Past exam of the number theory course of the University of Oxford / 2011 / A (Galois theory)

A splitting field of a nonconstant polynomial $f$ over a field $K$ is an field extension $L$ in which $f$ factors into linear factors and which is generated over $K$ by those roots. We prove existence and the divisibility simultaneously by induction on $n=\mathrm{deg}f$; constant nonzero polynomials have splitting field $K$.

If $f$ is reducible, write $f=gh$ with positive degrees $r,s<n$. By induction first split $g$, obtaining $E/K$ of degree dividing $r!$, then split $h$ over $E$, obtaining $L/E$ of degree dividing $s!$. The tower law for field extensions gives $[L:K]\mid r!s!\mid (r+s)!$. All adjoined elements are roots of $f$, so $L$ has the required minimality.

If $f$ is irreducible, take $E=K[x]/(f)$ and let $\alpha$ be the image of $x$. Then $[E:K]=n$ and $f=(x-\alpha )h$ in $E[x]$. Apply induction to $h$, of degree $n-1$, to obtain $[L:E]\mid (n-1)!$. ThereforeThis proof also covers inseparable polynomials; it does not incorrectly identify extension degree with automorphism count in that case.

For a splitting field $L$ of $f$ over $K$, define $\mathrm{Gal}(f)={\mathrm{Aut}}_K(L)$. Each automorphism permutes the distinct roots, since it fixes the coefficients. An automorphism fixing every root fixes the field they generate, so this action is faithful. If there are $r\le n$ distinct roots, it gives an embedding into $S_r$, and hence into $S_n$ by fixing $n-r$ additional symbols.

Factorization over the rational numbers givesThe rational roots add nothing to the splitting field. By Eisenstein's criterion at $7$, $x^3-7$ is irreducible. Put $a=\sqrt[3]{7}$ and $\zeta =e^{2\pi i/3}$. Its splitting field is $L=\mathbb{Q}(a,\zeta )$. The real field $\mathbb{Q}(a)$ has degree $3$; the nonreal element $\zeta$ has degree $2$ over it. Thus $[L:\mathbb{Q}]=6$, and its faithful action on the three nonrational roots identifies

The polynomial is irreducible by Eisenstein's criterion at $2$. Its Galois group therefore acts transitively on five roots, so its order is divisible by $5$ by the orbit-stabilizer theorem.

To count the real roots, differentiate:It is positive on $(-\infty ,0)$, negative on $(0,\sqrt[3]{4})$, and positive thereafter. The local maximum is $f(0)=2>0$ and the local minimum is $2-6(\sqrt[3]{4}{)}^2<0$. The limits at infinity have opposite signs, so there are exactly three real roots and one conjugate pair of nonreal roots. Complex conjugation fixes the three real roots and interchanges the other two; its permutation is a transposition. The prime-degree subgroup result supplied in the paper now gives

Over the finite field ${\mathbb{F}}_7$,The nonzero squares modulo $7$ are $1,2,4$, so $-1$ is not a square. Adjoining a root of $x^2+1$ produces ${\mathbb{F}}_{49}$ and already gives both roots. Its Frobenius automorphism has order $2$ and interchanges them, whence

Over the finite field ${\mathbb{F}}_{13}$,The discriminant of the quadratic is $9+24\equiv 7(\mathrm{mod}13)$. The nonzero squares are $1,3,4,9,10,12$, so the quadratic is irreducible. Its two roots lie in ${\mathbb{F}}_{13^2}$, with the Frobenius automorphism interchanging them. Hence

The fixed field isIt is a subfield because every field automorphism preserves sums, products and inverses.

Set $F=K^G$. If $[K:F]$ is infinite the desired finite lower bound is immediate. Otherwise let $b_1,\dots ,b_d$ be an $F$-basis of $K$. If $|G|>d$, the homogeneous system over $K$has a nonzero solution. For $a={\sum }_j{a}_j{b}_j$, with $a_j\in F$, every $\sigma$ fixes $a_j$, so the same coefficients give ${\sum }_{\sigma }{c}_{\sigma }\sigma (a)=0$. This contradicts linear independence of distinct field embeddings. Consequently

The Fundamental theorem of Galois theory says that for a finite Galois extension $L/F$ with group $G$, the maps $H\mapsto L^H$ and $E\mapsto \mathrm{Gal}(L/E)$ are mutually inverse inclusion-reversing correspondences between subgroups and intermediate fields. They satisfy $[L:L^H]=|H|$ and $[L^H:F]=[G:H]$. An intermediate extension $E/F$ is Galois exactly when its subgroup is normal, in which case its group is the quotient by that subgroup.

Let $a=\sqrt{2}$, $b=\sqrt{3}$, $c=i$. The real field $\mathbb{Q}(a,b)$ has degree $4$: if $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$, squaring forces $uv=0$ and either $u^2=3$ or $v^2=3/2$, both impossible. Adjoining the nonreal $i$ doubles the degree. All three square-root signs can be changed independently, givingWrite ${\sigma }_{uvw}$ for the automorphism sending $(a,b,c)$ to $((-1{)}^{u}a,(-1{)}^{v}b,(-1{)}^{w}c)$, where $(u,v,w)\in {\mathbb{F}}_2^3$. Every nonzero vector generates an order-two subgroup, so there are $7$ such subgroups. Their fixed fields are:

Each listed field has degree $4$ and is fixed by the indicated sign change, which proves equality with the fixed field.

There are also $7$ order-four subgroups: they are the kernels of the seven nonzero linear functionals on ${\mathbb{F}}_2^3$. For $e=(e_1,e_2,e_3)\ne 0$, define $H_e=\{{\sigma }_{uvw}:e_{1}u+e_{2}v+e_{3}w=0\}$. ThenThus the seven quadratic fields have radicands $2,3,-1,6,-2,-3,-6$. Their square classes are distinct, so all are different. Finally the trivial subgroup fixes $L$, and all of $G$ fixes $\mathbb{Q}$. This accounts for every subgroup and intermediate field.

The Fundamental theorem of symmetric polynomials states that every symmetric polynomial in $n$ variables over a field is a polynomial in the elementary symmetric polynomials, uniquely so. For the three roots, Vieta's formulas give $e_1=0$, $e_2=p$, $e_3=-q$.

The squared product of root differences is symmetric. Expansion in elementary symmetric polynomials givesOne can verify the displayed identity simply by substituting the definitions of $e_1,e_2,e_3$ and expanding; it is the cubic polynomial discriminant.

If the cubic is irreducible, its roots are distinct in characteristic zero, so $\delta \ne 0$. Its Galois group is a transitive subgroup of $S_3$, hence either $A_3$ or $S_3$. A permutation of the roots multiplies $\delta$ by its sign. The fixed-field assertion of the Fundamental theorem of Galois theory therefore gives $\delta \in \mathbb{Q}$ exactly when all group elements are even. ThusFor example, $f=x^3-3x-1$ has no rational root: neither $1$ nor $-1$ is a root, and these are the only possibilities by the rational root theorem. Its discriminant is $108-27=81$, so $\delta =\pm 9\in \mathbb{Q}$ and its group is $A_3$.

A finite solvable group is a group admitting a finite chain of normal subgroups with abelian successive quotients; equivalently its derived series eventually becomes trivial. The chainhas quotients $C_2,C_3,V_4$, all abelian. Thus $S_4$ is solvable. Every subgroup of a solvable group is solvable, because intersecting the chain with the subgroup gives factors isomorphic to subgroups of the original abelian factors. A Galois group of a polynomial of degree at most $4$ embeds in $S_4$ by its action on the roots, so it is solvable even if the polynomial is reducible or has repeated roots.

For the substitution, the characteristic-zero solvability by radicals theorem says that a polynomial has solvable Galois group if and only if its splitting field is contained in a finite tower formed by adjoining successive roots of elements of the preceding field. First place all roots $\alpha$ of $f$ in such a tower. Adjoin a primitive $m$th root of unity and, for every nonzero $\alpha$, an $m$th root ${\beta }_{\alpha }$. These are radical adjunctions, and all roots of $x^m-\alpha$ are ${\beta }_{\alpha }{\zeta }_m^j$. Zero roots require no adjunction. The resulting radical tower therefore contains the splitting field of $f(x^m)$. The theorem implies