Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 18I / c / Solution

Let $y_1,\dots ,y_r$ be the distinct roots of $g$ in its splitting field $L$. For each $i$, let $x_i$ be the unique $p$th root of $y_i$ in an algebraic closure. Thenso the roots ofare precisely the $x_i$. HenceBut $y_i=x_i^p$, so $L=K(y_1,\dots ,y_r)\subseteq K(x_1,\dots ,x_r)$. It follows thatwhich proves that $M$ is also the splitting field of $f$ over $K$, as in the splitting field of a polynomial obtained by Frobenius substitution.

Moreover $x_i^p=y_i\in L$, so every root of $f$ is purely inseparable over $L$. In fact $M/L$ is purely inseparable.

Now take $\sigma \in \mathrm{Aut}(L/K)$. The extension theorem for field embeddings extends $\sigma$ to a $K$-embedding of $M$ into an algebraic closure. Since $M$ is a splitting field of $f$ over $K$, its image is again $M$, so this extension is an automorphism $\tau$ of $M$. It is unique: for every root $x_i$,and the Frobenius endomorphism gives only one possible $p$th root. Since the $x_i$ generate $M$ over $L$, $\tau$ is uniquely determined. This is the unique embedding extension through a purely inseparable extension.