Let $X$ be a scheme locally of finite type over an arbitrary field $k$. I want to prove that the residue field at any non-closed point of $X$ is not algebraically closed.
We know that under our assumptions (from e.g. Qing Liu's book), a point closed in an affine open subscheme is closed in the entire scheme. Since stalks (and therefore residue fields) can be computed inside any open subscheme, the question is local, so we may assume that $X$ is the spectrum of a finitely generated $k$-algebra.
For a finitely generated $k$-algebra, the residue field at any prime ideal is a finitely generated field extension of $k$. Since our point is not closed, it is not a finite extension. A finitely generated non-finite field extension has to be transcendental.
Pick a transcendental element $x$. If the extension were to be algebraically closed, then $x^{\frac{1}{p}}$ belongs to it for any prime number $p$. Only finitely many elements of this form can enter our minimal generating set, and multiplication/division of elements $x^{\frac{n_i}{p_i}}$ with irreducible fraction in the power can not produce an element $x^{\frac{n}{p}}$ with $p>p_i$. We also can not have a rational expression $\frac{f(x^{\frac{n_i}{p_i}})}{g(x^{\frac{n_i}{p_i}})}=x^{\frac{n}{p}}$ because we can clear denominators and reduce to previous case. Since there are infinitely many primes, we win.
- Is my argument correct?
- In general, for what base scheme $S$ is it true that a scheme locally of finite type over $S$ the residue field at any non-closed point is not algebraically closed?
Best Answer
I just rewrite your idea. It should be correct.
If I well remember, for $k$-schemes $X$ locally of finite type, the set of closed points is exactly $X(\bar{k})$ ($\bar{k}$ being an algebraic closure of $k$).
So if $x \in X$ is not closed, then $k(x)/k$ is a finite type field extension of $k$, not algebraic. Then there is $(T_1,\cdots,T_n)$ algebraic independent elements of $k(x)$ s.t. $k(x)/F$ ($F=k(T_1, \cdots,T_n)$) is finite (its like the Noether normalisation lemma but easier for fields and transcendantal dimension).
If $\operatorname{char}(k)$ doesn't divide $m$, then $F(T_1^{1/m})/F$ is of degree $m$, so for $m$sufficiently large, $F(T_1^{1/m})$ is not in $k(x)$, so $k(x)$ is not algebraically closed.