Theorem 11.25 For any irreducible variety $V$ over $k$ the local dimension of $V$ at any point is equal to $\dim V$.
Proof: By the Normalization Lemma (Chapter 5, Exercise 16) we can find a polynomial ring $B=k[x_{1},\cdots, x_{d}]$ contained in $A(V)$ such that $d=\dim V$ and $A(V)$ is integral over $B$…..
In this case $A(V)$ is the coordinate ring of a variety $V$ over an algebraically closed field $k$. And $\dim V$ is defined as the transcendental degree of $k(V)$, which is a field of fraction of $A(V)$. Noether normalization lemma in this book is
Noether's normalization lemma: Let $k$ be a field and let $A\neq 0$ be a finitely generated $k$-algebra. Then there exists elements $y_{1},\cdots, y_{r} \in A$ which are algebraically independent over $k$ and such that $A$ is integral over $k[y_{1},\cdots, y_{r}]$
My question is, how can we say that $d= \dim V$? From the lemma, we know that there is such $k$-algebra $B$ inside of $A(V)$ and $A(V)$ is integral over $B$. Now, to show $d= \dim V$, we need to show that transcendence degree of the field of fraction $k(V)$ is equal to that of $k(x_{1},\cdots, x_{n})$, which is a subring. However, I don't know why this is true. Could you help me?
Best Answer
Edit: around any point we may choose an affine chart and compute dimension locally, so it suffices to assume $V$ is affine. In this case, the below argument works, since the function field of an (irreducible) affine variety is simply localization of the coordinate ring.
The fraction field of $A(V)$ is $k(V)$ and the fraction field of $k[x_1,\dots,x_d]$ is $k(x_1,\dots,x_d)$. Since $A(V)$ is integral over $k[x_1,\dots,x_d]$, $k(V)$ will be integral over $k(x_1,\dots,x_d)$ and hence has the same transcendence degree $d$.