Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 81 (Algebraic Number Theory)

Let $L/K$ be an extension of number fields, let $\alpha \in {\mathcal{O}}_L$ be a primitive element of a field extension, and let $f\in {\mathcal{O}}_K[X]$ be its monic minimal polynomial of an algebraic element. Put $R={\mathcal{O}}_K[\alpha ]$ and $S={\mathcal{O}}_L$. The Kummer-Dedekind theorem applies at a nonzero prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_K$ when $R_{\mathfrak{p}}=S_{\mathfrak{p}}$: equivalently, $\mathfrak{p}$ is outside the support of the finite-index quotient $S/R$. Write the factorization over the residue field $k={\mathcal{O}}_K/\mathfrak{p}$ aswith distinct monic irreducible polynomials ${\overline{g}}_j$, and choose monic lifts $g_j\in {\mathcal{O}}_K[X]$. ThenHere $f({\mathfrak{P}}_j/\mathfrak{p})$ denotes the residue-field degree, and the ramification index of a prime ideal is $m_j$.

To prove this, the local index condition givesThe final isomorphism is the Chinese remainder theorem. Each factor is a local ring with unique maximal ideal generated by ${\overline{g}}_j$, so its inverse image in $S$ is exactly ${\mathfrak{P}}_j$, and its residue field is $k[X]/({\overline{g}}_j)$. This proves the list of primes and the residue-field degrees. For the exponents, localize at ${\mathfrak{P}}_j$. The ring of integers of a number field is a Dedekind domain, so $S_{{\mathfrak{P}}_j}$ is a discrete valuation ring. If $\mathfrak{p}{S}_{{\mathfrak{P}}_j}={\mathfrak{P}}_j^{e_j}{S}_{{\mathfrak{P}}_j}$, the quotient has $e_j$ successive layers isomorphic to $S/{\mathfrak{P}}_j$, and hence $k$-dimension $e_j\mathrm{deg}{\overline{g}}_j$. The corresponding polynomial factor has $k$-dimension $m_j\mathrm{deg}{\overline{g}}_j$. Thus $e_j=m_j$, completing the proof rather than merely identifying the radicals.

For the calculation, set $K=\mathbb{Q}(\sqrt{5})$ and $t=(\sqrt{5}-1)/2$. The ring of integers of a quadratic field gives ${\mathcal{O}}_K=\mathbb{Z}[t]$, with $t^2+t-1=0$. The algebraic norm of $4+\sqrt{5}$ is $11$, so its principal ideal has ideal norm $11$. Its quotient ring has prime order and is therefore the finite field ${\mathbb{F}}_{11}$; consequently $(4+\sqrt{5})$ is a prime ideal. In that residue field, $\sqrt{5}=-4=7$ and $t=3$.

Let $L=\mathbb{Q}({\zeta }_5)$. Since $t={\zeta }_5+{\zeta }_5^{-1}$ lies in $\mathbb{Z}[{\zeta }_5]$, the permitted identity ${\mathcal{O}}_L=\mathbb{Z}[{\zeta }_5]$ gives ${\mathcal{O}}_L={\mathcal{O}}_K[{\zeta }_5]$. Thus the local index condition holds at every prime. The relative minimal polynomial of an algebraic element is $X^2-tX+1$, whose reduction isThe roots are distinct. Applying the Kummer-Dedekind theorem givesFor the last equality, evaluation at $5$ or $9$ in ${\mathbb{F}}_{11}$ gives ${\zeta }_5+{\zeta }_5^{-1}=3$ and therefore $4+\sqrt{5}=5+2t=0$. Both descriptions are the same kernels of these evaluation maps. Each prime has ideal norm $11$, residue-field degree one and ramification index of a prime ideal one.

Let $F/K$ be a finite Galois extension, choose a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$, and write $k_{\mathfrak{q}}$ and $k_{\mathfrak{p}}$ for the residue fields. The decomposition group and inertia group areThe Frobenius automorphism on the finite field $k_{\mathfrak{q}}$ is $x\mapsto x^{N\mathfrak{p}}$. Its inverse image defines a distinguished coset in $D_{\mathfrak{q}}/I_{\mathfrak{q}}$; a Frobenius element is a lift of that coset. At an unramified prime, $I_{\mathfrak{q}}=1$ and the lift is unique. Changing the prime above $\mathfrak{p}$ conjugates it, so a Frobenius conjugacy class is intrinsically attached to $\mathfrak{p}$. These are arithmetic Frobenius conventions.

Write $e=e(\mathfrak{q}/\mathfrak{p})$, $f=f(\mathfrak{q}/\mathfrak{p})$ and $g$ for the number of primes above $\mathfrak{p}$. The Galois group acts transitively on those primes, with stabilizer $D_{\mathfrak{q}}$. The fundamental identity for prime decomposition and the orbit-stabilizer theorem therefore giveIt remains to establish that the residue action has image of order $f$, rather than assume this in the desired equality.

Use the decomposition group of a prime and local Galois group identification $D_{\mathfrak{q}}=\mathrm{Gal}(F_{\mathfrak{q}}/K_{\mathfrak{p}})$. A generator of the finite field extension $k_{\mathfrak{q}}/k_{\mathfrak{p}}$ has a separable monic minimal polynomial of an algebraic element. Lift its coefficients to ${\mathcal{O}}_{K_{\mathfrak{p}}}$. The Hensel lemma lifts each of its distinct residue roots to $F_{\mathfrak{q}}$. One such lift has field degree at most $f$, while its residue generates the degree-$f$ finite field extension, so its field degree is exactly $f$ and its ramification index of a prime ideal is one. It therefore generates an unramified extension of degree $f$, and sending it to each of the other lifts gives its $f$ embeddings. Because $F_{\mathfrak{q}}/K_{\mathfrak{p}}$ is a Galois extension, these embeddings extend to its automorphisms. Their reductions realize every automorphism of $k_{\mathfrak{q}}/k_{\mathfrak{p}}$. Thus the residue action is surjective, and

Now let $G$ be cyclic of order ${\ell }^2$, and let $H=\mathrm{Gal}(F/L)$ be its unique subgroup of order $\ell$. In a tower of Galois extensions, inertia maps onto inertia in the intermediate extension. This also follows from the just-proved order formula: its image has order $e(F/K)/e(F/L)=e(L/K)$, using multiplicativity of the ramification index of a prime ideal in a tower. If $\mathfrak{p}$ ramifies in $L/K$, this image in $G/H$ is nontrivial. Every proper subgroup of $G$ is contained in $H$, so inertia upstairs must be $G$. HenceThe fundamental identity for prime decomposition then also forces a single prime above $\mathfrak{p}$ and residue-field degree one.

For a direct check of unramified behavior, the cyclotomic polynomial ${\Phi }_{25}$ divides $X^{25}-1$. Modulo $7$, the latter has derivative $25X^{24}=4X^{24}$ and has no common root with that derivative. Thus ${\overline{\Phi }}_{25}$ is separable, so $7$ does not divide its polynomial discriminant. The field discriminant divides this polynomial discriminant, and therefore $7$ is unramified in $\mathbb{Q}({\zeta }_{25})$. The Frobenius automorphism satisfiesIndeed its residue action is seventh powering. The reduced $25$th roots of unity are distinct because $7\nmid 25$, so the residue congruence identifies the exponent uniquely modulo $25$.

The Galois group is $(\mathbb{Z}/25\mathbb{Z}{)}^{\times }$, of order $\phi (25)=20$. Since $7^2=49\equiv -1(\mathrm{mod}25)$, $7^4\equiv 1$ but $7^2\not\equiv 1$. Consequently the Frobenius automorphism has order four. An unramified decomposition group is generated by that element, and the orbit-stabilizer theorem givesThere are therefore five primes above $7$, each with residue field ${\mathbb{F}}_{7^4}$.

For a number field $K$ and a modulus of a number field $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, let $I_K(\mathfrak{m})$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{K,1}(\mathfrak{m})$ consist of principal fractional ideals $(a)$ with $a\equiv 1$ at the finite modulus and $a>0$ at the real places in ${\mathfrak{m}}_{\infty }$. The ray class group is ${\mathrm{Cl}}_{\mathfrak{m}}(K)=I_K(\mathfrak{m})/P_{K,1}(\mathfrak{m})$.

The Main theorem of class field theory states that there is a unique finite abelian ray class field $K_{\mathfrak{m}}/K$ for which the Artin reciprocity map inducesIt is unramified outside the finite modulus and has no ramification at real places omitted from ${\mathfrak{m}}_{\infty }$. More generally, subgroups $H$ with $P_{K,1}(\mathfrak{m})\subseteq H\subseteq I_K(\mathfrak{m})$ correspond, with inclusions reversed, to intermediate abelian fields, with Galois group $I_K(\mathfrak{m})/H$. Every finite abelian extension occurs for some modulus. Equivalently, $K_{\mathfrak{m}}$ is the maximal abelian extension with conductor of an abelian extension dividing $\mathfrak{m}$. This states both reciprocity and the existence theorem of global class field theory.

There is no infinite component in the modulus $(5)$. The ordinary ideal class group of $\mathbb{Q}$ is trivial, and its unit group is $\{\pm 1\}$. The ray class exact sequence therefore givesConcretely, a fractional ideal prime to $5$ has a rational generator $r=a/b$ with $5\nmid ab$. Reducing $r$ modulo $5$ is well-defined up to multiplication by the units $\pm 1$. Its ray class is trivial precisely when one of the two generators $r,-r$ is congruent to one. Thus the quotient has the two classes $\{1,4\}$ and $\{2,3\}$. No positivity restriction is imposed on a generator.

To identify the corresponding field, put $E=\mathbb{Q}({\zeta }_5+{\zeta }_5^{-1})=\mathbb{Q}(\sqrt{5})$. The maximal real subfield of the fifth cyclotomic field is fixed by complex conjugation, the element $-1$ of $(\mathbb{Z}/5\mathbb{Z}{)}^{\times }$. For a prime $\ell \ne 5$, the Frobenius automorphism of $\mathbb{Q}({\zeta }_5)/\mathbb{Q}$ is ${\zeta }_5\mapsto {\zeta }_5^{\ell }$. Its restriction to $E$ thus corresponds exactly to the residue class of $\ell$ modulo $\{\pm 1\}$. Multiplicativity extends this to the Artin reciprocity map on all fractional ideals prime to $5$, whose kernel is exactly $P_{\mathbb{Q},1}((5))$. Its field discriminant $5$ also confirms that $E$ is unramified outside $5$; moreover $E$ is a totally real number field. The uniqueness in the Main theorem of class field theory now gives

Including the real place changes the ray class group: a principal generator must then be positive as well as congruent to one. The residue and signature group of a modulus givesHere the isomorphism sends $(u,\epsilon )$ to $\epsilon u$. Consequently ${\mathbb{Q}}_{(5)\infty }=\mathbb{Q}({\zeta }_5)$, of degree four. The finite modulus gives the real quadratic field; the modulus with the real place gives the full cyclotomic field. The printed finite-modulus assertion needs no correction.

Set $K=\mathbb{Q}(\sqrt{pq})$ and $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$. The distinct primes give independent square classes, so $[E:\mathbb{Q}]=4$ and $[E:K]=2$. In particular $K(\sqrt{p})=K(\sqrt{q})=E$: these are the same quadratic extension of $K$. We prove it is unramified by an explicit field discriminant calculation.

Since $p,q\equiv 1(\mathrm{mod}4)$, put $a=(1+\sqrt{p})/2$, $b=(1+\sqrt{q})/2$. Both are algebraic integers. The ring $R=\mathbb{Z}[a,b]\subseteq {\mathcal{O}}_E$ has basis $1,a,b,ab$. Up to the ordering of the basis, its trace pairing matrix is the tensor product of the quadratic trace matrices, because the four embeddings of $E$ choose the two radical signs independently. The two quadratic matrices areThusOn the other hand, $pq$ is squarefree and congruent to one modulo four, so $d_K=pq$. Let $j=[{\mathcal{O}}_E:R]$. The discriminant-index formula for an integral lattice and the relative discriminant tower formula giveThe ideal norm on the right is a positive integer, while $j$ is a positive integer. Hence $N({\mathfrak{d}}_{E/K})=1/j^2$ forces $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Its prime divisors are exactly the ramified finite primes, so $E/K$ is unramified at every finite prime, including primes above $2,p,q$. This also proves $d_E=p^2{q}^2$, agreeing with the discriminant of a biquadratic field formula.

Every embedding of $E$ is real, so the real places of $K$ split as well. The extension is quadratic, hence abelian. The ordinary Hilbert class field characterization therefore impliesFinally, the Artin reciprocity map identifies $\mathrm{Cl}(K)$ with $\mathrm{Gal}(H_K/K)$. Restriction onto $\mathrm{Gal}(E/K)\simeq C_2$ is surjective, so the ideal class group has a quotient of order two. Therefore $\boxed{|\mathrm{Cl}(K)|\text{ is even}}$.

We prove the Dirichlet density form of the Chebotarev density theorem. Let $L/K$ be a finite Galois extension with Galois group $G$, let $C$ be a conjugacy class in $G$, and let $S_C$ consist of the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. The assertion isThe same result holds for any union of conjugacy classes, by addition. The following proof uses exactly the allowed one-dimensional Artin L-function behavior at $1$ and the permitted induction and Euler-product properties.

First determine the order of vanishing at $1$ for an arbitrary character of a representation $\chi$ of $G$. The Brauer induction theorem supplies subgroups $H_j\le G$, linear characters ${\psi }_j$ of $H_j$, and integers $n_j$, possibly negative, such thatPut $K_j=L^{H_j}$. Compatibility of Artin L-functions with induced representations and direct sums gives, initially for $\mathrm{Re}s>1$,Each ${\psi }_j$ is a one-dimensional Galois representation for $L/K_j$, so the stated analytic assumptions apply to every factor. A trivial ${\psi }_j$ contributes a simple pole; a nontrivial ${\psi }_j$ is holomorphic and nonzero at one. Thus the product is meromorphic near one andThe character inner product is $⟨\alpha ,\beta {⟩}_G=|G{|}^{-1}{\sum }_{g\in G}\alpha (g)\overline{\beta (g)}$. Applying Frobenius reciprocity to each induced term givesIn particular, a nontrivial irreducible character has order zero, so its Artin L-function is holomorphic and nonzero near one. For the trivial character, $L_K(1_G,s)={\zeta }_K(s)$, the Dedekind zeta function of the base field $K$, and its pole is simple. This argument establishes the behavior needed at one, not holomorphy everywhere for arbitrary Artin L-functions.

Next convert that order calculation into a prime sum. For an irreducible character $\chi$, choose a unitary representation of dimension $d_{\chi }$ affording it, and choose a Frobenius automorphism $g_{\mathfrak{p}}$ at each unramified prime. The logarithm specified by the absolutely convergent Euler product gives, for real $s>1$,Only finitely many ramified primes occur. Their Euler-factor logarithms $R_{\chi }(s)$ stay bounded near one, since the relevant eigenvalues have modulus one and $(N\mathfrak{p}{)}^{-1}<1$. Also $|\chi (g)|\le d_{\chi }$, so the contribution with $m\ge 2$ is uniformly bounded for $s\ge 1$ byConvergence follows, for example, from the convergent ideal series defining ${\zeta }_K(2)$. ThereforeWriting $a_{\chi }=⟨\chi ,1_G{⟩}_G$, the order computation means $L_K(\chi ,s)=(s-1{)}^{-a_{\chi }}{U}_{\chi }(s)$, where $U_{\chi }$ is holomorphic and nonzero near one. Choose its local logarithm. Along the real interval approaching one, this and the Euler-product logarithm differ only by a fixed multiple of $2\pi i$. ConsequentlyOnly the trivial irreducible character has $a_{\chi }=1$; all the others have $a_{\chi }=0$.

Finally, character orthogonality makes the irreducible characters an orthonormal basis of class functions. For $c\in C$, the coefficient of $\chi$ in the class indicator is $⟨1_C,\chi {⟩}_G=(|C|/|G|)\overline{\chi (c)}$. HenceSum this identity over the unramified primes with weight $(N\mathfrak{p}{)}^{-s}$. The finite character sum and the boxed estimates giveDividing by $\log (1/(s-1))$ proves the asserted Dirichlet density. The same Euler-product calculation with ${\zeta }_K$ gives ${\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-s}=\log (1/(s-1))+O(1)$, so using all primes as the denominator is equivalent. Finitely many ramified primes do not affect either limit. No prime-counting asymptotic or natural-density Tauberian step is required by this proof.