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.



1 Answer 1


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.

  • $\begingroup$ Thanks for the answer. Is the discourse concerning the first question a general fact? Could you please write it down in general? Maybe the notations are simpler and I can apply it to other observations. $\endgroup$
    – user805324
    Jul 13, 2021 at 9:05
  • $\begingroup$ Which part? The first sentence is just the universal property of localization/fraction fields, the next claim is the structure of $F(t)$, and then finally we just use the fact that if we want to show some combination of elements is in the image of our homomorphism, it suffices to show each element individually is. $\endgroup$
    – KReiser
    Jul 13, 2021 at 9:08
  • $\begingroup$ $K((S(Y)_{x_i})_0)(x_i)$ I don't understand who this field is in general $\endgroup$
    – user805324
    Jul 13, 2021 at 9:11
  • $\begingroup$ Well, luckily you don't have to understand much about it here. The key point is that it's of the form $F(t)$ for $F$ some field (in this case $K((S(Y)_{x_i})_0)$) and some indeterminate $t$ which is transcendental over $F$ (in this case $x_i$). Elements of $F(t)$ have a particularly nice description: they're rational functions in $t$ with coefficients from $F$. This is all we use in the argument. $\endgroup$
    – KReiser
    Jul 13, 2021 at 9:13
  • $\begingroup$ a last question, why $$(S(Y)_{x_i})_0[x_i,x_i^{-1}]\hookrightarrow K((S(Y)_{x_i})_0)(x_i)$$? $\endgroup$
    – user805324
    Jul 13, 2021 at 9:22

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