Past exam of the number theory course of the University of Oxford / 2008 (B9, DMAB 2648, DMPB 2648)

Source: Mathematics release. The five-page paper is shared by Mathematics and Mathematics & Philosophy Part B, dated 7 June 2008. Bundle Part B 2007 11.pdf, pages 114–118.

Bounded exact-paper checks found no matching advertisement of worked answers. Both printed sections and all eight questions, including every unnumbered task, are covered. Source statements remain in the private editable transcription.

For an irreducible polynomial $f\in K[X]$, take $L=K[X]/(f)$ and let $\alpha$ be the class of $X$. The quotient is a field, is generated by $\alpha$ over $K$, and satisfies $f(\alpha )=0$. Its degree is $\boxed{[L:K]=\mathrm{deg}f}$, with basis $1,\alpha ,\dots ,{\alpha }^{\mathrm{deg}f-1}$.

A splitting field is a field generated over $K$ by all roots of the polynomial and in which the polynomial is a product of linear factors. Prove existence and the factorial bound together by induction on the degree. Constant nonzero polynomials and linear polynomials require no extension. For a reducible degree-$n$ polynomial write $g=hk$ with positive degrees $r,n-r$. Split $h$ by induction, then split $k$ over that field. The resulting field is generated by all roots of $g$, and its degree over $K$ is at most $r!(n-r)!\le n!$. If $g$ is irreducible, adjoin one root in degree $n$, divide out its linear factor, and split the degree-$(n-1)$ quotient over that extension. The tower formula gives degree at most $n(n-1)!=n!$. This also works in inseparable characteristic and proves the required bound without assuming Galoisness.

For the sextic with constant term three, Eisenstein's criterion at three gives degree six. Choose a root $\alpha$, so ${\alpha }^3$ is a square root of $-3$. Thensatisfies ${\zeta }^2-\zeta +1=0$, and is a primitive sixth root of unity: ${\zeta }^3=-1$ and $\zeta \ne -1$. The six distinct roots are $\alpha {\zeta }^j$ for $0\le j<6$, all in $\mathbb{Q}(\alpha )$. Hence the simple extension is itself the splitting field and $\boxed{[L:\mathbb{Q}]=6}$.

For the constant term two, Eisenstein at two gives $[\mathbb{Q}(\gamma ):\mathbb{Q}]=6$ for a root $\gamma$. Its splitting field is $\mathbb{Q}(\gamma ,{\zeta }_6)$. The supplied noncontainment of a root of $X^2+3$ means that adjoining ${\zeta }_6=(1+\sqrt{-3})/2$ has degree two. Thus

Every automorphism fixes zero and one, and preserves sums, products, negatives and multiplicative inverses of nonzero elements. Consequently elements fixed by every member of $G$ are closed under these operations and form a field. Artin fixed-field theorem gives $\boxed{[L:L^G]=|G|}$ for a finite automorphism group.

Here $\mathbb{Q}(t)$ is the fraction field of $\mathbb{Q}[t]$, with $t$ transcendental; its elements are fractions of rational-coefficient polynomials, with nonzero denominator and the usual equality by cross-multiplication. Substitution of $1-t$ or $1/t$ defines an automorphism, each its own inverse. With $\sigma \tau$ meaning first apply $\tau$ to a field element and then $\sigma$, its image on $t$ is $1/(1-t)$. Iterating gives $(t-1)/t$ and then $t$, so its order is three. The relations reduce every word to at most six possibilities, and the six resulting images areThey are distinct rational functions, so the group has six elements. Its faithful permutation action on $\{0,1,\infty \}$ identifies it with $\boxed{S_3}$.

For the first fixed field put $u=t(1-t)$. It is fixed by $\sigma$, and $t$ satisfies $X^2-X+u=0$ over $\mathbb{Q}(u)$. The degree is exactly two because the nontrivial automorphism $\sigma$ fixes that field. The fixed-field degree is also two, soFor the inversion put $v=t+1/t$. It is fixed and $t$ satisfies $X^2-vX+1=0$. The same degree argument givesThe source requests polynomial generators in both cases. The second requirement is impossible. If a nonconstant polynomial $p(t)=a_d{t}^d+\cdots +a_0$ were fixed by inversion, $p(t)=p(1/t)$ would equate a Laurent polynomial with highest exponent $d>0$ to one with no positive exponent, a contradiction. Thus the only polynomial invariants of inversion are constants, which cannot generate its nonconstant fixed field. The fixed field of inversion in a rational function field therefore has the second displayed rational-function generator.

A finite extension is Galois when it is both normal and separable; over $\mathbb{Q}$ separability is automatic, so it is equivalently a splitting field of a polynomial with distinct roots. Its Galois group consists of the field automorphisms fixing $\mathbb{Q}$. If $G=\mathrm{Gal}(L/\mathbb{Q})$, the Fundamental theorem of Galois theory gives mutually inverse inclusion-reversing bijectionsThey satisfy $[L:L^H]=|H|$ and $[L^H:\mathbb{Q}]=[G:H]$. The intermediate extension $K/\mathbb{Q}$ is Galois exactly when $H=\mathrm{Gal}(L/K)$ is normal, and then $\boxed{\mathrm{Gal}(K/\mathbb{Q})\cong G/H}$ via restriction.

Every permutation of the four roots permutes the three unordered partitions into two pairs and hence permutes the three ${\beta }_j$. Thus each coefficient of $g(X)={\prod }_{j=1}^3(X-{\beta }_j)$ is a symmetric polynomial in the four roots. The Fundamental theorem of symmetric polynomials expresses it in the elementary symmetric functions, which are rational coefficients of $f$. Hence $g\in \mathbb{Q}[X]$. Each product ${\alpha }_i{\alpha }_j$, $i<j$, occurs exactly twice in ${\beta }_1+{\beta }_2+{\beta }_3$, so that sum is $2q$. Therefore the quadratic coefficient is $\boxed{-2q}$.

For the given cubic write $g(X)=(X-2{)}^3+12$. The translated cubic $Y^3+12$ is Eisenstein at three, hence irreducible. It has one real root and a nonreal conjugate pair. Its Galois group is transitive on three roots, so its order is divisible by three; complex conjugation is a transposition. A subgroup of $S_3$ with both a three-cycle and a transposition is all of $S_3$. HenceLet $G=\mathrm{Gal}(L/\mathbb{Q})\le S_4$. Irreducibility of the quartic makes the root action transitive, so four divides $|G|$. The cubic splitting field $K$ lies inside $L$ by the definitions of the ${\beta }_j$, and restriction onto its Galois group gives a quotient of order six. Thus six also divides $|G|$. Since $|G|$ divides 24, it is 12 or 24. If it were 12, the supplied index-two result would give $G=A_4$. The restriction kernel would then be a normal subgroup of $A_4$ of index six, contrary to the other supplied result. Thus

A straightedge and compass construction uses finitely many steps drawing lines through known points and circles with known centres and constructible radii, and adjoining their intersection points. The coordinates obtained by line–line intersections require only field operations; line–circle and circle–circle intersections require at most a quadratic extension after eliminating a linear equation. Thus all coordinates lie in a tower of quadratic extensions of the initial coordinate field. The degree of any finite intermediate coordinate field containing that initial field divides a power of two and is therefore $\boxed{2^r}$ for some $r\ge 0$.

For the source's notation this assumes $K\supseteq K_0$, as is customary when previously constructed points are retained. If only the coordinates of a final subset are called $K$, replace it by the compositum $K_{0}K$; otherwise the expression $[K:K_0]$ need not be defined. The power-of-two condition is necessary and is not being asserted as sufficient for an arbitrary algebraic number.

Starting from a unit cube, doubling its volume would construct a length $\sqrt[3]{2}$. Its polynomial $X^3-2$ is Eisenstein at two, so its degree over $\mathbb{Q}$ is three, which cannot divide a power of two. Thus doubling the cube is impossible by ruler and compass.

Set $c=\cos \theta$, $s=\sin \theta$. The real part of $(c+is{)}^5$ is $c^5-10c^3{s}^2+5c{s}^4$. Replacing $s^2$ by $1-c^2$ givesChoose the given angle to have cosine $5/6$. It is constructible, since its unit circle point has coordinates $(5/6,\sqrt{11}/6)$. If a fifth of that angle were constructible and had cosine $c$, the identity would implyThis polynomial is Eisenstein at five: all nonleading coefficients are divisible by five and the constant term is not divisible by 25, while 96 is not divisible by five. Hence $[\mathbb{Q}(c):\mathbb{Q}]=5$. But the initial angle field is $\mathbb{Q}(\sqrt{11})$, itself of degree two, so adjoining any finite collection of constructible coordinates to it would still give total degree over $\mathbb{Q}$ a power of two. It could not contain a degree-five element. Thus general angle quintisection is impossible with straightedge and compass.

For the basis $w_1,\dots ,w_d$, its discriminant isEquivalently it is the square of the determinant of the matrix of the $d$ embeddings applied to the basis vectors. It is nonzero because number fields are separable. An integral basis is a basis of the ring of integers as a free abelian group; in particular it is a rational basis of the field and expresses every algebraic integer uniquely with integer coefficients.

To prove existence, start with any rational basis and multiply each vector by a suitable nonzero integer to make it integral. This is possible by clearing coefficients in its monic rational minimal polynomial of an algebraic element and rescaling the variable. Let $M$ be the lattice spanned over $\mathbb{Z}$ by these integral vectors, and let $D\ne 0$ be its discriminant. For any algebraic integer $\alpha ={\sum }_j{c}_j{w}_j$, each trace $\mathrm{Tr}(\alpha w_i)$ is an integer. The resulting linear system has integral coefficient matrix with determinant $D$. Its adjugate matrix shows $c_j\in D^{-1}\mathbb{Z}$. ThusA subgroup of a finitely generated free abelian group is free and finitely generated. Here its rank is $d$, since it contains $M$ and is contained in a rank-$d$ lattice. Its free generators are the required integral basis.

For squarefree $d\ne 1$, the quadratic integral basis isThe source allows squarefree nonzero integers, so the degenerate value $d=1$ gives $K=\mathbb{Q}$ and basis $\{1\}$ instead of a two-element basis.

In the biquadratic field define the three automorphisms by changing respectively the sign of $\sqrt{r}$, the sign of $\sqrt{2}$, and both signs. Their actions areHence the three sums are $2a+2b\sqrt{2}$, $2a+2c\sqrt{r}$ and $2a+2d\sqrt{2r}$. If $\mu$ is integral, each sum is integral and lies in its displayed quadratic subfield. Their radicands are squarefree and respectively congruent to two, three and two modulo four. The quadratic basis formula therefore forces all their two coefficients to be integers, proving

The Dedekind factorization theorem says: if $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and the rational prime $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. Then the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, with integer lifts $f_i$, andThe index condition is part of the theorem and must be checked before applying it at a ramified prime.

For this cubic, $X^3-6$ is Eisenstein and has discriminant $-972$. We first verify ${\mathcal{O}}_K=\mathbb{Z}[\theta ]$. The square of the index divides 972, so only two and three can divide the index. If either prime $p$ divided it, the finite quotient group would contain an element of order $p$, giving an integral element $(a+b\theta +c{\theta }^2)/p$ with digits $0\le a,b,c<p$, not all zero. Its field norm isFor $p=2$, the numerator has valuation zero if $a=1$, valuation one if $a=0,b=1$, and valuation two if $a=b=0,c=1$. For $p=3$ the same three cases, according to the first nonzero digit, give valuations zero, one and two. None gives an integer field norm after dividing by $p^3$, contradicting integrality. Thus the index is one.

Multiplication by $a+b\theta$ in the power basis, or the product over the three conjugates, gives $\boxed{N(a+b\theta )=a^3+6b^3}$. DefineReduction of the cubic gives $X^3$ at two and three, $(X-1)(X^2+X+1)$ at five, and $(X-3)(X-5)(X-6)$ at seven. The quadratic factor at five is irreducible since its discriminant is two, a nonsquare modulo five. HenceThe norms are respectively $2,3,5,25,7,7,7$. Norms and reduction of the elements at those primes giveFor example their absolute element field norms are $6,5,2,21,70$; each indicated prime divides the element by substitution in its residue field, and the product of the indicated ideal norms already equals that absolute field norm. Thus no additional factor or exponent is possible.

In the ideal of the specified quotient every prime exponent cancels:Equality with the unit ideal implies that both $\mu$ and its inverse are integral, so $\boxed{\mu \text{ is a unit}}$. For the real root, $1<\theta <2$. The numerator is negative and has absolute value less than two, while the denominator is greater than five. Thus $-2/5<\mu <0$, proving $\boxed{\mu \ne \pm 1}$. Its field norm is $-1$, consistent with the ideal calculation.

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one can multiply the fractional inverse $A^{-1}$ by a suitable nonzero integer to obtain such a $B$. By the given bound choose $0\ne \beta \in B$ with $|N(\beta )|\le c_{K}N(B)$. Put $I_A=(\beta )B^{-1}$. Since $(\beta )\subseteq B$, multiplication by $B^{-1}$ makes $I_A$ an integral ideal. Its class is $[B{]}^{-1}=[A]$, and multiplicativity of ideal norms givesThis proves the class representative bound without accidentally constructing a representative of the inverse class instead.

To determine the Ideal class group of Q of square root minus thirty-four, write $\eta =\sqrt{-34}$. The given integral basis has discriminant $-136$, so the imaginary-quadratic bound is $4\sqrt{34}/\pi <8$. Every class therefore has a representative of ideal norm at most seven, and factoring such representatives shows that the ideal class group is generated by prime ideals of ideal norm at most seven.

Two ramifies: let $P_2=(2,\eta )$, with $P_2^2=(2)$. Three is inert because $X^2+34\equiv X^2+1$ is irreducible modulo three, so its prime ideal ideal norm is nine and supplies no small generator. At five and seven the roots are $\pm 1$, so the primes of ideal norm five and seven are conjugate pairs. Set $P_5=(5,\eta -1)$ and $P_7=(7,\eta +1)$. Their conjugates have inverse ideal classes. No new prime ideal has ideal norm four, since the only prime above two has residue degree one; six is not a prime power. Thus $[P_2],[P_5],[P_7]$ generate the ideal class group.

The element field norms $N(4+\eta )=50$ and $N(1+\eta )=35$, together with reduction at the split primes, giveThe ideal norm products verify both exponents. Consequently $[P_5{]}^2=[P_2]$ and $[P_7]=[P_5]$, so the whole group is generated by $[P_5]$ and its order divides four. The class $[P_2]$ is not trivial: a generator would be an element of the ring of integers $a+b\eta$ of field norm two, but $a^2+34b^2=2$ has no integer solution. Since $[P_2{]}^2=1$, its order is exactly two. Hence $[P_5]$ has order four and

For integral solutions of x squared minus two equals y cubed, the equation implies $y\ge -1$, so $y+2>0$. If $y$ were even, then $x$ would be even and $x^2-2$ would be two or six modulo eight, whereas $y^3$ would be zero modulo eight. Thus $y$ and $x$ are odd. Also $3\nmid x$: otherwise $x^2-2\equiv 7(\mathrm{mod}9)$, which is not a cubic residue modulo nine. Reducing modulo three then gives $y\equiv 2(\mathrm{mod}3)$. ConsequentlyPut $\eta =\sqrt{-6}$ and $\alpha =x+(y+1)\eta$. Its field norm is $(y+2{)}^3$. The principal ideals $(\alpha )$ and $(\bar{\alpha })$ are coprime. Indeed a common prime ideal has underlying rational prime $p$ dividing $y+2$. The displayed gcd excludes two and three, so both two and $\eta$ are invertible modulo that prime ideal. The difference $\alpha -\bar{\alpha }=2(y+1)\eta$ therefore implies $p\mid y+1$, contradicting $p\mid y+2$.

Their product is $(y+2{)}^3$, so unique factorization of ideals in a number field and coprimality imply $(\alpha )=A^3$ for an integral ideal $A$. In the given ideal class group of order two, $[A{]}^3=1$ implies $[A]=1$. Hence $A=(u+v\eta )$ with integers $u,v$. Equality of principal ideals gives $\alpha =\epsilon (u+v\eta {)}^3$. The only units are $\pm 1$, and both are cubes, so the sign can be absorbed into $u,v$. Taking norms, and using $y+2>0$, givesThe coefficient of $\eta$ gives $y+1=3u^{2}v-6v^3$. Eliminating $y$ with the field norm identity yieldsThe denominator is nonzero for integral $v$. Polynomial division makes the integrality obstruction especially clear:If $u$ is integral, the reciprocal must be an integer, forcing $3v-1=\pm 1$. The only integral possibility is $v=0$. Then $u^2=1$, the field norm gives $y=-1$, and the real coefficient in the cube gives $x=u^3=\pm 1$. Both pairs directly satisfy the original equation. Therefore all integral solutions are