Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 32 (Algebraic Number Theory)

Let $L=\mathbb{Q}(\alpha )$, where $\alpha$ is an algebraic integer with monic minimal polynomial $f\in \mathbb{Z}[X]$, and put $I=[{\mathcal{O}}_L:\mathbb{Z}[\alpha ]]$. The Kummer-Dedekind theorem applies to a prime number $p$ with $p\nmid I$. Write its reduction as a product of polynomial factors over ${\mathbb{F}}_p$:where the $g_j$ are distinct monic irreducible polynomials, choose monic integral lifts $G_j$. Then the distinct prime ideals above $p$ are ${\mathfrak{p}}_j=(p,G_j(\alpha ))$, andThus the multiplicity of a factor is its ramification index of a prime ideal, and its degree is the residue degree. The hypothesis on the order in a number field is essential: one cannot infer prime ideal factorization from an arbitrary defining polynomial at a prime dividing its index. There is also a relative version over a base number field $K$: if ${\mathcal{O}}_L$ and ${\mathcal{O}}_K[\alpha ]$ agree after localization at $\mathfrak{p}$, factor the minimal polynomial over ${\mathcal{O}}_K/\mathfrak{p}$ and use $(\mathfrak{p},G_j(\alpha ))$ with the same conclusions.

The polynomial discriminant satisfiesA polynomial has no repeated roots over the algebraic closure exactly when it is coprime to its derivative. Consequently the assumed squarefree reduction has nonzero resultant with its derivative, so $p\nmid \mathrm{disc}(f)$. On the other hand the power basis is a basis of the order in a number field $\mathbb{Z}[\alpha ]$, and the discriminant-index formula for an integral lattice givesHere the field discriminant $d_L$ is an integer. ThereforeThis is squarefree reduction excludes a prime from the power-basis index; its argument establishes the index condition before applying the Kummer-Dedekind theorem.

Put $\alpha =44^{1/11}$ and $L=\mathbb{Q}(\alpha )$. The Eisenstein criterion at $11$ shows that $X^{11}-44$ is an irreducible polynomial, so $[L:\mathbb{Q}]=11$. Its derivative is $11X^{10}$, and the polynomial discriminant isThe discriminant-index formula for an integral lattice implies that the field discriminant has no prime divisor outside $\{2,11\}$. By the discriminant obstruction to ramification, every other prime is unramified.

To prove that neither candidate disappears through an index correction, let $\mathfrak{p}$ lie above $p\in \{2,11\}$ and normalize its valuation to have value group $\mathbb{Z}$. If $e=e_{\mathfrak{p}/p}$, the relation ${\alpha }^{11}=44$ givesIn both cases $11\mid e$. The fundamental identity for prime decomposition, ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}/p}{f}_{\mathfrak{p}/p}=11$, now forces exactly one prime ideal above each candidate, with $e=11$ and $f=1$. This is ramification forced by a coprime root valuation; at $2$ it avoids an invalid direct application of the Eisenstein criterion, since $4\mid 44$. Hence

Write $G=\mathrm{Gal}(F/K)$ and fix a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$. Suppose there is another prime $\mathfrak{r}$ above $\mathfrak{p}$ outside the $G$-orbit of $\mathfrak{q}$. Distinct nonzero prime ideals in the ring of integers are distinct maximal ideals, hence pairwise comaximal. The Chinese remainder theorem supplies $a\in {\mathcal{O}}_F$ satisfying $a\equiv 0(\mathrm{mod}\mathfrak{r})$ and $a\equiv 1(\mathrm{mod}\sigma \mathfrak{q})$ for every $\sigma \in G$. Its field norm isThe factor $a$ puts $b$ in $\mathfrak{r}$, hence in $\mathfrak{r}\cap {\mathcal{O}}_K=\mathfrak{p}$. But every $\sigma (a)$ is congruent to $1$ modulo $\mathfrak{q}$, since $a\equiv 1(\mathrm{mod}{\sigma }^{-1}\mathfrak{q})$. Thus $b\equiv 1(\mathrm{mod}\mathfrak{q})$, contradicting $b\in \mathfrak{p}\subset \mathfrak{q}$. This proves transitivity of the Galois action on primes without any unramified hypothesis.

Let $D={\mathrm{Stab}}_G(\mathfrak{q})$ be the decomposition group and $H=\mathrm{Gal}(F/L)$. Transitivity identifies the primes of $F$ above $\mathfrak{p}$ with the coset space $G/D$. Two such primes contract to the same prime in $L$ exactly when they are in the same $H$-orbit: apply the transitivity just proved to the Galois extension $F/L$. Therefore primes in an intermediate field as double cosets givesThe second equality follows by inversion of representatives. It allows one to count $D$-orbits on the embeddings of $L$ into $F$, represented by $G/H$. If $H$ is normal, this simplifies to $[G:HD]$; the double-coset formulation also handles nonnormal intermediate fields.

Write $\alpha ={\lambda }^{1/5}$, $M=\mathbb{Q}(\alpha )$ and $E=\mathbb{Q}({\zeta }_5)$. The Eisenstein criterion at $\lambda$ gives $[M:\mathbb{Q}]=5$, whereas the prime cyclotomic field degree gives $[E:\mathbb{Q}]=4$. The splitting field $F=E(\alpha )$ is a Galois extension of $E$, and an automorphism fixing $E$ sends $\alpha$ to ${\zeta }_5^b\alpha$. Thus its Galois group embeds in $C_5$. It is nontrivial, since $M$ cannot be contained in the degree-four field $E$. Therefore $[F:E]=5$ and $[F:\mathbb{Q}]=20$.

Index the five roots by $j\in {\mathbb{F}}_5$, with root ${\zeta }_5^j\alpha$. Every automorphism has the form ${\zeta }_5\mapsto {\zeta }_5^u$, $\alpha \mapsto {\zeta }_5^b\alpha$ and acts on indices as $j\mapsto uj+b$, with $u\in {\mathbb{F}}_5^{\times }$. This is a faithful inclusion of $G=\mathrm{Gal}(F/\mathbb{Q})$ into the affine group over a finite field, and both groups have order $20$, so it is an isomorphism. This is the affine Galois group of a prime-radical splitting field.

Let $D$ be the specified decomposition group. Its nonidentity element is an involution. If its linear coefficient were $1$, it would be a nontrivial translation, which has order $5$. Hence it acts as $j\mapsto -j+b$. There is exactly one fixed root, at $j=b/2$, and the other four roots form two pairs. By involution orbits in an affine group of odd prime degree, $D$ has three orbits on the five roots. Those roots identify with $G/\mathrm{Gal}(F/M)$, so the double-coset formula gives three primes in $M$.

For $E$, the embeddings identify with $G/\mathrm{Gal}(F/E)\simeq {\mathbb{F}}_5^{\times }$. The image of $D$ is $\{1,-1\}$ and acts by multiplication, giving the two orbits $\{1,-1\}$ and $\{2,-2\}$. ThereforeThe orbit argument counts primes whether or not $p$ is ramified; cyclicity of the given order-two decomposition group does not require that it be generated by an unramified Frobenius element.

For a positive integer $N$, a Dirichlet character $\psi$ modulo $N$ is a group homomorphism $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }\to {\mathbb{C}}^{\times }$, extended periodically to the integers and set equal to zero on integers not coprime to $N$. Its nonzero values are roots of unity. For $\mathrm{Re}s>1$, its Dirichlet L-function is the absolutely convergent Dirichlet seriesComplete multiplicativity of $\psi$ proves the Euler product by expanding each factor as a geometric series and applying unique integer factorization; absolute convergence justifies these rearrangements.

Suppose $\psi$ is a nonprincipal Dirichlet character. Choose a unit $u$ with $\psi (u)\ne 1$. Multiplication by $u$ permutes a complete residue system, so ${\sum }_{a=1}^N\psi (a)=\psi (u){\sum }_{a=1}^N\psi (a)$, and the sum is zero. Consequently $A(x)={\sum }_{n\le x}\psi (n)$ is bounded: complete periods contribute zero and the remaining incomplete period has at most $N$ terms. The Abel summation formula gives, initially for $\mathrm{Re}s>1$,For $\mathrm{Re}s>0$ the integral converges, locally uniformly together with its derivatives in $s$, because on a compact set with $\mathrm{Re}s\ge \delta >0$ the differentiated integrands are bounded by constants times $x^{-1-\delta }(1+\log x{)}^k$. Thus holomorphy of a Dirichlet series from bounded partial sums proves

For a nonreal $\psi$, the character ${\psi }^2$ is also nonprincipal. The supplied logarithmic Euler product yields, for real $\sigma >1$,At a prime dividing $N$ the numerator is $3$. At every other prime put $\psi (p{)}^k=e^{i\theta }$; then the numerator is $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. This Euler product positivity for L-function nonvanishing proves that the product inside the logarithm is at least one. If $L_N(\psi ,1)=0$, its holomorphy would give $L_N(\psi ,\sigma )=O(\sigma -1)$. Meanwhile $L_N({\psi }^2,\sigma )$ is bounded near one and the Riemann zeta function has a simple pole there. The product would therefore be $O(\sigma -1)$ and tend to zero, a contradiction.

For a real nonprincipal $\psi$, suppose instead that $L_N(\psi ,1)=0$. The product $H(s)=\zeta (s)L_N(\psi ,s)$ would then be holomorphic on the whole half-plane $\mathrm{Re}s>0$, because the zero cancels the only zeta pole there. In $\mathrm{Re}s>1$ it has coefficientsThese are nonnegative zeta-times-real-L coefficients. They are multiplicative, and their prime-power values are $k+1$ if $\psi (p)=1$, one or zero according as $k$ is even or odd if $\psi (p)=-1$, and one if $\psi (p)=0$. In particular $b_n\ge 0$ and $b_{m^2}\ge 1$ for every $m$.

The Taylor series of the assumed holomorphic $H$ about $2$ converges in the disk $|s-2|<2$. Termwise differentiation of its absolutely convergent Dirichlet series at $2$ gives $(-1{)}^k{H}^{(k)}(2)={\sum }_n{b}_n(\log n{)}^k/n^2$. Evaluate the Taylor series at $s=1/2$ and apply the Tonelli theorem to the resulting nonnegative terms:This contradicts holomorphy at $1/2$. The argument is the positive-coefficient mechanism of the Landau theorem for a Dirichlet series with nonnegative coefficients, with the needed step proved explicitly. Combining the real and nonreal cases gives the nonvanishing of a nonprincipal Dirichlet L-function at one:

Now let $a$ be coprime to $N$. The Dirichlet theorem on primes in arithmetic progressions asserts that infinitely many prime numbers satisfy $p\equiv a(\mathrm{mod}N)$. To prove it, separate the first-power terms in the given logarithmic Euler product:For $\sigma \ge 1$, the remainder is bounded uniformly byFor a nonprincipal character, holomorphy and nonvanishing near one give a local holomorphic logarithm. Along a short real interval to the right of one, this logarithm and the supplied continuous logarithm differ by a fixed integer multiple of $2\pi i$. Hence the prime-character sum near one is bounded:For the principal Dirichlet character ${\psi }_0$,The latter follows from the simple zeta pole and the bounded remainder. Finally Orthogonality of Dirichlet characters givesHere $\phi$ is the Euler totient function. The right-hand side tends to infinity as $\sigma \downarrow 1$, whereas a finite set of primes would give a bounded left-hand side. Therefore

Realize the two-dimensional irreducible representation $\rho$ as the standard representation of the symmetric group onwith $S_3$ permuting coordinates. It is the complement of the trivial summand in the three-point permutation representation. For each prime the local factor of an Artin L-function iswhere $I_p$ is the inertia group and Frobenius means any lift of the residue-field Frobenius automorphism. On $V^{I_p}$ the lift is independent of that choice. At an unramified prime, put $t=p^{-s}$. The Local Artin factors of the standard representation of S3 areThe polynomial discriminant of $X^3-3$ is $-243=-3^5$, so only $3$ can ramify in its splitting field. For $p\ne 3$, the squarefree factorization of $X^3-3$ modulo $p$ determines the Frobenius cycle type on its roots.

At $p=2$,The quadratic is irreducible, so Frobenius is a transposition andThus $a_2=0$, $a_4=1$ and $a_8=0$. At $p=5$,The quadratic discriminant is $3(\mathrm{mod}5)$, a nonsquare, so this is again a transposition. Consequently $L_5(\rho ,s)=(1-5^{-2s}{)}^{-1}$ and $a_5=0$.

At $p=7$, the possible cubes are $0,1,-1$, so $X^3-3$ has no root modulo $7$. Being cubic, it is irreducible; Frobenius is a three-cycle. Thereforeand $a_7=-1$.

The ramified factor at $3$ must be computed on inertia group invariants. Write $\alpha =3^{1/3}$ and set $\pi =({\zeta }_3-1)/\alpha$. Using $({\zeta }_3-1{)}^2=-3{\zeta }_3$ givesThe Eisenstein polynomial $X^6+3$ over ${\mathbb{Q}}_3$ has degree six, so ${\mathbb{Q}}_3(\pi )$ is a totally ramified degree-six subfield of ${\mathbb{Q}}_3({\zeta }_3,\alpha )$. The latter has degree at most $[F:\mathbb{Q}]=6$, so the fields agree. This Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3 shows $D_3=I_3=S_3$. Since the only permutation-invariant vectors in ${\mathbb{C}}^3$ have all coordinates equal, their intersection with $V$ is zero. ThusIn particular $a_3=a_9=0$; an unramified degree-two factor at $3$ would give the wrong coefficients.

Finally the Euler product makes its Dirichlet series coefficients multiplicative for coprime indices. Its constant coefficient is $a_1=1$, and $a_6=a_2{a}_3=0$, $a_{10}=a_2{a}_5=0$. Every integer up to ten is now covered, giving