Let $X$ be a $K$scheme of finite type over a field $K$, let $L$ be an extension field of $K$, let $X_L := L \times_K X$, and let $p:X_L \rightarrow X$ be the projection. For each $x \in X_L$ we get an extension $k(x) \supset k(p(x))$. Can I choose $L$ such that $k(x)$ contains an algebraic closure of $k(p(x))$? What about $L$ being the algebraic closure of $K$? All this holds obviously in case $x$ and $p(x)$ are closed, but I don't know about the general situation.

$\begingroup$ Why did I get a 1? $\endgroup$ – Georg S. Sep 5 '13 at 18:12
I think the answer depends on what you mean by "contains".
Suppose $K = \mathbb{Q}$, $X = \text{Spec }\mathbb{Q}[t]$. For any extension field $L$, $X_L = \text{Spec }L[t]$. Let $x\in X_L$ be the generic point, which must map to the generic point of $X$. Here, $k(x) = L(t)$, and $k(p(x)) = K(t)$. So of course if $L = \overline{K(t)}$, then you could say that $k(x)$ contains $k(p(x))$, but this containment isn't compatible with the natural extension of residue fields induced from the map $p$.
If you allow all these containments, then you can even pick a single $L$ (at least in the case of char 0) that will work for all $x$ (assuming $K$ isn't too big). Namely, set $L = \mathbb{C}$, and simply note that $\mathbb{C}$ contains every field of characteristic zero who's transcendence degree over $\mathbb{Q}$ is at most uncountable.
The example above easily extends to the general affine case, and since your question is local, it extends to all schemes of finite type over $K$.

$\begingroup$ You are of course right about the containment! I probably want the embedding $k(p(x)) \hookrightarrow k(x)$ to factor through an algebraic closure of $k(p(x))$. Is this possible? $\endgroup$ – Georg S. Sep 5 '13 at 20:27

$\begingroup$ It's not (except maybe in some trivial situations like when your scheme is Spec $K$ or something). Basically, you should ask yourself, where can the $t$ go? The natural embedding $K(t) = k(p(x))\hookrightarrow k(x) L(t)$ has to send the coordinate $t$ to, essentially, $t$. Any such embedding $k(p(x)) \hookrightarrow k(x)$ coming from a change of base morphism $p$ will essentially only be realized on the coefficient field. The variables (in our case $t$) will still be mapped to $t$, and is necessarily algebraically independent from the elements of the coefficient field. $\endgroup$ – Will Chen Sep 5 '13 at 22:33

$\begingroup$ If you want something more formal, we could argue as follows. A scheme of finite type over $K$ is given locally by a finitely generated $K$algebra. Ie, something like $K[x_1,...,x_n]/(f_1,...,f_m)$. Consider the generic point of some irreducible component. This point has residue field equal to the function field of the irreducible component, which is just the fraction field of some integral finitely generated $K$algebra. These fraction fields basically look like $K(x_1,...,x_n)$, where the $x_i$'s may satisfy some equation with coefficients in $K$. $\endgroup$ – Will Chen Sep 5 '13 at 22:44

$\begingroup$ extending the base from $K$ to any extension field $L$ will result in the residue field of that point becoming $L(x_1,..,x_n)$, where the $x_i$'s again satisfy the same relation. The residue field embedding then is just defined by sending $x_i\mapsto x_i$, and sending $K\hookrightarrow L$. This residue field extension can't possibly factor through the algebraic closure of $K(x_1,...,x_n)$. Why? Well, first note that $K(x_1,...,x_n)$ cannot be algebraically closed (this follows from the fact that it's the fraction field of a FINITELY generated $K$algebra). $\endgroup$ – Will Chen Sep 5 '13 at 22:47

$\begingroup$ In particular, using the same argument you can show that there is some integer $n$ for which $X^n  x_1$ has no root in $K(x_1,...,x_n)$. Now you just have to convince yourself that if $X^n  x_1$ had a root in $L(x_1,...,x_n)$, then it would have to have a root in $K(x_1,...,x_n)$. This should be pretty clear from our setup. $\endgroup$ – Will Chen Sep 5 '13 at 22:54