Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 29 (Local Fields)

The structural facts used about a non-Archimedean local field are that it is complete for a normalized discrete valuation $v:K^{\times }\to \mathbb{Z}$, its valuation ring ${\mathcal{O}}_K=\{x:v(x)\ge 0\}$ is a discrete valuation ring, its maximal ideal is ${\mathfrak{m}}_K=\pi {\mathcal{O}}_K$ for a uniformizer $\pi$, and its residue field is a finite field $k={\mathbb{F}}_q$. Polynomial operations are continuous for this valuation topology.

A simple-root version of Hensel lemma is: if $f\in {\mathcal{O}}_K[X]$, $a_1\in {\mathcal{O}}_K$, $f(a_1)\in {\mathfrak{m}}_K$, and $f^{\prime }(a_1)\notin {\mathfrak{m}}_K$, then there is a unique $a\in {\mathcal{O}}_K$ with $f(a)=0$ and $a\equiv a_1(\mathrm{mod}{\mathfrak{m}}_K)$.

Construct $a_n$ with $f(a_n)\in {\pi }^n{\mathcal{O}}_K$. Given $a_n$, put $a_{n+1}=a_n+{\pi }^n{b}_n$. Taylor expansion over the valuation ring givesThe omitted terms contain ${\pi }^{2n}$, and $2n\ge n+1$ for $n\ge 1$. The residue of $f^{\prime }(a_n)$ remains the nonzero residue of $f^{\prime }(a_1)$, so choose $b_n$ withThis makes $f(a_{n+1})\in {\pi }^{n+1}{\mathcal{O}}_K$. The congruences $a_{n+1}\equiv a_n(\mathrm{mod}{\pi }^n)$ make $(a_n)$ a Cauchy sequence; completeness gives its limit $a\in {\mathcal{O}}_K$. Continuity and $v(f(a_n))\ge n$ give $f(a)=0$, and the first congruence gives its specified residue class.

For uniqueness, if $a,b$ are two roots in that class, polynomial division of their difference givesThe second factor is a unit, since $f^{\prime }(b)$ is a unit and $a-b\in {\mathfrak{m}}_K$. Thus $a=b$. This proves both existence and uniqueness in the stated Hensel lemma.

Write the finite residue field as $k={\mathbb{F}}_q$, with $q=p^f$ and residue characteristic $p$. Every $\bar{a}\in k$ satisfies ${\bar{a}}^q=\bar{a}$. The polynomial $F(X)=X^q-X$ has derivative $q{X}^{q-1}-1$, whose reduction is $-1$. Hence Hensel lemma gives a unique lift $[\bar{a}]\in {\mathcal{O}}_K$ such thatThis includes $[0]=0$ and $[1]=1$. If $\bar{a}\ne 0$, its lift is a unit and satisfies $[\bar{a}{]}^{q-1}=1$.

The product $[\bar{a}][\bar{b}]$ also satisfies $X^q=X$ and has residue $\bar{a}\bar{b}$. Uniqueness therefore gives $[\bar{a}\bar{b}]=[\bar{a}][\bar{b}]$. Reduction is a left inverse of this map, so the map is injective. Restricting to nonzero residues proves the requested multiplicative Teichmuller lift:Its image is the group of $(q-1)$st roots of unity in $K$. Indeed a root of unity of that order is a unit satisfying $X^q=X$, hence is the unique lift of its residue. The only finite-field fact used here is $a^q=a$; for $a\ne 0$ it follows from Lagrange's theorem in the multiplicative group of order $q-1$.

Now suppose $\mathrm{char}K=p>0$. The Teichmuller lifts constructed in (b) are additive as well as multiplicative: if $a^q=a$ and $b^q=b$, then the characteristic-$p$ Frobenius identity gives $(a+b{)}^q=a^q+b^q=a+b$. Uniqueness of the lift with prescribed residue yields $[\bar{a}+\bar{b}]=[\bar{a}]+[\bar{b}]$. Their set is therefore a finite coefficient field in positive-characteristic local fields, and reduction identifies it with $k$.

Choose a uniformizer $\pi$. For $x\in {\mathcal{O}}_K$, let $a_0$ be the lift of its residue and write $x=a_0+\pi x_1$ with $x_1\in {\mathcal{O}}_K$. Repeat with $x_1$, then $x_2$, and so on. This givesThe remainder tends to zero, and completeness gives $x={\sum }_{j\ge 0}{a}_j{\pi }^j$. The expansion is unique: in a nonzero difference of two series the first different coefficient is a nonzero element of the coefficient field, hence a unit, and so gives exactly the valuation of that difference.

Consequently substitution $t\mapsto \pi$ defines a bijection $k[[t]]\to {\mathcal{O}}_K$. It is a ring homomorphism because the coefficient lift is a field homomorphism, polynomial operations agree before passage to limits, and multiplication is continuous. Every element of $K$ becomes integral after multiplication by a sufficiently large power of $\pi$. Allowing finitely many negative powers therefore extends substitution to the field isomorphismIt preserves the normalized discrete valuation, since the first nonzero coefficient determines the valuation on each side. The results used are the structural facts stated in (a), its proved Hensel lemma, the proved multiplicative lifting in (b), and the characteristic-$p$ identity $(a+b{)}^q=a^q+b^q$. No classification theorem for local fields is needed.

A Non-Archimedean absolute value on a field is a function $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ with $|x|=0$ exactly for $x=0$, $|xy|=|x||y|$, and the ultrametric inequality $|x+y|\le \max (|x|,|y|)$. It is nontrivial if some nonzero element has value different from one. Equivalent absolute values induce the same topology; for nontrivial absolute values this is equivalent to $|x{|}_2=|x{|}_1^c$ for some $c>0$.

Here is also a proof of the latter equivalence in the case needed. A positive power plainly preserves the open neighborhoods of zero. Conversely, if the topologies coincide, then $|x{|}_1<1$ if and only if $x^n\to 0$ in the first topology, hence if and only if $|x{|}_2<1$. Choose $a$ with $0<|a{|}_1<1$. For every $x\ne 0$, $m\in \mathbb{Z}$ and $n\ge 1$, whether $|x^n/a^m|<1$ agrees for the two values. Comparison with all rational numbers $m/n$ therefore identifies the two ratios $\log |x{|}_j/\log |a{|}_j$. Thus $|x{|}_2=|x{|}_1^c$ with $c=\log |a{|}_2/\log |a{|}_1>0$.

Now let $|\cdot |$ be nontrivial and non-Archimedean on $\mathbb{Q}$. Since $|1|=1$ and every positive integer is a sum of ones, $|n|\le 1$ for every integer. If all nonzero integers had value one, multiplicativity would make the absolute value trivial on all rationals. Hence some integer has value less than one; its prime factorization shows that some prime $p$ has $|p|<1$.

There cannot be two distinct such primes $p,q$. A Bezout identity $ap+bq=1$ with $a,b\in \mathbb{Z}$ would giveThus every other prime has value one. Prime factorization of the numerator and denominator of $x\in {\mathbb{Q}}^{\times }$ now givesConversely $|x{|}_p=p^{-v_p(x)}$ is multiplicative and satisfies the ultrametric inequality, since $v_p(x+y)\ge \min (v_p(x),v_p(y))$. Distinct primes give inequivalent values: $p^m\to 0$ for the $p$-adic absolute value but $|p^m{|}_q=1$ for $q\ne p$. We have proved the non-Archimedean part of Ostrowski theorem:There is exactly one equivalence class for each prime.

A basis of open sets in the p-adic integers isIndeed this is the set of $x$ with $v_p(x-a)\ge n$, equivalently $|x-a{|}_p<p^{-(n-1)}$; these neighborhoods become arbitrarily small. Reduction identifies ${\mathbb{Z}}_p/p^n{\mathbb{Z}}_p$ with $\mathbb{Z}/p^n\mathbb{Z}$, so each has index $p^n$.

If a subgroup $H\le ({\mathbb{Z}}_p,+)$ is open, a neighborhood of zero of the displayed form lies inside $H$. Thus $p^n{\mathbb{Z}}_p\subseteq H$ for some $n$, and $[{\mathbb{Z}}_p:H]\le p^n$ is finite.

Conversely suppose the abstract index is the finite integer $m$. The quotient is a finite abelian group of order $m$, so Lagrange's theorem gives $mx\in H$ for every $x\in {\mathbb{Z}}_p$. Write $m=p^{r}u$ with $p\nmid u$. Multiplication by the p-adic unit $u$ is a bijection of ${\mathbb{Z}}_p$, its inverse being multiplication by $u^{-1}\in {\mathbb{Z}}_p$. ConsequentlyThe subgroup $H$ is a union of open cosets of $p^r{\mathbb{Z}}_p$, and is therefore open. This proves the characterization of open subgroups of the p-adic integers:No prior assumption of closedness is required for the finite-index direction.

For an odd integer $x=2a+1$, $x^2=1+4a(a+1)\equiv 1(\mathrm{mod}8)$. The same conclusion holds for every 2-adic unit by reduction modulo eight. Thus an odd square must be congruent to one modulo eight.

Conversely let $u\in {\mathbb{Z}}_2^{\times }$ with $u\equiv 1(\mathrm{mod}8)$. For digit lifting for square roots of 2-adic units, start with $x_3=1$. If $x_n$ is odd and $x_n^2\equiv u(\mathrm{mod}{2}^n)$, considerFor $n\ge 3$,since the remaining term is divisible by $2^{2n-2}$ and $2n-2\ge n+1$. The odd multiplier $x_n$ lets the choice of ${\epsilon }_n$ correct exactly the next binary digit. Thus $x_{n+1}^2\equiv u(\mathrm{mod}{2}^{n+1})$. The corrections have valuations tending to infinity, so completeness gives $x=\lim x_n\in {\mathbb{Z}}_2^{\times }$ with $x^2=u$.

The unit $3$ is congruent to one modulo two but is not a square, and $5$ is congruent to one modulo four but is not a square, by the necessary modulo-eight condition. Therefore

Every nonzero P-adic number in ${\mathbb{Q}}_2$ is uniquely $2^{m}u$ with $m\in \mathbb{Z}$ and $u$ a 2-adic unit. A square has even valuation, and the unit square criterion shows that the unit class is determined by its odd residue modulo eight. The four residues form $C_2\times C_2$, generated by $-1$ and $5$; the valuation parity supplies the third factor, generated by $2$. Hence the square-class group of the 2-adic numbers isIts eight classes are represented by $1,-1,5,-5,2,-2,10,-10$. They are distinct: the first four have even valuation and distinct unit residues modulo eight, while multiplication by $2$ gives odd valuation and the same four distinct unit residues.

In characteristic different from two, completing the square in a quadratic minimal polynomial makes every quadratic extension $K(\sqrt{d})$ for a nonsquare $d\in K$. If $K(\sqrt{d})=K(\sqrt{e})$, write $\sqrt{e}=a+b\sqrt{d}$ and square. The coefficient $2ab$ must vanish. Since $e$ is nonsquare, $b\ne 0$, so $a=0$ and $e/d=b^2$. Conversely a square ratio plainly gives the same field. Thus the seven quadratic extensions of the 2-adic numbers are exactly

The unique unramified extension in this list is ${\mathbb{Q}}_2(\sqrt{5})$. To verify this, $\alpha =(1+\sqrt{5})/2$ is integral and satisfies $X^2-X-1=0$. Its reduction $X^2+X+1$ is irreducible over ${\mathbb{F}}_2$, so its residue field contains ${\mathbb{F}}_4$. The standard degree formula for finite extensions of complete discretely valued fields with perfect residue field is $[L:K]=e(L/K)f(L/K)$. Here the field degree is two, so the residue degree is two and the ramification index is one.

Conversely an unramified quadratic extension has residue field ${\mathbb{F}}_4$. A root there of $X^2+X+1$ lifts by Hensel lemma to a root of $X^2-X-1$, whose derivative is a unit. For this lift $\beta$, $(2\beta -1{)}^2=5$, so the extension contains ${\mathbb{Q}}_2(\sqrt{5})$ and must equal it. Every other field in the list is therefore ramified.

For the requested ramification groups, choose $L={\mathbb{Q}}_2(\sqrt{2})$ and $\pi =\sqrt{2}$. We use uniqueness of the extension of the valuation to a finite extension, together with the degree formula just stated. Since $v_2(\pi )=1/2$, the extension has ramification index two and residue degree one. Normalize $v_L(\pi )=1$, giving $v_L(2)=2$. Every element is $a+b\pi$ with $a,b\in {\mathbb{Q}}_2$, and the two summands have valuations of opposite parity:Thus $a+b\pi$ is integral exactly when $a,b\in {\mathbb{Z}}_2$, proving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$ directly.

The nonidentity automorphism $\sigma$ negates $\pi$. For any integral $a+b\pi$,of valuation at least three, with equality when $b$ is a unit. By the definition $G_i=\{\sigma :v_L(\sigma x-x)\ge i+1\text{ for every }x\in {\mathcal{O}}_L\}$, the ramification in the quadratic extension generated by the square root of two is thereforeThe lower ramification break is two; the valuation used here is the integral normalization on $L$.

For the finite Galois extension, normalize the discrete valuation $v_L$ to have value group $\mathbb{Z}$. The inertia group isThe lower ramification numbering isThe wild inertia group is $P=G_1$, and the tame inertia quotient is $I/P$. We will prove that $P$ is a $p$-group, where $p=\mathrm{char}{k}_K$, and that $I/P$ is cyclic of order prime to $p$. Tame inertia here is a quotient, not a specified canonical subgroup complementary to $P$.

First reduce the inertia calculation to a totally ramified extension. Choose $\bar{b}$ generating the finite residue extension $k_L/k_K$. Its minimal polynomial is separable. Lift that monic polynomial to $f\in {\mathcal{O}}_K[X]$, and use Hensel lemma in ${\mathcal{O}}_L$ to obtain a root $b$ with residue $\bar{b}$. Irreducibility of the reduction implies irreducibility of $f$ over $K$, by the monic form of the Gauss lemma for polynomials over a discrete valuation ring. Thus $K_0=K(b)$ has degree $[k_L:k_K]$ and residue field $k_L$; the degree formula makes $K_0/K$ unramified. An automorphism fixes $b$ precisely when it fixes $\bar{b}$: the nontrivial direction is uniqueness in Hensel lemma. Hence $I=\mathrm{Gal}(L/K_0)$, and $L/K_0$ is totally ramified.

Use the permitted integral-generation fact ${\mathcal{O}}_L={\mathcal{O}}_{K_0}[{\pi }_L]$. Every element of $I$ fixes the coefficients in this expression. The identity $\sigma ({\pi }_L^j)-{\pi }_L^j=(\sigma {\pi }_L-{\pi }_L){\sum }_{h=0}^{j-1}(\sigma {\pi }_L{)}^h{\pi }_L^{j-1-h}$ consequently proves the uniformizer criterion for lower ramification groups:Necessity follows by testing $x={\pi }_L$, and the displayed power identity proves sufficiency for every integral polynomial in ${\pi }_L$.

Define the uniformizer character of inertiaIt is multiplicative: writing $u_{\tau }=\tau {\pi }_L/{\pi }_L$, one has $\sigma \tau {\pi }_L/{\pi }_L=\sigma (u_{\tau })(\sigma {\pi }_L/{\pi }_L)$, and $\sigma$ acts trivially on residue units. Its kernel is exactly $v_L(\sigma {\pi }_L-{\pi }_L)\ge 2$, namely $G_1$. Changing the uniformizer to $u{\pi }_L$ leaves the residue character unchanged, since $\overline{\sigma (u)/u}=1$. ThereforeEvery finite subgroup of a field's multiplicative group is cyclic: if its exponent is $m$, every element is a root of $X^m-1$, so its order is at most $m$; a finite abelian group has an element of order its exponent, constructed by multiplying elements of maximal prime-power orders, so its order is at least $m$. Equality makes it cyclic. Here $k_L^{\times }$ has order $|k_L|-1$, prime to $p$. We have proved

To verify the assertion about wild inertia as well, for $i\ge 1$ defineWrite $\sigma {\pi }_L={\pi }_L+a{\pi }_L^{i+1}$ and $\tau {\pi }_L={\pi }_L+b{\pi }_L^{i+1}$. Composition gives $\sigma \tau {\pi }_L\equiv {\pi }_L+(a+b){\pi }_L^{i+1}(\mathrm{mod}{\pi }_L^{i+2})$, because inertia fixes residues and $i\ge 1$. Thus ${\theta }_i$ is additive with kernel $G_{i+1}$. Thus positive ramification quotients are elementary abelian: each $G_i/G_{i+1}$ is a subgroup of the additive finite field, hence an elementary abelian $p$-group.

Finally the filtration terminates: a nonidentity element of $I$ cannot fix ${\pi }_L$, since it already fixes $K_0$ and ${\mathcal{O}}_L={\mathcal{O}}_{K_0}[{\pi }_L]$. Its displacement has finite valuation, and there are only finitely many automorphisms. Iterating the finite $p$-group quotients gives that $P$ is a $p$-group. It is normal in $I$ and has prime-to-$p$ quotient, so it is its unique Sylow $p$-subgroup, justifying the wild inertia terminology.

Put $K=\mathbb{C}((t))$. This is a complete discretely valued field, with valuation ring $\mathbb{C}[[t]]$ and residue field $\mathbb{C}$; its infinite residue field means it is not a locally compact local field. The finite-extension facts needed here are uniqueness of the extended valuation, completeness of the extension, and $[L:K]=ef$ for complete discretely valued fields with perfect residue field. They apply in this characteristic-zero setting.

Let $[L:K]=n$. The finite residue extension of $\mathbb{C}$ must be $\mathbb{C}$ itself, so $f=1$ and $e=n$. Normalize $v_L$ integrally and choose a uniformizer $\pi$. Since $v_L(t)=n$, write $t=u{\pi }^n$ with $u\in {\mathcal{O}}_L^{\times }$.

The residue of $u$ has an $n$th root $b_0\in {\mathbb{C}}^{\times }$. Apply Hensel lemma to $X^n-u$: its derivative at $b_0$ is the unit $n{b}_0^{n-1}$ because the residue characteristic is zero. Hence there is $v\in {\mathcal{O}}_L^{\times }$ with $v^n=u$. Setting $s=v\pi$ givesThe elements $1,s,\dots ,s^{n-1}$ are linearly independent over $K$. In a nontrivial relation ${\sum }_{i=0}^{n-1}{a}_i{s}^i=0$, the nonzero summands have valuations $n{v}_K(a_i)+i$, distinct modulo $n$, so exactly one has least valuation and cannot cancel. Since $L$ has dimension $n$, this proves $L=K(s)$.

The constants already embed $\mathbb{C}$ into ${\mathcal{O}}_L$ with residue field $\mathbb{C}$. Subtracting the residue coefficient and dividing by $s$ repeatedly, exactly as in Question 1(c), expresses every integral element uniquely as ${\sum }_{j\ge 0}{c}_j{s}^j$ with $c_j\in \mathbb{C}$. Completeness and the first-nonzero-coefficient valuation show that allowing finite negative powers givesThis proof of finite extensions of the complex Laurent series field did not need a Galois hypothesis. All $n$ distinct $n$th roots of unity lie in the constant field, and the substitutions $s\mapsto \zeta s$ preserve $s^n=t$. They give all $n$ automorphisms, so every such finite extension is cyclic Galois:

Choose compatible roots $s_n=t^{1/n}$ in an algebraic closure, first choosing them successively along the factorial-degree tower and then taking powers to define the other roots, with $s_n=s_m^{m/n}$ whenever $n\mid m$. The fields $K_n=K(s_n)$ are nested under divisibility. Every finite extension inside the closure equals one of them: its constructed root $s$ of $X^n-t$ differs from $s_n$ by a constant root of unity. Therefore their union is the entire algebraic closure. Explicitly, every algebraic element generates a finite extension, so lies in the union; any polynomial over the union has its coefficients in one finite extension, and its roots are still algebraic over $K$, hence also lie there.

For $n\mid m$, restriction of automorphisms sends ${\zeta }_m\in {\mu }_m$ to ${\zeta }_m^{m/n}\in {\mu }_n$. Using the compatible primitive roots $\exp (2\pi i/n)$, these are the reduction maps on exponent groups. Thus the absolute Galois group of the complex Laurent series field isThe inverse limit carries its profinite topology; choosing compatible roots gives the displayed identification with the profinite completion of the integers.