Past exam of the number theory course of the University of Oxford / 2009 / A / 2

A splitting field is an extension in which the polynomial factors into linear factors and which is generated over the base field by its roots. Prove existence together with the factorial divisibility by strong induction on $d$, with degrees zero and one immediate for a nonzero polynomial.

If $f$ is reducible, write $f=gh$ with $1\le e=\mathrm{deg}g<d$, $\mathrm{deg}h=d-e$. By induction first split $g$ in $K_1/K$ with degree dividing $e!$, then split $h$ over $K_1$ with degree dividing $(d-e)!$. The resulting field is generated over $K$ by all roots of $f$, hence is a splitting field. Its degree divides $e!(d-e)!$, which divides $d!$ because the binomial coefficient $d!/[e!(d-e)!]$ is an integer.

If $f$ is irreducible, adjoin a root $\alpha$, obtaining degree $[K(\alpha ):K]=d$. Over this field divide out $X-\alpha$ and split the degree-$(d-1)$ quotient by induction. That splitting field together with the base already contains every root and has degree over $K$ dividing $d(d-1)!=d!$. ThusThe reducible/irreducible distinction proves divisibility, not just an upper bound on the degree, and works without assuming separability.

In the distinct-root convention, a polynomial is separable if it has no repeated roots in its splitting field, equivalently $\gcd (f,f^{\prime })=1$. A common factorwise convention instead requires each distinct irreducible factor to have no repeated roots; repeated occurrences of an already separable factor do not affect the splitting field. Under either separability convention, the splitting field is a finite normal separable extension andFor failure in the inseparable case, take $K={\mathbb{F}}_p(t)$ and $f=X^p-t$. It is irreducible, for example by Eisenstein at the prime $t$ in ${\mathbb{F}}_p[t]$. Its splitting field is $L=K(t^{1/p})$ of degree $p$, but the polynomial has just one distinct root. Every $K$-automorphism fixes that root, so the Galois group is trivial rather than having order $p$.

Factor $f_1=(X^2-3)(X^2+2)$. Its splitting field is $L=\mathbb{Q}(\sqrt{3},\sqrt{-2})$. The first quadratic field is real, so it cannot contain the nonreal $\sqrt{-2}$. Thus adjoining the second square root doubles the degree again. Independent sign changes of the two square roots give all four automorphisms, so

Eisenstein's criterion at two makes $X^5-4X+2$ irreducible: all nonleading coefficients are divisible by two and the constant is not divisible by four. Its Galois group is therefore transitive on five roots, and its order is divisible by five, so Cauchy's theorem for finite groups supplies a five-cycle.

The derivative is $5X^4-4$, with real critical points $\pm c$, $c=(4/5{)}^{1/4}$. The values are $f(-c)=2+16c/5>0$ and $f(c)=2-16c/5<0$. The derivative is positive outside these two points and negative between them, so there are exactly three real roots and one nonreal conjugate pair. Complex conjugation therefore acts as a single transposition.

A five-cycle together with any transposition generates $S_5$: conjugating that transposition by powers of the cycle gives edges joining labels with a fixed nonzero difference modulo five. These edges form a connected graph, and transpositions on a connected graph generate the symmetric group. HenceThe equality of degree and group order uses separability in characteristic zero.

The nonzero cubes modulo thirteen are $\{1,5,8,12\}$, which excludes three. Thus this cubic has no root in its base field and is irreducible. For a root $\alpha$, the extension ${\mathbb{F}}_{13}(\alpha )$ has degree three. The base field already contains the primitive cube root of unity $3$, since $3^3=1$ and $3\ne 1$. Hence the other two roots are $3\alpha$ and $9\alpha$, already in this extension, making it the full splitting field.

The Frobenius endomorphism satisfies ${\alpha }^{13}=\alpha ({\alpha }^3{)}^4=3\alpha$, so it cycles all three roots. Therefore