_i})_0[x_i,x_i^{-1}]\hookrightarrow K((S(Y)_{x_i})_0)(x_i)$ and the latter is a field, we get an injection $\iota$ from the fraction field of the LHS to the RHS. We now need to show that this is surjective. Any element of the RHS can be written as $$\frac{\sum p_jx_i^j}{\sum q_kx_i^k}\in K((S(Y)_{x_i})_0)(x_i)$$ where $p_j,q_k\in K((S(Y)_{x_i})_0)$, so it suffices to show that $\sum p_jx_i^j$ and $\sum q_kx_i^k$ are in the image of the map from $K((S(Y)_{x_i})_0[x_i,x_i^{-1}])$. But this is clear: each $p_j$ and $q_k$ is in the image of the map from $K((S(Y)_{x_i})_0)\subset K((S(Y)_{x_i})_0[x_i,x_i^{-1}])$, and $x_i$ is clearly in the image of $\iota$ too.
Question 2: This is just the fact that transcendence degree adds over field extensions. Given a tower of field extensions $F_1\subset F_2\subset F_3$, we have $\operatorname{trdeg}_{F_1} F_3=\operatorname{trdeg}_{F_1} F_2 + \operatorname{trdeg}_{F_2} F_3$. Take $F_1=k$, $F_2=K((S(Y)_{x_i})_0)$, and $F_3=K((S(Y)_{x_i})_0)(x_i)$: as $x_i$ is transcendental over $K((S(Y)_{x_i})_0)$, we have that $F_3$ is a transcendence degree one extension of $F_2$ and we're done.