Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 123 (Algebraic Number Theory)

Use the simple-root form of the Hensel lemma: for $F\in {\mathcal{O}}_K[T]$ and $\bar{a}\in k_K$ satisfying $\overline{F}(\bar{a})=0$ and $\overline{F^{\prime }}(\bar{a})\ne 0$, there is a unique $a\in {\mathcal{O}}_K$ with $\bar{a}$ as its reduction and $F(a)=0$.

Let $d=[\ell :k_K]$. The residue field $k_K$ is finite, so $\ell /k_K$ is separable, and the primitive element theorem gives $\ell =k_K(\bar{\alpha })$. Let $\bar{F}$ be the monic irreducible polynomial of $\bar{\alpha }$, of degree $d$, and choose a monic coefficient lift $F\in {\mathcal{O}}_K[T]$. Irreducible reduction implies that $F$ is irreducible over $K$: any monic factorization over $K$ has integral coefficients, since all its roots are integral over the valuation ring, and would reduce to a factorization of $\bar{F}$.

Adjoin a root $\alpha$ and put $L=K(\alpha )$. Then $[L:K]=d$ and $\alpha \in {\mathcal{O}}_L$. Its reduction satisfies $\bar{F}$, so $k_L$ contains a copy of $\ell$ over $k_K$. For finite extensions of non-Archimedean local fields, the fundamental degree relation isThe residue degree is at least $d$, while the left side equals $d$, henceThis constructs the required unramified extension.

For uniqueness, let $L^{\prime }/K$ be another unramified extension with a chosen identification $k_{L^{\prime }}\cong \ell$. The element corresponding to $\bar{\alpha }$ is a simple residue root of $F$, because finite fields are perfect. The Hensel lemma in $L^{\prime }$ gives a root ${\alpha }^{\prime }\in {\mathcal{O}}_{L^{\prime }}$ with this reduction. Since $F$ is irreducible of degree $d$, $[K({\alpha }^{\prime }):K]=d=[L^{\prime }:K]$, so $L^{\prime }=K({\alpha }^{\prime })$. Sending $\alpha$ to ${\alpha }^{\prime }$ yields a $K$-isomorphism $L\to L^{\prime }$, compatible with the prescribed residue identification. ThusThis proves the existence and uniqueness of unramified local extensions using the stated version of Hensel's lemma, without assuming the classification in advance.

Write $q=\#k_K=p^f$. For every $a\in k_K$, the polynomial $F(T)=T^q-T$ has $a$ as a residue root and $\overline{F^{\prime }}=-1$. By the Hensel lemma there is a unique element $[a]\in {\mathcal{O}}_K$ such thatThis defines the Teichmuller lift, including $[0]=0$. The two displayed properties determine each value uniquely by Hensel's uniqueness statement.

They also imply the standard Teichmuller lift properties. The elements $0,1$ already satisfy their defining equations, so $[0]=0$ and $[1]=1$. The product $[a][b]$ reduces to $ab$ and satisfies $([a][b]{)}^q=[a][b]$, so uniqueness gives $[ab]=[a][b]$. Likewise $[a{]}^p$ reduces to $a^p$ and is fixed by its $q$th power, givingFor $a\ne 0$, division of $[a{]}^q=[a]$ by $[a]$ gives $[a{]}^{q-1}=1$. Thus the nonzero Teichmuller representatives are precisely the prime-to-$p$ roots of unity lifting the nonzero residue classes. This is a multiplicative section of reduction; additivity is not asserted in mixed characteristic.

For the convergence, use the normalized discrete valuation $v_K$ with $v_K(\pi )=1$, and put $e=v_K(p)\ge 1$. A useful binomial estimate isIndeed, writing $u=v+\delta$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$, so all terms of $(v+\delta {)}^p-v^p$ have valuation at least $\min \{e+r,pr\}$. The ultrametric inequality gives the bound for their sum.

Finite fields are perfect, so the successive $p$th roots $x_n$ exist uniquely and satisfy $x_n^{p^n}=x$. Let $a_n=[x_n]$. Since $y_n-a_n$ lies in the maximal ideal, iterating the estimate $n$ times givesMultiplicativity, or the Frobenius identity above, gives $a_n^{p^n}=[x_n^{p^n}]=[x]$. ThereforeThis proves the approximation of Teichmuller lifts by iterated pth powers for completely arbitrary choices of the lifts $y_n$, including $x=0$. The original PDF confirms $x\in k_K$, correcting the capital $K_K$ in the TeX transcription.

Give ${\overline{K}}_{\mathfrak{p}}$ the unique extension of the absolute value of $K_{\mathfrak{p}}$. We use two standard facts: a complete non-Archimedean absolute value extends uniquely to every finite algebraic extension, and any two embeddings of a finite separable extension into an algebraic closure are conjugate by an automorphism of that algebraic closure.

A $K$-embedding $\tau :L\to {\overline{K}}_{\mathfrak{p}}$ pulls back this absolute value to a finite place of a number field on $L$ extending the given place of $K$. By the permitted equivalence of places and primes, it determines a prime ideal ${\mathfrak{q}}_{\tau }$ above $\mathfrak{p}$. Every element of $\mathrm{Gal}({\overline{K}}_{\mathfrak{p}}/K_{\mathfrak{p}})$ preserves the absolute value by uniqueness, so ${\mathfrak{q}}_{\tau }$ is constant on each orbit.

Every prime $\mathfrak{q}$ above $\mathfrak{p}$ occurs: its completion of a number field at a prime ideal $L_{\mathfrak{q}}$ is a finite extension of $K_{\mathfrak{p}}$ and admits a $K_{\mathfrak{p}}$-embedding into ${\overline{K}}_{\mathfrak{p}}$. Composing with $L\hookrightarrow L_{\mathfrak{q}}$ gives an embedding inducing $\mathfrak{q}$.

Conversely, suppose two embeddings induce the same prime. Their pulled-back absolute values have the same normalization on $K$, so they extend continuously to embeddings of the same field $L_{\mathfrak{q}}$. To see that their images are still inside the algebraic closure rather than only its completion, the closure of $\tau (L)$ is exactly the finite complete field $K_{\mathfrak{p}}\tau (L)$. The two resulting embeddings of $L_{\mathfrak{q}}$ are conjugate by the stated separable-extension fact. Restricting the conjugacy to $L$ proves that the original embeddings are in the same orbit. Thus the prime ideals and local embedding orbits are naturally identified:

If $L=K(\alpha )$, an embedding is determined by the image of $\alpha$, which may be any root of its minimal polynomial $f$. The orbits on these roots are exactly the root sets of the monic irreducible factors of $f$ over $K_{\mathfrak{p}}$. HenceIf $f_j$ corresponds to ${\mathfrak{q}}_j$, then $L_{{\mathfrak{q}}_j}\cong K_{\mathfrak{p}}[T]/(f_j)$ and $[L_{{\mathfrak{q}}_j}:K_{\mathfrak{p}}]=\mathrm{deg}{f}_j$. This is the local factorization and extended absolute values correspondence and does not require an integral power basis.

The polynomial is Eisenstein at $2$: every nonleading coefficient is even, and its constant term $126$ is not divisible by $4$. Therefore it is irreducible over $\mathbb{Q}$ andBy part (a), the number of primes above a rational prime is the number of irreducible factors over the corresponding $p$-adic field.

We use the Newton polygon theorem in this form: a segment of slope $-r$ and horizontal length $d$ accounts for $d$ roots of valuation $r$, counted with multiplicity, in an algebraic closure with $v_p(p)=1$. We also use the denominator criterion for Newton polygon irreducibility: if all roots have valuation $a/n$ with $\gcd (a,n)=1$, any monic factor of degree $d$ has constant-term valuation $da/n\in \mathbb{Z}$, hence $n\mid d$.

At $3$, the nonzero coefficient points areThe lower hull is the single segment from $(0,2)$ to $(5,0)$, with slope $-2/5$. All five roots therefore have valuation $2/5$. A proper factor of degree $d$ would require $5\mid d$, which is impossible for $1\le d<5$. Thus $g$ is irreducible over ${\mathbb{Q}}_3$, givingThe completion has ramification index $5$ and residue degree $1$: the value $2/5$ forces its ramification index to be divisible by $5$, and its degree is $5$.

At $7$, the nonzero coefficient points areThe lower hull consists of a length-three segment of slope $-1/3$ and a length-two horizontal segment. Thus three roots have valuation $1/3$, and two have valuation zero. Reduction givesBoth $1$ and $-1$ are simple roots, so the Hensel lemma gives two distinct linear factors over ${\mathbb{Q}}_7$. The remaining cubic accounts for all the valuation-$1/3$ roots and is irreducible by the same denominator argument. Hence the local factor degrees are $1,1,3$, andThe two linear factors give $(e,f)=(1,1)$, and the cubic gives $(e,f)=(3,1)$. These data sum to degree five, as required. This computes the local prime decomposition of T5 plus 6T3 plus 252T2 plus 126. Counting distinct factors merely modulo $3$ would not justify the result, since that reduction is $T^5$.

The only possible rational roots of the monic cubic are $1,-1$, and neither is a root. Thus $f$ is irreducible. The discriminant of a depressed cubic givesIt is not a rational square. By the permitted cubic discriminant criterion, the splitting field has Galois group $S_3$, degree six, and contains its quadratic subfield generated by $\sqrt{\Delta }$. Consequently

For a rational prime $r\ne 59$, the reduction of $f$ is separable because its discriminant is a unit at $r$. Choose a finite extension of ${\mathbb{F}}_r$ in which that reduction splits. Part 1(a) lifts it to an unramified extension of ${\mathbb{Q}}_r$, and the Hensel lemma lifts all the simple roots into that extension. The local splitting field is therefore unramified over ${\mathbb{Q}}_r$. In particular all primes of $L/K$ above such $r$ are unramified.

At $59$, the supplied simple residue root $-28$ lifts to a root in ${\mathbb{Q}}_{59}$. The remaining factor is quadratic, so the local splitting field has degree at most two over ${\mathbb{Q}}_{59}$. It contains $\sqrt{-59}$, which is not a square in ${\mathbb{Q}}_{59}$ because its valuation is odd. It is therefore exactly ${\mathbb{Q}}_{59}(\sqrt{-59})$. This is also the completion of $K$ at its unique prime above $59$: the polynomial $T^2+59$ is Eisenstein there. Thusfor a prime $\mathfrak{P}$ above $\mathfrak{p}$. All the other primes above it give the same local conclusion by Galois conjugacy. The relative extension is therefore unramified at $59$ as well; indeed that prime splits completely in $L/K$.

The field $K$ has only complex places, so there is no infinite-place ramification. Hence $\boxed{L/K\text{ is unramified everywhere}}$.

We use the Hilbert class field theorem: for a number field $E$, its maximal everywhere unramified abelian extension $H_E$ is finite, contains every such extension, and satisfies $\mathrm{Gal}(H_E/E)\cong {\mathrm{Cl}}_E$, so $[H_E:E]=h_E$. The cyclic degree-three extension just constructed is contained in $H_K$. ThereforeThis is the unramified cubic extension of Q of square root minus fifty-nine; only divisibility of the class number, rather than its exact value, is needed here.

Use the Hilbert class field theorem stated in part (a), and let $H_L$ be the Hilbert class field of $L$. Put $F=M\cap H_L$ inside a common algebraic closure.

Any intermediate extension of $M/L$ is totally ramified at the given prime. To verify this even when $M/L$ is not Galois, there is a unique prime of $F$ above $\mathfrak{p}$, its residue field embeds into the residue field at the unique prime of $M$ and therefore equals $k_{\mathfrak{p}}$, and the fundamental identity for prime decomposition gives ramification index $[F:L]$. On the other hand $F/L$ is a subextension of $H_L/L$, so it is unramified there. Its ramification index is consequently both $[F:L]$ and $1$, forcingBecause $H_L/L$ is Galois, this intersection identity givesThus $M{H}_L/M$ is abelian. We use the standard local fact that unramified extensions remain unramified under finite base extension: after base change each factor is obtained by lifting an extension of the residue field. Since $H_L/L$ is unramified at every finite prime, $M{H}_L/M$ is also unramified at every finite prime. There are no real places of $M$, since every embedding of $M$ restricts to an embedding of the totally imaginary field $L$. Hence there is no infinite-place obstruction.

The maximality of $H_M$ now implies $M{H}_L\subseteq H_M$. Taking degrees over $M$ proves the class number divisibility under total ramification:The argument uses the degree of an unramified class-field extension, and does not assume that extension of ideals induces an injection of ideal class groups.

We use the following standard facts about cyclotomic fields. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ isAlso $\mathbb{Q}({\zeta }_{{\ell }^r})$ has degree $\phi ({\ell }^r)={\ell }^{r-1}(\ell -1)$, and $\ell$ is totally ramified, with a unique prime and residue field ${\mathbb{F}}_{\ell }$. These are the quadratic subfield of a prime cyclotomic field and total ramification in a prime-power cyclotomic field results.

TakeSince $59\equiv 3(\mathrm{mod}4)$, every $E_r$ contains $K=\mathbb{Q}(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot 59^{r-1}$, the index in $K/\mathbb{Q}$ is $2$, and the relative index is thereforeThe field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. ThusThe ring of integers of a number field is a Dedekind domain, and is a principal ideal domain exactly when its ideal class group is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot 59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of nonprincipal cyclotomic fields of conductor a prime power is

Normalize the discrete valuation by $v_L({\pi }_L)=1$, and put $v_L(0)=+\infty$. For $\sigma \ne 1$, define its ramification numberand set $i_G(1)=+\infty$. The minimum exists: all finite values are nonnegative integers, and a nonidentity automorphism moves some integral element. The lower ramification numbering isThus $G_s=G_{\lceil s\rceil }$, $G_0$ is the inertia group, and $G_1$ is the wild inertia group. This states the convention at integer endpoints as well as between them. Automorphisms preserve $v_L$, so these are normal subgroups of $G$.

Define the Herbrand function and its inverse byThe integrand is positive and piecewise constant. Since $G_t=1$ for sufficiently large $t$, the function is continuous, strictly increasing and unbounded, and its inverse is defined on all nonnegative reals. The upper ramification numbering isIn particular, the denominator in the Herbrand integral involves $G_0$, even when $L/K$ is not totally ramified. These conventions are also given in Milne's Class Field Theory, Chapter I, §4.

Now use the monogenic hypothesis ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$. Since $L/K$ is Galois, its minimal polynomial has the distinct conjugates $\sigma (\alpha )$ as roots, soTaking valuations proves the first equality.

For every polynomial $P\in {\mathcal{O}}_K[T]$, the difference $P(\sigma (\alpha ))-P(\alpha )$ is divisible in ${\mathcal{O}}_L$ by $\sigma (\alpha )-\alpha$. Every element of ${\mathcal{O}}_L$ is such a polynomial value, and taking the element $\alpha$ attains the resulting lower bound. HenceFor an integer $i_G(\sigma )=m\ge 0$, the element $\sigma$ lies in exactly the groups $G_0,\dots ,G_{m-1}$; if $m=0$, it lies in none. Summing this count over the nonidentity elements and interchanging the two finite sums givesThis proves the ramification-group sum for a monogenic integer ring. Equivalently it is the different exponent from ramification groups, since the different ideal is generated by $f^{\prime }(\alpha )$ under the same monogenic hypothesis.

The polynomial is Eisenstein at $3$, so a root $\alpha$ generates a totally ramified cubic field $F={\mathbb{Q}}_3(\alpha )$ and is a uniformizer of $F$. Its cubic discriminant isIts $3$-adic valuation is $3$, so it is not a square. The cubic discriminant criterion therefore givesThe unique quadratic subfield is generated by $\sqrt{\Delta }$. Since the permitted square root $\sqrt{-5}$ is in ${\mathbb{Q}}_3$, this subfield isIt is totally ramified of degree two, by the Eisenstein polynomial $T^2-3$. The ramification index of $M/{\mathbb{Q}}_3$ is divisible by both the index $3$ of $F$ and the index $2$ of $E$. Since the whole degree is $6$, it follows thatIn particular $G_0=G$.

We use the following standard facts about totally ramified extensions: every uniformizer generates the valuation ring over the base valuation ring, and for a totally ramified Galois extensionThus we can use the uniformizer criterion for lower ramification groups. Put $s=\sqrt{3}$. Since $v_M(\alpha )=2$ and $v_M(s)=3$,is a uniformizer of $M$.

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3$ be the three roots. The Galois group acts transitively on their unordered pairs and preserves the valuation, so all differences between distinct roots have the same valuation $d$. The discriminant identity giveshence $d=3$. Thus, for any $\sigma$, either $\sigma (\alpha )=\alpha$ or $\sigma (\alpha )-\alpha$ has valuation $3$.

A nontrivial three-cycle fixes $s$, since it fixes the quadratic subfield, and sends $\alpha$ to a different root. Also $v_M(\sigma (\alpha )+\alpha )=2$: write the sum as $2\alpha +(\sigma (\alpha )-\alpha )$, whose two terms have unequal valuations $2,3$. ThereforeA transposition changes the sign of $s$. ConsequentlyThe numerator has valuation $4$: it is $2{\alpha }^2$ plus terms of strictly higher valuation if the root changes, and exactly $2{\alpha }^2$ if the root is fixed. Thus every transposition has $v_M(\sigma (\pi )-\pi )=1$. We conclude that the ramification groups of the splitting field of T3 minus 3T plus 3 over Q3 are

As an independent check, the cubic field has ${\mathcal{O}}_F={\mathbb{Z}}_3[\alpha ]$ and different exponent $v_F(g^{\prime }(\alpha ))=v_F(3({\alpha }^2-1))=3$. The extension $M/F$ is tamely totally ramified of degree two, so its different exponent is $1$. The different in a tower gives $d(M/{\mathbb{Q}}_3)=1+2\cdot 3=7$, agreeing with the ramification sum $(6-1)+(3-1)=7$.

The decomposition group of the chosen prime is its stabilizer under the Galois group:The transitivity of the Galois action on primes identifies the primes of $M$ above $\mathfrak{p}$ with $G/D$ through $gD\mapsto g\mathfrak{P}$.

Put $H=\mathrm{Gal}(M/L)$. Contraction of prime ideals to ${\mathcal{O}}_L$ is constant on every $H$-orbit, because every element of $H$ fixes $L$ pointwise. It is surjective onto the primes of $L$ above $\mathfrak{p}$, since prime ideals extend in the integral extension ${\mathcal{O}}_M/{\mathcal{O}}_L$. Moreover, two primes of $M$ contracting to the same prime of $L$ are conjugate by $H$: the extension $M/L$ is Galois and the same transitivity theorem applies to it.

Thus the fibers of contraction are exactly the $H$-orbits on $G/D$. These orbits are the double cosets $H\backslash G/D$, proving the primes in an intermediate field as double cosets correspondenceExplicitly, replacing $g$ by $hgd$ does not change the contracted prime. Conversely, if $g\mathfrak{P}$ and $g^{\prime }\mathfrak{P}$ contract to the same prime, some $h\in H$ sends the former to the latter, so $g^{\prime -1}hg\in D$ and $H{g}^{\prime }D=HgD$. This also verifies the stated order of the two sides of the double coset.

A prime is completely split in an extension of degree $n$ exactly when there are $n$ primes above it. Indeed, the fundamental identity for prime decomposition is ${\sum }_{\mathfrak{q}\mid \mathfrak{p}}{e}_{\mathfrak{q}}{f}_{\mathfrak{q}}=n$, and every summand is a positive integer. Thus $n$ primes means that every ramification index and residue degree equals one, and conversely.

By part (a), the number of primes of $L$ above $\mathfrak{p}$ is the number of right $D$-orbits on $H\backslash G$. This set has $[G:H]=[L:K]$ elements. It has that many orbits exactly when every orbit is a singleton, equivalently whenThis condition says $gD{g}^{-1}\subseteq H$ for every $g$, orthe subgroup core.

Because $M$ is the Galois closure of $L/K$, it is generated by all the fields $g(L)$. The fixed field of $gH{g}^{-1}$ is $g(L)$, so the fixed field of the core is their compositum, namely $M$. The Galois correspondence therefore gives ${\mathrm{Core}}_G(H)=1$. HenceIn the Galois extension $M/K$, the number of primes above $\mathfrak{p}$ is $[G:D]$, so $D=1$ is equivalent to complete splitting in $M$. This proves that complete splitting is preserved by Galois closure:No prior unramifiedness assumption is needed: complete splitting already forces all ramification indices to be one, and the decomposition-group argument includes the ramified case.