1

I want to show that $k(x)$ is not a finitely generated $k$-algebra. Here is my attempt:

Suppose by contradiction that $k(x)$ is a finitely generated $k$-algebra. Suppose $k[a_1,\dots,a_d]$ is a Noetherian Normalization. Since $k[a_1,\dots,a_d]\subseteq k(x)$ is an integral extension and $k(x)$ is a field, so as $k[a_1,\dots,a_d]$, By Zariski lemma, $k[a_1,\dots,a_d]$ is a finite field extension over $k$.

Claim: any element in $k(x)\backslash k$ is not algebraic over $k$. To see this, assume by contradiction that there exists some algebraic $\frac{f}{g}$ with $\deg(f)>0$ or $\deg(g)>0$

Let $h(z):=z^n+\dots+a_0\in k[z]$ be its minimal polynomial. ($a_0\neq 0$) This implies: $$ f^n+a_{n-1} f^{n-1} g+\dots+a_0 g^n=0$$ Contradiction.

By this claim, we know $k[a_1,\dots,a_d]=k$ and $k(x)$ is integral over $k$, absurd.

Am I correct? I am also curious about any easier way to prove this. Thanks!

Zoudelong
  • 1,721

2 Answers2

3

Here is a proof that uses less machinery, although that doesn't make it any better or worse.

Consider the $k$-subalgebra $A$ of $k(x)$ generated by the rational functions $p_1(x)/q_1(x), \ldots, p_n(x)/q_n(x)$. Any element $f$ of $A$ has the property that there exists $N$ such that $(q_1q_2\ldots q_n)^N f \in k[x]$.

I claim that $k(x)$ does not have the following property: there exists a polynomial $q$ such that for any $g \in k(x)$, there exists $N$ with $q^N g \in k[x]$. Given the claim, it follows that $A$ can't be $k(x)$ and so $k(x)$ can't be finitely generated.

Indeed, take infinitely many irreducible elements $\alpha_i$ of $k[x]$. Then such a polynomial $q$ would have to satisfy $q^{N_i} \cdot (1 / \alpha_i) \in k[x]$ for all $x$. But by unique factorization, this implies $\alpha_i$ divides $q$, which then is divisible by infinitely many irreducible elements, a contradiction.

hunter
  • 32,629
2

You don't need to use Noether normalization; this statement is already close to being equivalent to Zariski's lemma. Zariski's lemma can be stated in the following form: if $K$ is a finitely generated field extension of $k$, then $K$ is algebraic (since $K$ is finitely generated this is equivalent to $K$ being finite). Taking the contrapositive, Zariski's lemma is equivalent to the statement that if $K$ contains an element transcendental over $k$, then $K$ cannot be finitely generated.

You can see here for a "cheap proof" of this result when $k$ is uncountable, as well as a link to a discussion on MO of how to deduce the general case from this case.

Qiaochu Yuan
  • 468,795