How to show rational function field of an affine subvariety with dim>0 is not algebraically closed?

Question

I do not know how to show the following statement.

If $X\subset A^n$ is an irreducible subvariety, $\dim X>0$, then the rational function field of $X$, $K(X)$ is not algebraic closed.

What I feel is like following: If $K(X)$ is alg closed, then I pick up some none constant rational function f on X, f $\in K(X)$, so it has n$^{th}$ root $g_n$for all n, and $g_n$ is also a rational function. so $f^n$=$g_n$ and after reduction of fractions, each gives an equation holds on X and for n to be different prime p, since A(X) is finitely generated. This seems like a contradiction.

But the above does not look like a proof. Can anyone give a clean and neat proof of it?

Thanks!

Answer 1

Take some $f\in k[X]$ that is neither zero nor a unit. We will show that we cannot take arbitrary roots of $f$ in $k(X)$.

Here is a sketch of the proof for curves:

Because $f$ is not a unit, there is some (regular) $p\in X$ with $f(p) = 0$. Because $f$ is not zero, it has some finite order of vanishing $n$ at $p$. But then $f$ cannot be an $(n+1)$-th power of a rational function, i.e. $T^{n+1}-f$ has no root in $k(X)$.

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The above is a decent intuitive argument, but I think it's problematic to talk about the order of vanishing at an arbitrary point when $X$ has dimension $>1$. We should pick the (scheme-theoretic) generic point of a codimension $1$ subvariety.

Here is a more detailed version, that works in all dimensions:

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First, we may assume $X$ is smooth—otherwise, restrict to an affine open away from the singular locus, which does not change the function field.

By Krull's Principal Ideal Theorem, $f$ is contained in a height $1$ prime $P \subset k[X]$. The localization $k[X]_P$ is a regular local ring of dimension $1$, so it is a DVR.

Suppose there exists some $g\in K(X)$ with $g^k = f$. Taking valuations, we see that $k\nu_P (g) = \nu_P (f)$ is a multiple of $k$.

Since $\nu_P (f)>0$, this is only possible for finitely many $k$, so there are infinitely many $k$ for which $T^k - f$ has no solution in $K(X)$ (for example, $k=n+1$).

Answer 2

A different route (independent of the dimension) is to start by recalling Emmy Noether's normalization lemma. It implies that the function field $K(X)$ is a finitely generated algebraic extension of the field of rational functions $L=K(y_1,y_2,\ldots,y_d)$ for some algebraically independent elements $y_1,y_2,\ldots,y_d, d\le n$ that form a transcendence basis of $K(X)/K$. Here $d=\dim X$, but that isn't actually needed in what follows.

A finitely generated algebraic extension is always finite, so we can conclude that $[K(X):L]=N$ for some positive integer $N(<\infty)$. On the other hand the field $L$ has algebraic extensions of degrees $>N$. For example $L'=L(y_1^{1/(N+1)})$ is an extension field of degree $[L':L]=N+1$. If the field $K(X)$ were algebraically closed, it should contain the field $L'$, but we just saw that this cannot be the case.

Therefore $K(X)$ cannot be algebraically closed.