Galois theory course of the University of Oxford (B3.1)

Source: released examination PDF, pages 239–240. Solutions follow the numbering of the 14 June 2023 Galois Theory paper. Part 2(c) needs a characteristic restriction absent from the original PDF; the counterexample and corrected proof are given below. Part 2(b)'s Vandermonde formula uses monic normalization.

For $K=\mathbb{Q}(\zeta )$, every Galois automorphism sends the primitive root $\zeta$ to another primitive $m$th root of unity, uniquely ${\zeta }^a$ for $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$. The Cyclotomic character of a finite cyclotomic extension isIf ${\chi }_m(\sigma )=a$ and ${\chi }_m(\tau )=b$, then $(\sigma \tau )(\zeta )=\sigma ({\zeta }^b)={\zeta }^{ab}$, proving the group homomorphism property. Its kernel is trivial: an automorphism fixing $\zeta$ fixes the field it generates. Thus the character is injective. Here its domain is the finite cyclotomic Galois group; the corresponding map from an absolute Galois group need not be injective.

The element $\zeta$ satisfies $T^2-\eta T+1=0$, so $[\mathbb{Q}(\zeta ):\mathbb{Q}(\eta )]\le 2$. Since $\eta =\zeta +\overline{\zeta }$ is real, $\mathbb{Q}(\eta )\subseteq \mathbb{R}$. A primitive root of order $m\ge 3$ is nonreal: the only real roots of unity are $\pm 1$, of orders at most two. Hence $\zeta \notin \mathbb{Q}(\eta )$ andEquivalently, nontrivial complex conjugation fixes precisely the maximal real subfield of a cyclotomic field in this degree-two extension.

Irreducibility of the cyclotomic polynomial gives $[K:\mathbb{Q}]=\phi (15)=8$. The extension is Galois, since $K$ is the splitting field of $T^{15}-1$ in characteristic zero. The injection in part (a) is consequently an isomorphism, and the Chinese remainder theorem givesThe Fundamental theorem of Galois theory identifies proper intermediate fields strictly above $\mathbb{Q}$ with proper nontrivial subgroups. Such subgroups have order two or four. The two-torsion subgroup is $C_2\times C_2$, with three nonidentity elements and hence three order-two subgroups. There are also three order-four subgroups: they are the kernels of the three nonzero group homomorphisms $C_2\times C_4\to C_2$. Distinct such maps have distinct kernels because the target has only one nonidentity element. Therefore

Put $\omega ={\zeta }^5$, $\tau ={\zeta }^3$, $a=1+2\omega$, and $b=1+2(\tau +{\tau }^{-1})$. The primitive cubic and quintic root identities give $a^2=-3$, $b^2=5$. These elements therefore provide the following six generators:

The three quadratic fields are distinct because $-3,5,-15$ are distinct rational square classes. To justify the fifth entry, the two quadratic fields $\mathbb{Q}(a)$ and $\mathbb{Q}(b)$ are distinct, so their compositum has degree four. Moreover $a+b\ne 0$ andThus their sum alone generates the compositum.

The field $\mathbb{Q}(\tau )$ has degree four by the quintic cyclotomic polynomial. Its Galois group is $C_4$ and its unique quadratic subfield is $\mathbb{Q}(b)$, so it cannot contain $a$. It is therefore different from $\mathbb{Q}(a,b)$. The final field has degree four by part (b) and is real; the preceding two quartic fields contain nonreal $\tau$ and $a$, respectively. Thus all three quartics are distinct, and the count in part (i) proves the list exhaustive. This gives the subfields of the fifteenth cyclotomic field without relying on a choice of square-root signs.

The Galois group acts transitively on the $n$ roots because $f$ is irreducible and separable. The orbit-stabilizer theorem givesso $n$ divides $|G|$. Equivalently, the tower law gives $[K:F]=[K:F({\alpha }_1)]n$, while the finite Galois extension has group order $[K:F]$.

For the root-discriminant convention in the requested identity, defineThe Vandermonde determinant is $\det V(\alpha )={\prod }_{i<j}({\alpha }_j-{\alpha }_i)$. Squaring removes the signs, provingIt is nonzero because the roots are distinct. It lies in $F$, since its root expression is fixed by every Galois automorphism and $K^G=F$.

If $f$ is not monic and has leading coefficient $c$, the usual polynomial discriminant is $c^{2n-2}{\prod }_{i<j}({\alpha }_i-{\alpha }_j{)}^2$. Thus the printed Vandermonde equality uses the discriminant of the monic normalization $f/c$. The extra factor is itself a nonzero square in $F$, so it does not alter the square-class criterion in the next part.

First assume $\mathrm{char}F\ne 2$. Set $d=\det V(\alpha )\ne 0$. If ${\Delta }_f=b^2$ with $b\in F$, then $(d-b)(d+b)=0$ in the field $K$, so $d=\pm b\in F$. For an automorphism inducing the root permutation $\pi$,Since $\sigma$ fixes $d$ and $1\ne -1$ in $F$, every induced permutation has positive sign. The root action is faithful because the roots generate $K$, givingThis is the discriminant criterion for an alternating Galois group.

As printed for arbitrary fields, the assertion is false in characteristic two. Over ${\mathbb{F}}_2$, the polynomial $T^2+T+1$ is irreducible and separable, with roots $\alpha ,\alpha +1$ in ${\mathbb{F}}_4$. Their difference is one, so the discriminant is the square $1$. Nevertheless its Galois group is $C_2$, whereas $A_2$ is trivial. In characteristic two the permutation signs become indistinguishable as field elements; square discriminants in characteristic two therefore cannot detect the alternating subgroup in this way.

Denote the three stated pair sums by $r_1,r_2,r_3$. They are pairwise distinct: for example,and the other differences factor similarly. Every Galois automorphism fixes each $r_i\in F$, and hence fixes each of the three corresponding partitions of four roots into two pairs. The kernel of the permutation action $S_4\to S_3$ on these partitions isThus $G$ embeds in the Klein four-group. By part (a), $4$ divides $|G|$, and the containing group has order four. ConsequentlyThis Split resolvent criterion for a Klein four Galois group uses the cubic resolvent of a quartic. The pair-sum differences remain nonzero in characteristic two, so this proof does not need the missing characteristic assumption from part (c).

A nonconstant polynomial is separable when it has distinct roots in a splitting field, equivalently when it is coprime to its formal derivative. For such $f$, Bézout's identity for polynomials gives $a,b\in F[T]$ with $af+b{f}^{\prime }=1$. This identity remains valid over any extension $E/F$. A repeated root $\gamma \in E$ would satisfy $f(\gamma )=f^{\prime }(\gamma )=0$, giving the contradiction $1=0$. Thus there are no repeated roots in any field extension. The argument in fact applies to every separable polynomial, whether irreducible or not.

For nonconstant irreducible $f$, its greatest common divisor with $f^{\prime }$ is either $1$ or $f$. Since $\mathrm{deg}{f}^{\prime }<\mathrm{deg}f$ when $f^{\prime }\ne 0$, inseparability is equivalent to $f^{\prime }=0$. In characteristic $p$,exactly when $a_j=0$ for every exponent $j$ not divisible by $p$. ThusNo assumption that the coefficient field is perfect is required: vanishing derivative concerns the exponents, not whether the coefficients have $p$th roots in $F$.

Let $E=F({\alpha }^p)\subseteq F(\alpha )$. If $\alpha$ is a separable algebraic element over $F$, its minimal polynomial over $E$ divides the original separable minimal polynomial. It also divides $T^p-{\alpha }^p=(T-\alpha {)}^p$, which has only one distinct root in any extension. Being separable with only one distinct root, that minimal polynomial has degree one. Hence $\alpha \in E$ and $F(\alpha )=F({\alpha }^p)$.

Conversely, if the degree-$n$ minimal polynomial $m$ over $F$ is inseparable, part (b) writes it as $m(T)=g(T^p)$ with $\mathrm{deg}g=n/p$. Then $g({\alpha }^p)=0$, soThe fields cannot be equal. This proves the equivalencethe separability criterion using a pth power.

Put $\gamma =\alpha +\beta$, $E=F(\gamma )$ and $E_0=F({\gamma }^p)$. By the preceding separability criterion using a pth power, it suffices to prove $E=E_0$.

Since $\alpha$ is separable, $\alpha \in F({\alpha }^p)$. But ${\alpha }^p={\gamma }^p-{\beta }^p\in E_0(\beta )$, so $\alpha \in E_0(\beta )$. ConsequentlyLet $h$ and $m$ be the monic minimal polynomials of $\beta$ over $E_0$ and $E$, respectively. Both divide its separable minimal polynomial over $F$, so both are separable. Also $m\mid h$ in $E[T]$.

Every $c\in E=F(\gamma )$ has $c^p\in F({\gamma }^p)=E_0$. Hence $m(T{)}^p$ has coefficients in $E_0$ and vanishes at $\beta$, forcing $h\mid m^p$ in $E_0[T]$. Since $h$ remains a square-free polynomial over $E$, this divisibility implies $h\mid m$ in $E[T]$. Together with $m\mid h$, monicity gives $h=m$. ThereforeThe tower law makes $[E:E_0]=1$. Part (c) now proves that $\gamma =\alpha +\beta$ is separable over $F$. This establishes closure of separable algebraic elements under addition without presupposing that closure as a field theorem.