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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.

Assume now that $g$ is irreducible and separable. Its splitting field $L/K$ is Galois, and its Galois group acts transitively on the roots $y_i$ by the Galois group of a polynomial. Part (c) extends every such automorphism uniquely to $M$. If $x_i^p=y_i$, the extension sends $x_i$ to the unique $p$th root above the image of $y_i$. The transitivity lifted through unique pth roots therefore shows thatacts transitively on the roots of $f$.

Every automorphism of $M/K$ preserves $L$, because $L$ is the splitting field of $g$ over $K$. Transitivity therefore implies that either all roots $x_i$ lie in $L$ or none does. Since $x_i^p\in L$, the minimal polynomial of a purely inseparable element has degree either one or $p$. Consequently,

Let $h\in K[T]$ be a monic irreducible factor of $f$. If $h$ is inseparable, part (a) givesfor a nonconstant monic $q\in K[T]$. Since $q(T^p)$ divides $g(T^p)$, polynomial division in $K[U]$ and substitution $U=T^p$ show that $q$ divides $g$. Irreducibility of $g$ forces $q=g$, hence $h=f$. Thereforewhich is the irreducible factors after Frobenius substitution dichotomy.

If every coefficient of $g$ is a $p$th power, saythenso $f$ is reducible.

Conversely, suppose $f$ is reducible and factor it into distinct monic irreducibles:Every $h_i$ is then a proper factor, hence separable by the preceding dichotomy. Since $f^{\prime }=0$, unique factorization and $h_i^{\prime }\ne 0$ force $p\mid e_i$ for every $i$. Thus $f=q^p$ for some monic $q\in K[T]$. The Frobenius endomorphism raises coefficients to their $p$th powers, and the coefficient of $T^{ip}$ in $f=g(T^p)$ is the coefficient of $T^i$ in $g$. Hence every coefficient of $g$ is a $p$th power in $K$. We have proved the reducibility criterion after Frobenius substitution:

A field is perfect when every algebraic extension is separable. In positive characteristic $p$, this is equivalent to surjectivity of the Frobenius endomorphism $x\mapsto x^p$; every finite field is perfect.

For a prime number $p$ and integer $r\ge 1$, every interior binomial coefficient in row $p^r$ is divisible by $p$. The polynomial identity $(1+x{)}^p\equiv 1+x^p(\mathrm{mod}p)$ iterates under the Frobenius endomorphism to $(1+x{)}^{p^r}\equiv 1+x^{p^r}(\mathrm{mod}p)$. This concerns divisibility by $p$, not necessarily by $p^r$.

Coefficientwise reduction defines a ring homomorphismIt preserves sums, products, and divisibility. The Frobenius endomorphism gives $\overline{g(X^p)}=\bar{g}(X{)}^p$.

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 $L$.