Past exam of the number theory course of the University of Oxford / 2009 (B9, DMAB 2648, DMCN 2648, DMPB 2648, DMST 2648)

Source: Mathematics release, supplied bundle Part B 2007 11.pdf, pages 173–176. The four-page paper is dated 8 June 2009 and is shared by Mathematics, Mathematics & Computer Science, Mathematics & Philosophy and Mathematics & Statistics Part B.

Bounded exact-paper checks found no matching advertisement of worked answers. Both sections and all printed parts are answered; full statements remain in the private editable transcription.

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.

The set ${\mathcal{O}}_K$ consists of the elements of $K$ integral over $\mathbb{Z}$, meaning roots of monic polynomials in $\mathbb{Z}[X]$.

First prove the finite-module criterion for integrality in a slightly stronger form, for any nonzero finitely generated $\mathbb{Z}$-submodule $M\subseteq K$ stable under multiplication by $\alpha$. Choose generators $m_1,\dots ,m_r$ and write $\alpha m_i={\sum }_j{a}_{ij}{m}_j$ with integer coefficients. Multiply the vector equation $(\alpha I-A)m=0$ by its adjugate matrix. This gives $\det (\alpha I-A)m_i=0$ for every generator. Some generator is nonzero and $K$ is a field, so $\det (\alpha I-A)=0$. The polynomial $\det (XI-A)$ is monic with integer coefficients, proving integrality of $\alpha$. This implication needs no prior closure theorem for ${\mathcal{O}}_K$.

Conversely, if $\alpha$ satisfies a monic polynomial of degree $d$, then $M=\mathbb{Z}[\alpha ]$ is generated by $1,\alpha ,\dots ,{\alpha }^{d-1}$ and is stable under $\alpha$. Every $\beta \in M$ also maps $M$ into itself because it is an integer polynomial in $\alpha$; the finite-module criterion for integrality just proved makes every such $\beta$ integral. Thus this $M$ really lies in ${\mathcal{O}}_K$, proving the stated equivalence without circularity.

For two integral elements $\alpha ,\beta$, the module spanned by ${\alpha }^i{\beta }^j$ for $i<d_{\alpha }$, $j<d_{\beta }$ contains one, is finitely generated, and is stable under both elements by their monic equations. It is therefore stable under $\alpha +\beta$ and $\alpha \beta$. The finite-module criterion for integrality proves that both are integral. Negatives follow by replacing a multiplication matrix by its negative, and $0,1$ are integral. Hence $\boxed{{\mathcal{O}}_K\text{ is a ring}}$.

The integral basis of the cubic field of discriminant minus 140 can be obtained as follows. For the specified field, Eisenstein at two makes $X^3+2X+2$ irreducible, and $\theta$ is integral because the polynomial is monic. The discriminant of a monic cubic $X^3+rX+s$ is $-4r^3-27s^2$, soHere ${\Delta }^2$ is the printed notation for the signed discriminant, namely the square of a Vandermonde product of the conjugates; that product need not be real.

Suppose $\eta =(a+b\theta +c{\theta }^2)/2$ is integral with the prescribed binary coefficients. The root traces give $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\theta )=0$, $\mathrm{Tr}({\theta }^2)=-4$. Thus $\mathrm{Tr}(\eta )=(3a-4c)/2$ must be an integer, forcing $a=0$. If $b=0$, nonzeroness of the coefficient triple forces $c=1$, giving $\eta ={\theta }^2/2$. If $b=1$, multiply by $\theta$ and use ${\theta }^3=-2\theta -2$:Closure of the ring then again gives ${\theta }^2/2\in {\mathcal{O}}_K$.

But $N_{K/\mathbb{Q}}(\theta )=-2$, so the field norm of ${\theta }^{-1}$ is $-1/2$. The field norm of an algebraic integer is an integer, proving ${\theta }^{-1}\notin {\mathcal{O}}_K$. Dividing the defining cubic by $2\theta$ gives ${\theta }^2/2=-1-{\theta }^{-1}$. Thus ${\theta }^2/2$ cannot be integral either.

Finally the discriminant-index formula for an integral lattice says $-140=[{\mathcal{O}}_K:\mathbb{Z}[\theta ]{]}^2\mathrm{disc}(K)$. The square of the integer index divides $140$, so its only possibilities are one and four, giving index one or two. If the index were two, a nonzero coset would supply an integral element $(a+b\theta +c{\theta }^2)/2$ after reducing coefficients modulo two, with a nonzero binary triple. The preceding argument would force the impossible integrality of ${\theta }^2/2$. Hence

ideals are coprime when $A+B={\mathcal{O}}_K$, equivalently $1=a+b$ for some $a\in A$, $b\in B$. Always $AB\subseteq A\cap B$. If $x\in A\cap B$, then $x=xa+xb\in AB$, giving $\boxed{AB=A\cap B}$ in the coprime case. For nonzero integral ideals, the Chinese remainder theorem gives ${\mathcal{O}}_K/(AB)\cong {\mathcal{O}}_K/A\times {\mathcal{O}}_K/B$, and taking finite cardinalities yields $\boxed{N(AB)=N(A)N(B)}$.

For a nonzero prime ideal, the chain of finite-index inclusions givesby the given hypothesis. Distinct nonzero prime ideals are maximal and coprime, as are their positive powers. Unique factorization of ideals in a number field therefore gives $N(A)={\prod }_{P}N(P{)}^{v_P(A)}$. Exponents add under ideal multiplication, proving multiplicativity for arbitrary nonzero integral ideals. This is the standard domain of the finite ideal norm used here.

For the Dedekind factorization theorem, choose an integral primitive element $\theta$ of $K$ with monic minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and assume the rational prime $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$. Factorover ${\mathbb{F}}_p$, with distinct monic irreducible factors. For integer polynomial lifts $f_i$, the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, andThus the residue degree is $\mathrm{deg}{\bar{f}}_i$, the ramification index is $e_i$, and ${\sum }_i{e}_i\mathrm{deg}{\bar{f}}_i=[K:\mathbb{Q}]$. The index restriction is essential to this direct factorization formula.

For $K=\mathbb{Q}(\sqrt{-2})$, the ring of integers is $\mathbb{Z}[\sqrt{-2}]$, so the index is one and the theorem applies to every prime. Modulo two, $X^2+2=X^2$, giving one prime ideal of ideal norm two. For an odd rational prime, a prime ideal has ideal norm $p$ if the polynomial splits, and ideal norm $p^2$ if it is irreducible. In the former case the supplied condition forces $p\equiv 1$ or $3(\mathrm{mod}8)$; in the latter case every odd square is one modulo eight. ConsequentlyThe odd part of any ideal norm is thus one or three modulo eight, because $3^2\equiv 1$. Multiplication by a power of two leaves possible residues $1,3$ for exponent zero, $2,6$ for exponent one, $4$ for exponent two, and $0$ for exponent at least three. Hence $\boxed{N(I)\not\equiv 5,7(\mathrm{mod}8)}$ for every nonzero integral ideal.

For nonzero ideals, unique factorization of ideals in a number field converts $AB=C^3$ to $v_P(A)+v_P(B)=3v_P(C)$. Coprimality says at most one of the first two exponents is positive, so each is divisible by three. Dividing its exponents by three constructs ideals whose cubes are $A,B$. If a zero ideal is allowed, coprimality forces the other ideal to be the whole ring and the conclusion is immediate with the zero ideal as its own cube.

To compute the Ideal class group of Q of square root minus thirty, put $\theta =\sqrt{-30}$. The ring of integers is $\mathbb{Z}[\theta ]$, with discriminant $-120$. The given Minkowski bound for ideal classes is $2\sqrt{120}/\pi =4\sqrt{30}/\pi <7$. Hence the ideal class group is generated by prime ideals with ideal norm below seven. The rational primes two, three and five all ramify, giving unique prime ideals $P_2=(2,\theta )$, $P_3=(3,\theta )$, $P_5=(5,\theta )$ with $P_i^2=(i)$ and ideal norm $i$. There is no additional prime ideal of ideal norm four because two ramifies with residue degree one; six is not a prime power. Thus these three classes generate and each has order at most two.

As ${\theta }^2=-30$, unique factorization of ideals in a number field gives $(\theta )=P_2{P}_3{P}_5$, so their three classes multiply to one. None is principal: a generator of one would be an integral element $a+b\theta$ of field norm $a^2+30b^2$ equal to two, three or five, and none of those field norm equations has integer solutions. The three nontrivial classes are therefore distinct, with the third the product of the first two. Hence

To rule out integer solutions, suppose $x^2+30=y^3$. Necessarily $y>0$. If a prime $p\mid 30$ divided $y$, then reduction modulo $p$ would force $p\mid x$. Because thirty is squarefree, the valuation of $x^2+30$ at that prime would be exactly one, while that of $y^3$ is a positive multiple of three, a contradiction. Thus $\gcd (y,30)=1$.

If a prime ideal divided both $(x+\theta )$ and $(x-\theta )$, it would divide their difference $(2\theta )$ and their product $(y{)}^3$. Its underlying rational prime would divide $N(2\theta )=120$, hence divide thirty, and would also divide $y$. This contradicts the gcd just proved, so the two ideals are coprime. Their product is $(y{)}^3$, so the first part gives $(x+\theta )=A^3$ for an ideal $A$.

In a ideal class group of exponent two, $[A{]}^3=1$ forces $[A]=1$. Thus $A=(a+b\theta )$ with $a,b\in \mathbb{Z}$. Equality of principal ideals gives $x+\theta =\epsilon (a+b\theta {)}^3$ for a unit $\epsilon$. The only units are $\pm 1$, since $a^2+30b^2=1$ forces $b=0$, $a=\pm 1$; both signs are cubes and can be absorbed into the generator. Comparing the coefficient of $\theta$ after cubing now giveswhich is impossible modulo three. Therefore