Past exam of the number theory course of the University of Oxford / 2011 (B9, DMAB 2648, DMCN 2648, DMPB 2648, DMST 2648)

Source: Mathematics release, bundle Part B 2007 11.pdf, pages 302–305. The paper was sat on 10 June 2011, 2.30–5.30pm, and is shared by Mathematics, Mathematics and Computer Science, Mathematics and Philosophy, and Mathematics and Statistics Part B.

The department's answer-release policy excludes ordinary odd years. Bounded exact-paper checks found no matching advertised answers; all parts are answered below.

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

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; it consists of $n$ algebraic integers and is also a $\mathbb{Q}$-basis of $K$.

The lattice $M$ has finite index $h=[{\mathcal{O}}_K:M]>1$. Choose a prime $p\mid h$. The finite abelian group ${\mathcal{O}}_K/M$ has an element of order $p$, represented by $\alpha \in {\mathcal{O}}_K\setminus M$. ThusIf all $c_j$ were divisible by $p$, cancellation in the characteristic-zero field would put $\alpha$ in $M$, a contradiction. The discriminant-index formula for an integral lattice giveswhere ${\Delta }^2(K)$ is an integer. Hence $p^2\mid {\Delta }^2(M)$, and the required integral element is $\alpha =p^{-1}{\sum }_j{c}_j{w}_j$.

The polynomial is irreducible by Eisenstein's criterion at $q$, so $[K:\mathbb{Q}]=n$. Its constant coefficient gives the field normThe defining equation givesIf $q\nmid r$ and $0\le i<n$, thenhas $q$-adic valuation $i-n<0$. It is not an integer. The norm of an algebraic integer is an integer, so $r{\theta }^i/q$ is not integral.

Suppose now that $\gamma$ is integral. Multiply it by ${\theta }^{n-1}$. All terms except $r_0{\theta }^{n-1}/q$ are integral, since their powers of $\theta$ are at least $n$. Thus this remaining term is integral, and the norm obstruction forces $q\mid r_0$. Subtract the integer $r_0/q$ from $\gamma$. Multiply the remainder by ${\theta }^{n-2}$; the only term of exponent below $n$ is $r_1{\theta }^{n-1}/q$, so $q\mid r_1$. Continuing in this way proves $q\mid r_j$ for every $j$. ThereforeFor $n=1$ the same conclusion follows directly from the norm obstruction with $i=0$.

Let $\theta =\sqrt[3]{3}$. The power-basis discriminant isIf $\mathbb{Z}[\theta ]$ were a proper sublattice of ${\mathcal{O}}_K$, part (a) would produce a prime $p$ whose square divides $243$ and a new integral element with denominator $p$. The only possible prime is $3$. But $x^3-3$ is Eisenstein at $3$, and part (b) says that every integral element of $(1/3)\mathbb{Z}[\theta ]$ already belongs to $\mathbb{Z}[\theta ]$. This is a contradiction. Hence

The Kummer-Dedekind theorem states the following. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral and $p\nmid [{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor the minimal polynomial modulo $p$ as $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ into distinct monic irreducible factors. ThenThe index restriction is essential.

For the proposed ideal relation, reflexivity uses $\alpha =\beta =1$ and symmetry interchanges $\alpha ,\beta$. If $(\alpha )I_1=(\beta )I_2$ and $(\gamma )I_2=(\delta )I_3$, multiplication gives $(\alpha \gamma )I_1=(\beta \delta )I_3$. All multipliers remain nonzero. Thus it is an equivalence relation, namely equality of ideal classes.

Put $s=\sqrt{-2}$. The minimal polynomial is $x^2+2$ and the given full power basis makes every prime eligible for the theorem. Modulo $3$ its roots are $\pm 1$; modulo $5$ and $7$ it is irreducible; modulo $17$ its roots are $\pm 7$. ConsequentlyThe two inert ideals have norms $25$ and $49$; each split factor has norm equal to its rational prime.

For $p_1\equiv 1(\mathrm{mod}8)$, quadratic reciprocity and its supplementary laws give $(-2/p_1)=1$. Thus $x^2+2$ has two distinct roots modulo $p_1$, giving distinct prime ideals $P,P^{\prime }$ withThe given principal ideal domain property makes $P=(x+ys)$ for some integers $x,y$. Taking norms, which are positive here, gives $p_1=N(P)=x^2+2y^2$.

Let $\omega =(1+\sqrt{-3})/2$, whose minimal polynomial is $x^2-x+1$. For $p_2\equiv 1(\mathrm{mod}3)$, the cyclic group ${\mathbb{F}}_{p_2}^{\times }$ has a nontrivial cube root of unity $t$. Then $-t$ and $-t^2$ are distinct roots of $x^2-x+1$. By the Kummer-Dedekind theorem,Since the given ring of integers is a principal ideal domain, write $Q=(a+b\omega )$. Its norm isMultiplication by $4$ completes the square:Thus $x=2a+b$ and $y=b$ supply the required integers.

Put $s=\sqrt{-5}$ and $P=(2,1+s)$. The polynomial $x^2+5$ reduces to $(x+1{)}^2$ modulo $2$, so the Kummer-Dedekind theorem givesIf $P=(a+bs)$, its norm would imply $a^2+5b^2=2$, impossible for integers $a,b$. Thus $P$ represents the nontrivial ideal class. Since the class number is $2$, every nonprincipal integral ideal $I$ has $[I]=[P]$, and $[IP]=1$. Write $IP=(\gamma )$ with $\gamma \in {\mathcal{O}}_M$; integrality follows from $IP\subseteq {\mathcal{O}}_M$. Multiplying by $P$ givesNorms give $4N(I)=2N(\gamma )$, hence $N(\gamma )=2N(I)$. If $N(I)=p_3\equiv 1(\mathrm{mod}20)$, writing $\gamma =a+bs$ yieldscontradicting the list $0,1,4$ of squares modulo $5$. Therefore no nonprincipal ideal has norm $p_3$.

As $p_3\equiv 1(\mathrm{mod}4)$ and $p_3\equiv 1(\mathrm{mod}5)$, quadratic reciprocity gives $(-5/p_3)=1$. Thus $x^2+5$ has two distinct roots modulo $p_3$ andThe norm obstruction shows that $R$ must be principal, say $R=(x+ys)$. Taking its norm proves

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one obtains such an ideal by multiplying the fractional ideal $A^{-1}$ by a suitable nonzero integer. The assumed Minkowski bound for ideal classes gives $0\ne \beta \in B$ withBecause $(\beta )\subseteq B$ and nonzero ideals of the ring of integers are invertible,It is an integral ideal in class $[B{]}^{-1}=[A]$. By multiplicativity of the ideal norm,The same argument applies if the starting $A$ is a fractional ideal, since its class is unchanged by clearing denominators.

Write $s=\sqrt{-33}$. The discriminant is $-132$, and the bound of part (a) isEvery ideal class therefore has an integral representative of norm at most $7$. The Kummer-Dedekind theorem givesModulo $5$, $x^2+33=x^2+3$ is irreducible, so there is no ideal of norm $5$. Modulo $7$ it has roots $\pm 3$, giving $(7)=R\bar{R}$, where $R=(7,s+3)$. The ideal of $3+s$ has norm $42$, and reduction in the three residue fields shows that it lies in $P,Q,R$. Equality of norms therefore givesIt follows that $[R]=[P][Q]$, because $[P{]}^2=[Q{]}^2=1$. The other prime over $7$ has the inverse class, which is the same class here.

By unique factorization of nonzero ideals, all norms at most $7$ now give only the classes $1,[P],[Q],[PQ]$: norm $4$ gives $P^2=(2)$, norm $6$ gives $PQ$, and norm $7$ was handled above. None of $P,Q,PQ$ is principal, since their norms are $2,3,6$ andAlso $[P]\ne [Q]$, since equality would make $PQ$ principal. Thus there are exactly four classes, all of exponent two:

For degree $3$ there are either three real embeddings or one real embedding and one complex pair. The respective Minkowski bound for ideal classes areUnder the given discriminant bound they are at most $\frac{2}{9}\sqrt{45}<2$ and $\frac{8}{9\pi }\sqrt{45}<2$. Hence every ideal class contains an integral ideal with positive integer norm strictly below $2$, so with norm $1$. The only such ideal is ${\mathcal{O}}_L$. Therefore

Part (a), applied to any ideal class, produces an integral ideal of positive integer norm at most $c_K$, so $c_K\ge 1$. Writing $n=[K:\mathbb{Q}]$ and using $s\le n/2$ givesFor the last inequality, write $n^n/n!={\prod }_{j=1}^n(n/j)$; the first $\lfloor n/2\rfloor$ factors are at least $2$, and the remaining factors are at least $1$. Since $\lfloor n/2\rfloor \ge (n-1)/2$, squaring yieldsAs $\pi /2>1$, the assumed upper bound $D$ forcesFor $D\ge 1$, the ceiling of the right-hand side is a valid integer $M$. If $D=0$, no number field satisfies the premise, so any positive $M$ works.