Factorizarion criterium to decide whether $K$ is the full constant field of $F/K$

Question

I took the following statement from an answer in math.stackexchange:

If $G(x,y)$ is an irreducible polynomial in $K[x,y]$, and $F=K(x,y)$ is an algebraic extension of $K(x)$ defined by $G(x,y)=0$, then $F\cap \overline{K}=K\Leftrightarrow G$ is absolutely irreducible (i.e. irreducible in $\overline{K}[x,y]$).

This looks useful, but I couldn't find a proof of it anywhere, nor could I prove it myself, so I thought I better ask.

I had the following idea: define $\widetilde{K}:=F\cap \overline{K}$. Since $ G(x,y)$ is irreducible in $K[x,y]$, then $[F:K(x)]=\deg_y(G(x,y))=:d$ (degree of $G$ with respect to $y$).

If $G=G_1...G_\ell$ is the factorization in $\overline{K}[x,y]$, then $[\widetilde{K}(x,y):\widetilde{K}(x)]=\deg_y(G_i(x,y))$ $=:\widetilde{d}$ for some $i$ (I couldn't prove this equality, but I believe it's true).

For $n:=[\widetilde{K}:K]$ and $m:=[\widetilde{K}(x,y):K(x,y)]$, we have the following diagram:

We have $dm=\widetilde{d}n$. Now:

$G$ is not absolutely irreducible $\Rightarrow \widetilde{d}<d\Rightarrow n>1$, i.e., $K\subsetneq\widetilde{K}$. (ok)

Conversely, $G$ absolutely irreducible $\Rightarrow d=\widetilde{d}\Rightarrow n=m$ (now what?).

Any ideas or corrections will be welcome. Thanks!

Answer 1

Let $F$ be the function field of a $k$-variety of dimension $n$ : a finite extension of $k(T),T=t_1,\ldots,t_n$.

Let $K = F \cap\overline{k}$ its field of constants.

Let $L/K$ a finite simple extension $L = K[z]/(h(z))$

$F \otimes_K L= F[z]/(h(z))$

It is a field iff $h(z)$ is irreducible in $F[z]$.

If it is not then $h(z) = u(z)v(z) \in F[z]\cap \overline{K}[z]=K[z]$ which is a contradiction.

Thus $F \otimes_K L= F[z]/(h(z))$ is a field.

Moreover $F[z]/(h(z))\cap \overline{K} = L$.

Proof : assume we are in characteristic $0$ so that every finite extension is simple, with $M=F[z]/(h(z)) \cap \overline{K}$, assuming it is larger than $L$, then $M/K$ is a finite thus simple extension ie. $M = K(a)$, let $H(w)\in K[w]$ be $a$'s minimal polynomial, by the preceding $H(w)$ is irreducible in $F[w]$ so that $F(a) = F[w]/(H(w))$ is a $F$-vector space of dimension $\deg H=[M:K]> [L:K]=\deg h$ which contradicts that $F(a) \subset F[z]/(h(z))$.

Not so sure of the proof in characteristic $p$.

Replacing $K$ by $L$ we obtain that for any finite extension $E/K$ then $F \otimes_K E$ is a field which implies $F\otimes_K \overline{K}$ is a field.

When $F = K(t_1)[s_1]/(G(t_1,s_2))$ such that $K = F \cap \overline{K}$ you get that $F\otimes_K \overline{K} = \overline{K}(t_1)[s_1]/(G(t_1,s_2))$ is a field ie. $G(t_1,s_2)$ is irreducible in $\overline{K}(t_1)[s_1]$ and hence in $\overline{K}[t_1,s_1]$.