Exercise 5.18 of Atiyah and Macdonald

Question

Following is proof of Zariski’s lemma by Zariski himself:

Let $k$ be a field and let $B$ be a finitely generated $k$-algebra. Suppose that $B$ is a field. Then $B$ is a finite algebraic extension of $k$. (This is another version of Hilbert’s Nullstellensatz. The following proof is due to Zariski. For other proofs, see (5.24), (7.9).)

Let $x_1,…,x_n$ generate $B$ as a $k$-algebra. The proof is by induction on $n$. If $n=1$ the result is clearly true, so assume $n\gt 1$. Let $A=k[x_1]$ and let $K=k(x_1)$ be the field of fraction of $A$. By the inductive hypothesis, $B$ is finite algebraic extension of $K$, i.e. coefficient of the form $a/b$ where $a$ and $b$ are in $A$. If $f$ is product of the denominators of all these coefficient, then each of $x_2,…,x_n$ is integral over $A_f$. Hence $B$ and therefore $K$ is integral over $A_f$.

Suppose $x_1$ is transcendental over $k$. Then $A$ is integrally closed, because it is a unique factorization domain. Hence $A_f$ is integrally closed (5.12), and therefore $A_f=K$, which is clearly absurd. Hence $x_1$ is algebraic over $k$, hence $K$ (and therefore $B$) is a finite extension of $k$.

I am filling detail in given solution: If $n=1$ and $x_1$ is transcendental over $K$, then $B=k[x_1]\cong$ polynomial ring $k[X]$ in one indeterminate. Which contradict $B$ is a field.

Note $K=k(x_1)$ is smallest subfield of $B$ containing $A$, so $K\subseteq B$. We have $A\subseteq A_f\subseteq K$, first inclusion follows from natural homomorphism $A\to A_f$ is injective since $A$ is integral domain, and second from $K=(A-\{0\})^{-1}A$. So $B=A_f[x_2,…,x_n]$ is finitely generated $A_f$-algebra. Since $x_2,…,x_n$ are integral over $A_f$, we have $B$ is integral over $A_f$.

By this post, fraction field of $A_f$ is $K$. So $A_f$ is integrally closed by (5.12). Since $K$ is integral over $A_f$, we have $A_f=K$. In particular, $1/x_1=a/f^m$ for some $a\in A$ and $m\geq 0$. Then $x_1a-f^m=0$. Which contradict initial assumption of $x_1$ is transcendental over $k$.

Question: I am bit unsure with second and third paragraph of my filling detail. Is there an easier way to think of localization $A_f$ as subset of $B$? I think, we have $A_f\cong A[f^{-1}]$ and clearly $A[f^{-1}]\subseteq B$. In solution manual on net, proof of $A_f=K$ doesn’t look as straightforward as mine.

Answer 1

First, $B$ needs to be a finite $k$-algebra, but I think it is just a typo. I want to add some arguments to make things clearer.

For the case $n=1$, you do not need to proceed by contradiction. If $k[x_1]$ is a field, then $\frac{1}{x_1}$ is a polynomial in $x_1$. By multipying by $x_1$, we have that $x_1$ is a solution to a polynomial equation, thus is algebraic.

For the second paragraph, indeed $A_f$ is a subset of $B$, as $B$ is a field that contains $x_1,\dots,x_n$.

On your last paragraph, it is easy to see that the fraction field of $A_f$ is $K$ ; indeed you have the inclusions $A\subset A_f \subset K=\textrm{Frac}(A)$, hence $\textrm{Frac}(A_f)=K$.
The ring $A_f$ is indeed integrally closed, as $A$ is integrally closed. Then, as $K \subset B$ and $B$ is integral over $A_f$, we have that $K$ is integral over $A_f$. As $A_f$ is integrally closed, we have that $K=A_f$.

But then I think you are making a mistake. The polynomial in $x_1$ $a x_1 - f^m$ may be $0$, so that the equation $ax_1 - f^m=0$ says nothing about the algebraicity of $x_1$. Instead, you really want to show that it is not possible that $k[x_1][f^{-1}] = k(x_1)$ when $x_1$ is transcendental. Indeed, if $P_1,\dots,P_m$ are the prime factors of $f$, then a polynomial $P\in k[x_1]$ is invertible in $k[x_1][f^{-1}]$ if and only if its prime factors are $P_1,\dots,P_m$. But there is always infinitely irreducible elements ; if $k$ is infinite, then the $X-a, \ a\in k$ are irreducible. If $k$ is a finite field, the theory of finite field shows that there is infinitely many primes in $k[x]$.