Past exam of the number theory course of the University of Oxford / 2007 (B9, DMAB 2648, DMCN 2648, DMPB 2648)

Source: Oxford mathematics examination release, private bundle Part B 2007 11.pdf, pages 46–51. The printed cover identifies the shared Part B examination for mathematics, mathematics and computer science, and mathematics and philosophy. The full editable transcription is retained in the private examination archive.

By Vieta's formulas, the elementary symmetric polynomials in the roots giveFor each root, ${\alpha }_j^3+c_1{\alpha }_j^2+c_2{\alpha }_j+c_3=0$. Multiplying by ${\alpha }_j^k$ and summing gives the power sum recurrenceIts initial values are $s_0=3$, $s_1=-c_1$ and $s_2=c_1^2-2c_2$. They are polynomials with integer coefficients in $c_1,c_2,c_3$. The recurrence uses only addition and multiplication by those coefficients, so induction proves the same for every $s_k$. This is the cubic case of Newton's identities.

For $c_1=0$, put ${\alpha }_1=u$, ${\alpha }_2=v$, ${\alpha }_3=-u-v$. Then $c_2=-(u^2+uv+v^2)$ and $c_3=uv(u+v)$. Expanding the product of root differences givesThis is the discriminant of a depressed cubic in depressed form, and the displayed identity also proves its expression in integer-coefficient symmetric polynomials.

Scaling every root scales its elementary symmetric polynomials by the corresponding powers of $d$. Therefore the monic polynomial isThis remains valid for $d=0$, when all three roots are zero; deriving it by symmetric coefficients avoids dividing by $d$.

Reciprocal roots require $c_3\ne 0$, equivalently every original root is nonzero. Their monic polynomial is the normalized reciprocal polynomialIf $c_3=0$, at least one reciprocal in the source is undefined, so that part has no polynomial interpretation without the nonzero-root qualification.

The sum of squared roots is $c_1^2-2c_2$. The sum of their pairwise products isTheir product is $c_3^2$. Thus the required monic polynomial is

An algebraic element is a separable algebraic element over $K$ if its minimal polynomial of an algebraic element has distinct roots in a splitting field. A field extension is a separable extension if all its elements are separable algebraic elements.

The formal derivative is additive and satisfies $D(X^m)=m{X}^{m-1}$. For monomials, $D(X^r{X}^s)=(r+s)X^{r+s-1}=X^{r}D(X^s)+X^{s}D(X^r)$. Distributing over the finitely many coefficients proves the product rule $D(fg)=fDg+gDf$ over any field. If $f=(X-\alpha {)}^{2}h$ in an extension, the rule gives $Df=2(X-\alpha )h+(X-\alpha {)}^{2}Dh$, so $Df(\alpha )=0$, including in characteristic two. If $f,Df$ were relatively prime in $K[X]$, the Bézout's identity for polynomials would give $uf+vDf=1$, whose evaluation at $\alpha$ is impossible. Hence they are not relatively prime.

In characteristic zero, the formal derivative of a positive-degree irreducible polynomial $m$ is nonzero and has smaller degree. Thus $\gcd (m,Dm)=1$, since an irreducible polynomial has no proper nonconstant factor. By the previous argument, $m$ has no repeated root. Every element of a finite field extension is algebraic, so the extension is a separable extension.

For $K={\mathbb{F}}_p(t)$, let ${\alpha }^p=t$. The characteristic-$p$ identity gives $X^p-t=(X-\alpha {)}^p$, so $K(\alpha )$ is already a splitting field. It is a proper extension by the allowed no-root fact. More precisely, the minimal polynomial of an algebraic element $\alpha$ divides $X^p-t$ and, over $K(\alpha )$, equals $(X-\alpha {)}^r$ for some $1\le r\le p$. If $r<p$, its next-to-leading coefficient is $-r\alpha \in K$, forcing $\alpha \in K$ because $r\ne 0$ in ${\mathbb{F}}_p$, a contradiction. Hence its minimal polynomial of an algebraic element is $X^p-t$, with its one root repeated $p$ times. The extension is therefore inseparable, of degree $p$.

By contrast, every finite field extension $L/{\mathbb{F}}_p$ is a separable extension. If $[L:{\mathbb{F}}_p]=n$, then $|L|=p^n$. Lagrange's theorem in its nonzero multiplicative group gives $a^{p^n}=a$ for all $a\in L$. Thus every element is a root of $X^{p^n}-X$, whose formal derivative is $-1$; it has no repeated roots. Every minimal polynomial of an algebraic element divides this separable polynomial, proving the assertion.

Put $\theta =\sqrt{p+\sqrt{p}}$. Then $({\theta }^2-p{)}^2=p$, so $\theta$ is a root ofThis polynomial satisfies Eisenstein's criterion at $p$: every nonleading coefficient is divisible by $p$, while the constant coefficient $p(p-1)$ is not divisible by $p^2$. ThereforeNow suppose $p=n^2+1$, taking $n>0$ without loss of generality. Let $\beta =\sqrt{p-\sqrt{p}}$. Since $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$,Consequently all four distinct roots $\pm \theta ,\pm \beta$ lie in this field. This gives a cyclic quartic field from a nested quadratic radical. It is a splitting field in characteristic zero, hence a normal extension and a Galois extension.

There is a Galois automorphism $\sigma$ sending $\theta$ to $\beta$: a root image defines an embedding, and normality makes it an automorphism of the splitting field. It sends $\sqrt{p}={\theta }^2-p$ to $-\sqrt{p}$, soThus $\theta \mapsto \beta \mapsto -\theta \mapsto -\beta \mapsto \theta$ is a four-cycle. The Galois group has order four and contains an element of order four, givingFor the requested lower degrees, $(q,r)=(3,4)$ gives $\sqrt{q+\sqrt{r}}=\sqrt{5}$ and degree two; $(q,r)=(3,1)$ gives $2$ and degree one. Both pairs use positive integers.

A finite straightedge and compass construction repeatedly draws the line through two available distinct points or the circle with an available center and an available distance as radius, and adjoins isolated intersections of two such lines or circles. A finite set is constructible when its points occur after finitely many such steps, starting with the specified points.

If the current coordinate field is $F$, intersections of two lines solve linear equations over $F$. A line and a circle lead to a quadratic equation over $F$. For two distinct circles, subtraction of their equations gives a line, reducing to the previous case. Thus each step adjoins coordinates in an extension of degree one or two. The tower law gives a finite tower of quadratic-or-trivial field extensions containing all output coordinates. If the final coordinate field contains $K_0$, thenbecause its degree divides the power-of-two degree of that tower.

Two implicit qualifications are necessary for this degree statement. The output is obtained in finitely many steps; the aggregate of infinitely many individually constructible points need not generate a finite extension. Also, unless the output retains the original coordinates, $K$ need not contain $K_0$. In that case the degree statement applies to the compositum $K_{0}K$, rather than an undefined $[K:K_0]$. The heptagon application explicitly includes $S_0$, so neither issue obstructs it.

Here $K_0=\mathbb{Q}$, and a vertex adjacent to $(1,0)$ has first coordinate $c=\cos (2\pi /7)$, so $\mathbb{Q}(c)\subseteq K$. Let $\zeta =e^{2\pi i/7}$. The cyclotomic polynomial $1+X+\cdots +X^6$ is irreducible, giving $[\mathbb{Q}(\zeta ):\mathbb{Q}]=6$. Since $\zeta +{\zeta }^{-1}=2c$, it satisfies $X^2-2cX+1$ over $\mathbb{Q}(c)$. The element $\zeta$ is nonreal while $\mathbb{Q}(c)\subset \mathbb{R}$, so this relative degree is exactly two. By the tower law,If the regular regular heptagon were constructible, this degree would divide $[K:\mathbb{Q}]=2^r$, impossible. Therefore the regular heptagon is not constructible from the given unit segment.

For the $n$ distinct complex embeddings ${\sigma }_1,\dots ,{\sigma }_n$ of the number field, define the discriminant of elements of a number field byThe equality follows by multiplying the transpose of the embedding matrix by itself. Each field trace is rational: it is also the trace of the rational matrix of multiplication by $w_i{w}_j$ on $K$. Thus the determinant lies in $\mathbb{Q}$. If the $w_i$ are algebraic integers, the products and their conjugates are algebraic integers, and so are their sums. A rational algebraic integer is an integer, hence all trace entries and their determinant are integers.

Write $u_j={\sum }_i{a}_{ij}{v}_i$. The inclusion of the generated abelian groups makes $A=(a_{ij})$ an integer matrix. Its embedding matrix is the embedding matrix of the $v_i$ multiplied by $A$. Taking squared determinants givesFor independent algebraic integers $v_i$, choose an integral basis ${\omega }_i$ of ${\mathcal{O}}_K$. The same formula gives $D(v)=[{\mathcal{O}}_K:{\sum }_i\mathbb{Z}{v}_i{]}^{2}D(\omega )$. If $D(v)$ is nonzero and a square-free integer, the integer index must be one, proving that the $v_i$ form an integral basis.

Now put $\beta =3\alpha +4$. Since $\mathbb{Q}(\beta )=\mathbb{Q}(\alpha )$ has degree three, the monic polynomialis its minimal polynomial of an algebraic element. The field norm is the negative of its constant term:For the discriminant calculation, $f^{\prime }(\alpha )=3{\alpha }^2+4\alpha$. In the basis $1,\alpha ,{\alpha }^2$, using ${\alpha }^3=-2{\alpha }^2-1$, multiplication by this element has matrixThus the given derivative-norm formula gives $D(1,\alpha ,{\alpha }^2)=-59$. Since $\alpha$ satisfies a monic polynomial with integer coefficients, $1,\alpha ,{\alpha }^2$ are algebraic integers; the fact that a squarefree discriminant certifies an integral basis proves

Dedekind factorization theorem states that if $K=\mathbb{Q}(\theta )$ with integral $\theta$ and $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, then a factorization $\overline{f}={\prod }_i{\overline{f}}_i^{e_i}$ of its minimal polynomial of an algebraic element into distinct monic irreducible polynomials over ${\mathbb{F}}_p$ givesThe factors are distinct prime ideals; the residue degrees and ramification indices satisfy ${\sum }_i{e}_i\mathrm{deg}{f}_i=[K:\mathbb{Q}]$. Unique prime ideal factorization holds in the ring of integers.

For a square-free integer $d\equiv 1(\mathrm{mod}4)$, ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, so the index condition is automatic. If $p_i\mid d$, the polynomial $X^2+d$ reduces to $X^2$ modulo $p_i$. HenceTherefore $d{\mathcal{O}}_K=({\prod }_i{P}_i{)}^2=(\sqrt{-d}{)}^2$. By unique prime ideal factorization, ${\prod }_i{P}_i=(\sqrt{-d})$.

Let $S$ be a nonempty proper subset and put $m={\prod }_{i\in S}{p}_i$. For $d>1$, $1<m<d$ and $m$ is a square-free integer. If ${\prod }_{i\in S}{P}_i=(u+v\sqrt{-d})$, its ideal norm givesThe inequality $m<d$ forces $v=0$, leaving $m=u^2$, impossible for a square-free integer greater than one. Thus every such subproduct is nonprincipal.

These are ramified prime classes for discriminant minus four d. The ideal classes $[P_i]$ have order dividing two. Their product is the identity, and the preceding proof rules out every other nonempty subset relation. The homomorphism ${\mathbb{F}}_2^k\to \mathrm{Cl}(K)$ defined by these classes consequently has kernel exactly $\{0,(1,\dots ,1)\}$. Its image is thereforeThe boundary case $d=1$ has $k=0$ and makes the printed exponent $k-1$ meaningless; its field is $\mathbb{Q}(i)$ with trivial ideal class group. The subgroup statement is understood for $d>1$, while the prime-factor and proper-subset statements are empty for $d=1$.

Minkowski convex body theorem states that a centrally symmetric convex measurable subset of ${\mathbb{R}}^n$ with volume greater than $2^n$ times the covolume of a full Euclidean lattice contains a nonzero lattice point. The equivalent closed-boundary form follows by compactness for compact convex bodies.

The set $S_t$ is centrally symmetric. It is convex because for $0\le \lambda \le 1$, the triangle inequality gives $\Vert \lambda u+(1-\lambda )v{\Vert }_1\le \lambda \Vert u{\Vert }_1+(1-\lambda )\Vert v{\Vert }_1\le t$. It is a diamond of area $2t^2$. For the embedded nonzero integral ideal $I$, the given covolume is $2N(I)\sqrt{d}$. Hence Minkowski convex body theorem applies whenFor the resulting nonzero $\alpha \in I$, write $x={\sigma }_1(\alpha )$, $y={\sigma }_2(\alpha )$. ThenDecrease $t$ to $2\sqrt{N(I)\sqrt{d}}$. A fixed slightly larger diamond contains only finitely many lattice points, so one nonzero $\alpha$ recurs along a sequence of such $t$ values. Passing to the limit provesGiven a target ideal class $C$, choose an integral ideal $I$ in $C^{-1}$. Since $\alpha \in I$, the ideal $J=(\alpha )I^{-1}$ is integral, belongs to $C$, and has ideal normFor $K=\mathbb{Q}(\sqrt{14})$, every ideal class thus has an integral representative of norm at most $\sqrt{14}<4$. The only possibilities are norms one, two or three. The prime two ramifies, giving the unique ideal of norm two $P_2=(2,\sqrt{14})$. The element $4+\sqrt{14}$ has field norm $16-14=2$, so $(4+\sqrt{14})=P_2$, making it principal. The prime three is inert because $X^2-14\equiv X^2+1(\mathrm{mod}3)$ is an irreducible polynomial; its prime ideal has norm nine, so there is no ideal of norm three. Every possible small representative is principal, proving class number one for Q square root of fourteen. Therefore

Work in ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, since $d\equiv 2(\mathrm{mod}4)$. First $x$ must be odd: if it were even, coprimality would make $a$ odd, so $x^2+a^{2}d\equiv 2(\mathrm{mod}4)$, which cannot be an odd power $y^p$ for $p\ge 3$. Thus $y$ is odd too. Also no odd prime $\ell$ can divide both $x$ and $d$. Such a prime does not divide $a$, and square-freeness of $d$ gives $v_{\ell }(x^2+a^{2}d)=1$, whereas $v_{\ell }(y^p)$ is divisible by $p$. Hence $\gcd (x,d)=1$.

Put $\alpha =x+a\sqrt{-d}$. Suppose a prime ideal $P$ divides both $(\alpha )$ and $(\overline{\alpha })$, and let $\ell$ be the rational prime below it. It then contains $2x$ and $2a\sqrt{-d}$. The case $\ell =2$ is impossible because $N((\alpha ))=y^p$ is odd. If $\ell$ is odd, the first inclusion gives $\ell \mid x$, and $\gcd (a,x)=1$ makes $a$ nonzero in the residue field. The second inclusion therefore gives $\sqrt{-d}\in P$, so $\ell \mid d$, contradicting the previous paragraph. Thus $(\alpha )$ and $(\overline{\alpha })$ are coprime ideals. In particular they generate the unit ideal, the precise meaning of coprimality here; no unique factorization of elements is assumed.

Their product is $(y{)}^p$. Unique prime ideal factorization and disjoint prime support imply $(\alpha )=A^p$ for an integral ideal $A$. In the ideal class group, $[A{]}^p=1$. Since $p$ does not divide the class number, Lagrange's theorem forces $[A]=1$. Write $A=(u+v\sqrt{-d})$. Equality of the principal ideals gives $\alpha =\epsilon (u+v\sqrt{-d}{)}^p$. The only units are $\pm 1$, since a unit satisfies $u^2+d{v}^2=1$ and $d\ge 2$. Both signs are $p$th powers for odd $p$, so they can be absorbed into the generator:Apply the class-number obstruction from a ramified odd exponent. For the equation with $a=1$ and $p\mid d$, compare the coefficients of $\sqrt{-d}$ in this identity. The coefficient on the right isIts $j=1$ term is divisible by $p$, and every $j\ge 3$ term contains a factor $d$, so the whole coefficient is divisible by $p$. The coefficient on the left is one, a contradiction. Thus there are no positive integer solutions under the stated class-number and divisibility assumptions.

Finally,The integer $174=2\cdot 3\cdot 29$ is square-free, is two modulo four, and is divisible by three. If its imaginary quadratic class number were not divisible by three, the preceding result with $p=3$ would prohibit this solution. Consequently