Past exam of the galois theory course of the University of Oxford / 2023 / 3

A nonconstant polynomial is separable when it has distinct roots in a splitting field, equivalently when it is coprime to its formal derivative. For such $f$, Bézout's identity for polynomials gives $a,b\in F[T]$ with $af+b{f}^{\prime }=1$. This identity remains valid over any extension $E/F$. A repeated root $\gamma \in E$ would satisfy $f(\gamma )=f^{\prime }(\gamma )=0$, giving the contradiction $1=0$. Thus there are no repeated roots in any field extension. The argument in fact applies to every separable polynomial, whether irreducible or not.

For nonconstant irreducible $f$, its greatest common divisor with $f^{\prime }$ is either $1$ or $f$. Since $\mathrm{deg}{f}^{\prime }<\mathrm{deg}f$ when $f^{\prime }\ne 0$, inseparability is equivalent to $f^{\prime }=0$. In characteristic $p$,exactly when $a_j=0$ for every exponent $j$ not divisible by $p$. ThusNo assumption that the coefficient field is perfect is required: vanishing derivative concerns the exponents, not whether the coefficients have $p$th roots in $F$.

Let $E=F({\alpha }^p)\subseteq F(\alpha )$. If $\alpha$ is a separable algebraic element over $F$, its minimal polynomial over $E$ divides the original separable minimal polynomial. It also divides $T^p-{\alpha }^p=(T-\alpha {)}^p$, which has only one distinct root in any extension. Being separable with only one distinct root, that minimal polynomial has degree one. Hence $\alpha \in E$ and $F(\alpha )=F({\alpha }^p)$.

Conversely, if the degree-$n$ minimal polynomial $m$ over $F$ is inseparable, part (b) writes it as $m(T)=g(T^p)$ with $\mathrm{deg}g=n/p$. Then $g({\alpha }^p)=0$, soThe fields cannot be equal. This proves the equivalencethe separability criterion using a pth power.

Put $\gamma =\alpha +\beta$, $E=F(\gamma )$ and $E_0=F({\gamma }^p)$. By the preceding separability criterion using a pth power, it suffices to prove $E=E_0$.

Since $\alpha$ is separable, $\alpha \in F({\alpha }^p)$. But ${\alpha }^p={\gamma }^p-{\beta }^p\in E_0(\beta )$, so $\alpha \in E_0(\beta )$. ConsequentlyLet $h$ and $m$ be the monic minimal polynomials of $\beta$ over $E_0$ and $E$, respectively. Both divide its separable minimal polynomial over $F$, so both are separable. Also $m\mid h$ in $E[T]$.

Every $c\in E=F(\gamma )$ has $c^p\in F({\gamma }^p)=E_0$. Hence $m(T{)}^p$ has coefficients in $E_0$ and vanishes at $\beta$, forcing $h\mid m^p$ in $E_0[T]$. Since $h$ remains a square-free polynomial over $E$, this divisibility implies $h\mid m$ in $E[T]$. Together with $m\mid h$, monicity gives $h=m$. ThereforeThe tower law makes $[E:E_0]=1$. Part (c) now proves that $\gamma =\alpha +\beta$ is separable over $F$. This establishes closure of separable algebraic elements under addition without presupposing that closure as a field theorem.