Past exam of the number theory course of the University of Oxford / 2009 / A (Galois Theory)

The alternating polynomial given by the Vandermonde product is alternating: interchanging two variables changes its sign. One proof writes it as a Vandermonde determinant, up to a fixed sign, and interchanging two variables interchanges the corresponding determinant rows. Since transpositions generate $S_n$,Thus every even permutation fixes $\delta$, while a transposition negates the nonzero formal polynomial, so it is not invariant under all of $S_n$. This uses $n\ge 2$ and characteristic different from two, as in the subsequent rational setting. For $n=1$ both groups are trivial and the last distinction does not arise.

The discriminant is a symmetric polynomial in the four roots, so the Fundamental theorem of symmetric polynomials expresses it polynomially in the elementary symmetric polynomial. For this depressed quartic they are $e_1=e_2=0$, $e_3=-r$, $e_4=s$. Scaling each root by $t$ multiplies the discriminant by $t^{12}$, while $r,s$ scale by $t^3,t^4$. Thus only weighted-degree-twelve monomials can appear, giving ${\Delta }_f=A{r}^4+B{s}^3$.

For $r=0$, use $f^{\prime }=4X^3$ and the monic discriminant identity ${\Delta }_f=(-1{)}^{n(n-1)/2}{\prod }_i{f}^{\prime }({\alpha }_i)$. At $n=4$ this gives $4^4({\prod }_i{\alpha }_i{)}^3=256s^3$, so $B=256$. For $s=0$, the roots are zero together with those of $g=X^3+r$. The cubic's discriminant is $-\prod g^{\prime }({\beta }_i)=-27r^2$. The extra squared differences from zero multiply this by $(\prod {\beta }_i{)}^2=r^2$, giving $-27r^4$. HenceThese computations are polynomial identities, so they extend to parameter values with repeated roots as well.

For a polynomial with $n$ distinct roots, the criterion isWrite $\delta ={\prod }_{i<j}({\alpha }_i-{\alpha }_j)\ne 0$. Part (a) gives $\sigma \delta =\mathrm{sgn}(\sigma )\delta$. If every Galois permutation is even, $\delta$ is fixed by the entire Galois group and therefore belongs to its fixed field $\mathbb{Q}$. Thus ${\Delta }_f={\delta }^2$ is a rational square. Conversely, if ${\Delta }_f=q^2$ with $q\in {\mathbb{Q}}^{\times }$, then ${\delta }^2=q^2$ in a field implies $\delta =\pm q$. Every automorphism fixes this nonzero element, so its permutation sign must be one.

The distinct-root hypothesis is necessary in the statement's intended degree-$n$ permutation action. The source says “arbitrary degree” without stating that hypothesis. If repeated roots were allowed, zero discriminant would carry no sign information: $(X-1{)}^2(X^2-2)$ has discriminant zero, while its nontrivial automorphism swaps $\sqrt{2},-\sqrt{2}$ and fixes the two rational-root positions, an odd permutation. The criterion above applies to the squarefree polynomial, or to the squarefree part with its degree and root action adjusted accordingly.

Set $Y=X+1$. The polynomial becomes $Y^4+Y+1$, and translating all roots leaves their pairwise differences unchanged. Part (b) givesIt is nonzero and lies strictly between $15^2=225$ and $16^2=256$. An integer which is a rational square is an integer square, so this is not a rational square. The roots are distinct because the discriminant is nonzero. By (c), $\boxed{G_f⊈A_4}$.

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

A finite group is solvable if its derived series, defined by $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$, reaches the identity after finitely many steps. Equivalently it has a subnormal series with abelian factor groups. The factors need not all have prime order for this definition.

Let $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. This is an abelian subgroup, and conjugation preserves its nonidentity cycle type, so it is normal in $S_4$ and hence in $A_4$. The serieshas factors $V_4\cong C_2\times C_2$, $A_4/V_4\cong C_3$, and $S_4/A_4\cong C_2$, all abelian. Thus $\boxed{S_4\text{ is solvable}}$.

The rotations form a cyclic subgroup $R\cong C_n$ of index two, so it is normal. The quotient $D_{2n}/R$ has order two and is cyclic. Hence $1⊴R⊴D_{2n}$ has abelian factors, proving $\boxed{D_{2n}\text{ is solvable}}$. Its commutator subgroup is contained in the abelian rotation subgroup, so its second derived subgroup is already trivial.

Let $\alpha =p^{1/p}$ and let $\zeta$ be a primitive $p$th root of unity. The roots are $\alpha {\zeta }^j$, so $L=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the prime cyclotomic extension has degree $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. Both degrees divide $[L:\mathbb{Q}]$, and they are coprime. Conversely adjoining $\alpha$ to the cyclotomic field has degree at most $p$. These two bounds giveRestriction to $E=\mathbb{Q}(\zeta )$ gives a normal subgroup $N=\mathrm{Gal}(L/E)$ of order $p$, hence cyclic, with quotient $\mathrm{Gal}(E/\mathbb{Q})\cong C_{p-1}$. Restriction is onto because embeddings of the intermediate field extend in the finite Galois splitting field. Thus $1⊴N⊴G$ has abelian factors, proving $\boxed{G\text{ is solvable}}$. For $p=2$ the cyclotomic field is just $\mathbb{Q}$, and the same argument gives degree two and a cyclic group.