Question on Proof of Atiyah-Macdonald Theorem 11.25

Question

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?

Answer 1

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 $d$.

Answer 2

TomGrubb already gave a valid explanation. Here's some insight into the phenomenon at play:

Everything that follows is a generalization of [C]. Let $k\subset A$ be a ring extension, where $k$ is a field. If there exists a subset $\{x_i\}\subset A$ such that

1. $\{x_i\}$ is algebraically independent over $k$ and
2. $k[x_i]\subset A$ is an integral extension,

we say that $\{x_i\}$ is a transcendence system of $A$ over $k$. Whereas a transcendence basis of a field extension always exists, a transcendence system of a ring extension of $k$ may or may not exist. If it exists, it needs to be maximal in the algebraically independent subsets of $A$ over $k$.

Example 1. The field $k(x)$ has no transcendence system over $k$. To look for a contradiction, suppose it does. By [AM, Proposition 5.7], it is also a transcendence basis over $k$ (by [C, Proposition 1], it is made up of a single element $f/g$, $f,g\in k[x],g\neq 0$) and $k[f/g]$ is a field. We may assume that $\operatorname{gcd}(f,g)=1$ in $k[x]$. But then $k[f/g]$ is not a field (an equation $g/f=\sum^n_{i=0}a_i(f/g)^i$, $a_i\in k$, implies that $f|g^{n+1}$ in $k[x]$, which is not possible).

Proposition 1. All transcendence bases of $A$ over $k$ (if some exists) have the same cardinality (this cardinality is called the transcendence rank of $A$ over $k$, denoted $\operatorname{trank}_kA$).¹

Proof. For the general case, try to generalize the proof of 030F (the case where $A$ is a domain may be deduced from Lemma 2 below plus the result for fields). Here is a proof that works in the finite case. $\square$

Lemma 1. Suppose the set $\left\{x_1, x_2, \ldots, x_m\right\} \subset A$ is algebraically independent over $k$ and suppose the set $\left\{y_1, y_2, \dots, y_n\right\} \subset A$ has the property that $k[y_j]\subset A$ is an integral extension. Then $m \leq n$ and we may reorder the $y_j$'s so that $k[x_1, \ldots, x_m, y_{m+1}, \ldots, y_n] \subset A$ is an algebraic extension.

Reversing the roles of $\left\{x_i\right\}$ and $\{y_j\}$ in the lemma, you see that any two finite transcendence bases have the same cardinality. The lemma also implies that if one transcendence base is finite then so is any other.

Proof. Same as [C, Lemma 1]. $\square$

Noether normalization lemma can be synthesized as: every ring which is of finite type over a field $k$ has a finite transcendence system over $k$; moreover, in this case, $\operatorname{trank}_kA=\dim A$ (by 00OK and since $\dim B[x_1,\dots,x_n]=n+\dim B$ for $B$ Noetherian).

Lemma 2. Suppose $A$ is a domain and let $K$ be the fraction field of $A$. Any transcendence system of $A$ over $k$ is a transcendence basis of $K$ over $k$. In this case, $\operatorname{trank}_kA=\operatorname{trdeg}_kK$.

Proof. Let $\{x_i\}\subset A$ be a transcendence basis over $k$. We need to show that the field extension $k(x_i)\subset K$ is integral. Since (i) every element of $K$ is of the form $a/b$, $a,b\in A$, $b\neq 0$, and (ii) a product of algebraic elements is again algebraic, it suffices to show that $1/b$ is algebraic over $k(x_i)$. This follows from the fact that if $L\subset K$ is a field extension, then a nonzero element $\alpha\in K$ is algebraic over $L$ iff $\alpha^{-1}$ is. $\square$

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¹I made up the terminologies of “transcendence system” and “transcendence rank,” for I also made up the associated notions. If you know some reference in the literature of any of these notions, please, leave a comment below 🙂.

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References

[AM] Atiyah, Macdonald, Introduction to Commutative Algebra

[C] B. Conrad, Transcendence Bases and Noether Normalization (archived in the Wayback Machine).