Decomposition of a finite extension of $k(x_1,...,x_n) $ into separable/inseparable parts

127 Views Asked by At

I'm reading Janos Kolloar's Lecture on Resolution of Singularities and have an question about an argument used in the proof of Theorem 1.33 (page 25):

THEOREM 1.33 Let $S$ be an integral domain that is finitely generated over a field $k$, and let $F \subset Frac(S)$ be a finite field extensionof it's quotient field. Then the normalization of $S$ in $F$ is finite over $S$.

the proof strategy is: By Noether normalization theorem $S$ is finite over $R=k[x_1,...,x_n]$ thus it suffuce to consider the finite field extension $k(x_1,...,x_n) \subset F$.

Proof. [...] $F$ is a finite extension of $k(x_1,...,x_n)$ so there is a finite purely inseparable extension $E/k(x_1,...,x_n)$ such that $EF/E$ is separable. (???) Every finite purely inseparable extension of $k(x_1,...,x_n)$ is contained in a field $k'(x_1^{p^{-m}},...,x_n^{p^{-m}})$, where $k'/k$ is finite and purely inseparable (???) [...]

Questions:

Q_1: Why the author talks explicitly about finite purely inseparable extension $E/k(x_1,...,x_n)$ such that $EF/E$ is separable? Why he not simply says that $F$ contains an intermediate field $E$ such that extension $E/k(x_1,...,x_n)$ is finite purely inseparable and $F/E$ is separable? Or is this in general wrong claim? That is why we need the field $EF$ in the game?

Q_2: Why every finite purely inseparable extension of $k(x_1,...,x_n)$ is contained in a field $k'(x_1^{p^{-m}},...,x_n^{p^{-m}})$, where $k'/k$ is finite and purely inseparable?