Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 18G (Galois Theory)

The tower law for field extensions states that for finite field extensionsone hasTo prove it, let $(u_i{)}_{i=1}^m$ be a $K$-basis of $E$ and $(v_j{)}_{j=1}^n$ an $E$-basis of $L$. Every $x\in L$ can be written as ${\sum }_j{a}_j{v}_j$ with $a_j\in E$, and expanding each $a_j$ in the basis $(u_i)$ shows that the $mn$ products $u_i{v}_j$ span $L$ over $K$.

Ifthen independence of the $v_j$ over $E$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$, and independence of the $u_i$ over $K$ gives every $c_{ij}=0$. Thus $(u_i{v}_j)$ is a $K$-basis of $L$, proving the formula.

A splitting field of $f\in K[X]$ is an extension $L/K$ in whichfor roots ${\alpha }_r\in L$, and which is generated by those roots:The generation condition is the required minimality: no proper intermediate field containing $K$ can still contain every root.

Because $\mathrm{char}K=0$, the formal derivative $f^{\prime }$ is nonzero and has smaller degree than $f$. Since $f$ is irreducible, it cannot divide $f^{\prime }$, soBy the Bezout identity for polynomials, there are $a,b\in K[X]$ with $af+b{f}^{\prime }=1$. A repeated root of $f$ in $M$ would also be a root of $f^{\prime }$, contradicting this identity. Thus the result that an irreducible polynomial in characteristic zero is separable gives

Write $q=p^r$, where $p=\mathrm{char}K$. Forthe derivative isso $h$ is a separable polynomial and all its roots in its splitting field $L$ are distinct.

Let $R$ be the set of these roots. The Frobenius endomorphism and the identitiesshow that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne 0$, inversion. Thus $R$ is a finite field. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ andHence the polynomial characterization of a finite field gives

Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an algebraic closure. The minimal polynomial of an algebraic element givesso $K(\alpha )$ has $q^n$ elements. Its nonzero elements form a finite group under multiplication, of order $q^n-1$; hence ${\alpha }^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the irreducible factors of a finite-field Frobenius polynomial satisfy

For any prime $p$, takeThe Eisenstein criterion, applied at the prime element $t$ of ${\mathbb{F}}_p[t]$, shows that $f$ is irreducible over $K$ and therefore has degree $p$. If $\alpha$ is a root in any extension field, the Frobenius endomorphism givesIt follows that this purely inseparable polynomial over a rational function field has at most one distinct root in every extension, and that root has multiplicity $p$, as required.