= Solution
Let $y_1,\ldots,y_r$ 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. Then
$$
T^p-y_i=(T-x_i)^p,
$$
so the roots of
$$
f(T)=g(T^p)
$$
are precisely the $x_i$. Hence
$$
M=L(x_1,\ldots,x_r).
$$
But $y_i=x_i^p$, so $L=K(y_1,\ldots,y_r)\subseteq K(x_1,\ldots,x_r)$. It follows that
$$
M=K(x_1,\ldots,x_r),
$$
which 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 $\sigma\in\operatorname{Aut}(L/K)$. The <extension count for field embeddings>[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 $x_i$,
$$
\tau(x_i)^p=\sigma(y_i),
$$
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>.
Back to article page
