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

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:

Solved by gpt-5.6-sol high.