Past exam of the number theory course of the University of Oxford / 2011 / A / 1

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