Splitting field of a polynomial obtained by Frobenius substitution

Let $g\in K[T]$ in characteristic $p$, let $f(T)=g(T^p)$, let $L$ split $g$ over $K$, and let $M$ split $f$ over $L$. If $y_i$ are the roots of $g$ and $x_i^p=y_i$, then the Frobenius endomorphism makes each $x_i$ unique andbecause $y_i=x_i^p$. Thus $M$ is already the splitting field of $f$ over $K$, and every root of $f$ is purely inseparable over $L$.