Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 136 (Local Fields)

The strong form of Hensel lemma says the following. Let $v$ be a discrete valuation on a complete field $K$. If $f\in {\mathcal{O}}_K[X]$ andthen $f$ has a root $x\in {\mathcal{O}}_K$ satisfying $v(x-x_0)>v(f^{\prime }(x_0))$.

To prove it, apply Newton iteration over a valued field:The initial inequality says that the first correction has valuation greater than $v(f^{\prime }(x_0))$. Taylor expansion then shows inductively that $v(f^{\prime }(x_n))=v(f^{\prime }(x_0))$, while the valuations of the corrections tend to infinity. Hence $(x_n)$ is a Cauchy sequence. Completeness gives a limit $x$, and continuity gives $f(x)=0$.

Now decompose the multiplicative group asThe P-adic valuation givesand cubing is the identity on $\{\pm 1\}$. Put $U_2=1+9{\mathbb{Z}}_3$. Expansion gives $U_1^3\subseteq U_2$. Conversely, for $u=1+9a\in U_2$, choose $b\equiv a(\mathrm{mod}3)$ and put $x_0=1+3b$. For $f(X)=X^3-u$,so the strong form of Hensel lemma produces a cube root in $U_1$. Thus $U_1^3=U_2$. Finally,is an isomorphism. Combining the valuation and principal-unit factors proves the cube-class group of the 3-adic numbers identity

Assume $|a{|}_2>1$, equivalently $v_2(a)<0$, and writeFor $f(Y)=Y^2-(1+4/a^4)$ at $Y=1$,The strong form of Hensel lemma therefore gives $y\in {\mathbb{Q}}_2$ with $y^2=1+4/a^4$. Taking $X=a^{2}y$ yields $X^2=a^4+4$. Hence every $a$ with $|a{|}_2>1$, and in particular every sufficiently large $a$, works.

For a finite extension $L/K$ of local fields, let $e=e(L/K)$ be its ramification index and $f=f(L/K)=[k_L:k_K]$ its residue-field degree. An unramified extension has $e=1$. A totally ramified extension has $e=[L:K]$, equivalently $f=1$. A tamely ramified extension has separable residue extension and ramification index coprime to the residue characteristic.

Let $\bar{\theta }$ generate the finite extension $k_L/k_K$, and let $\bar{g}$ be its minimal polynomial. Lift $\bar{g}$ to a monic $g\in {\mathcal{O}}_K[X]$ and choose any lift $t\in {\mathcal{O}}_L$ of $\bar{\theta }$. Since finite fields are perfect fields, ${\bar{g}}^{\prime }(\bar{\theta })\ne 0$. The simple-root form of Hensel lemma, applied inside $L$, gives $\theta \in {\mathcal{O}}_L$ withSet $K_0=K(\theta )$. Its residue field contains $k_K(\bar{\theta })=k_L$, soThe equation $g(\theta )=0$ gives the reverse inequality. Thus $[K_0:K]=f$, its residue-field degree is $f$, and $e(K_0/K)=1$; hence $K_0/K$ is unramified. Since $k_{K_0}=k_L$, the extension $L/K_0$ has residue-field degree one and is totally ramified. This constructs the maximal unramified subextension of a local field extension.

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. In lower numbering,These are the ramification groups; $G_0$ is the inertia group and $G_1$ is the wild inertia group.

Because ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$, every $x\in {\mathcal{O}}_L$ is $P(\alpha )$ for some $P\in {\mathcal{O}}_K[X]$. The polynomial identity $P(Y)-P(X)=(Y-X)Q(X,Y)$ has integral coefficients. Consequently the inequality for $\alpha$ implies it for every $x$, and the converse follows by taking $x=\alpha$. Therefore

Since $L/K$ is a Finite Galois extension, the minimal polynomial factors asDifferentiating and evaluating at $\alpha$ givesand henceFor a fixed nonidentity $\sigma$, its valuation is exactly the number of integers $s\ge 0$ for which $\sigma \in G_s$. Interchanging the two finite sums proves the ramification-group sum for a monogenic integer ring:

The extension is unramified exactly when $G_0=1$, which by this nonnegative sum is equivalent to $v_L(f^{\prime }(\alpha ))=0$, or $f^{\prime }(\alpha )\in {\mathcal{O}}_L^{\times }$. Moreover $|G_0|=e(L/K)$, so the $s=0$ term is $e(L/K)-1$. Equalityholds exactly when $G_1=1$, which is exactly tame ramification.

A Lubin–Tate series for $\pi$ is a power series $f(X)\in {\mathcal{O}}_K[[X]]$ such thatThe key Lubin–Tate lemma is that if $f$ and $g$ are such series and $L(X_1,\dots ,X_r)={\sum }_i{a}_i{X}_i$ with $a_i\in {\mathcal{O}}_K$, there is a unique series $F\in {\mathcal{O}}_K[[X_1,\dots ,X_r]]$ satisfying

Taking $g=f$ and $L=X+Y$ defines ${\mathcal{F}}_f(X,Y)$. Uniqueness applied to the two sides of each identity proves the identity, associativity, and commutativity axioms, so ${\mathcal{F}}_f$ is a formal group law. Taking $L=aX$ defines endomorphisms $[a{]}_f$, and uniqueness givesThus ${\mathcal{F}}_f$ is the Lubin–Tate formal group as a formal ${\mathcal{O}}_K$-module, with $[\pi {]}_f=f$.

For another Lubin–Tate series $g$, apply the lemma with $L=X$ to obtainUniqueness shows that $h$ respects both formal addition and every scalar endomorphism. Its linear coefficient is the unit one, so it has a compositional inverse; equivalently, applying the lemma with $f$ and $g$ reversed supplies the inverse. Hence this Lubin–Tate change of series is an isomorphism ${\mathcal{F}}_f\cong {\mathcal{F}}_g$ of formal ${\mathcal{O}}_K$-modules.

Define the Lubin–Tate torsion byThe scalar endomorphisms make this a module over ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$.

The Newton polygon or Weierstrass preparation theorem applied to the Lubin–Tate congruences shows that $[{\pi }^n{]}_f$ has exactly $q^n$ distinct roots in $\overline{\mathfrak{m}}$. More precisely, the quotient of the distinguished factors for $[{\pi }^n{]}_f$ and $[{\pi }^{n-1}{]}_f$ has degree $q^{n-1}(q-1)$, and its roots are precisely the points killed by ${\pi }^n$ but not by ${\pi }^{n-1}$.

Choose such a point ${\omega }_n$. If $a\in {\mathcal{O}}_K$, write $a={\pi }^{r}u$ with $u$ a unit. Since $[u{]}_f$ is an automorphism, $[a{]}_f({\omega }_n)=0$ exactly when $r\ge n$. Thusis injective. Both sides have $q^n$ elements, so it is an isomorphism of modules. Therefore ${\mu }_{f,n}$ is a free module of rank one.

Over $K={\mathbb{Q}}_p$, consider the two Lubin–Tate seriesBoth have linear term $pX$ and reduce to $X^p$ modulo $p$. The nonzero roots of $f$ are ${\zeta }_p^i-1$, while the nonzero roots of $g$ satisfy

The Lubin–Tate change of series $h:{\mathcal{F}}_f\to {\mathcal{F}}_g$ and its inverse have coefficients in ${\mathbb{Z}}_p$ and converge on the maximal ideal. They therefore give mutually inverse bijections between the first torsion sets without changing the fields generated by them. Hence the first Lubin–Tate torsion fields for the p-adic numbers satisfySince ${\mathbb{Q}}_p({\zeta }_p-1)={\mathbb{Q}}_p({\zeta }_p)$, the required equality follows.

Because $k$ is a perfect field, for each $a\in k$ choose a compatible sequenceand choose arbitrary lifts $x_n\in {\mathcal{O}}_K$ of $a_n$. If $u\equiv v(\mathrm{mod}{\pi }^m)$, the binomial theorem and the fact that the residue characteristic is $p$ giveIt follows that $x_{n+1}^{p^{n+1}}\equiv x_n^{p^n}(\mathrm{mod}{\pi }^{n+1})$. Thus $(x_n^{p^n})$ is Cauchy, and completeness definesThe same congruence shows that the limit is independent of all lift choices. Taking products before passing to the limit proves $[ab]=[a][b]$, and reduction gives $[a]\equiv a(\mathrm{mod}\pi )$.

For uniqueness, let $s,t:k\to {\mathcal{O}}_K$ be two multiplicative lifts. Given $a$ and any $n$, choose $b\in k$ with $b^{p^n}=a$. Since $s(b)\equiv t(b)(\mathrm{mod}\pi )$, repeated powering yieldsCompleteness and separation force $s(a)=t(a)$. This is the unique Teichmuller lift.

For $x\in {\mathcal{O}}_K$, let $a_0$ be its residue and put $x_1=(x-[a_0])/\pi$. Repeat with $x_1,x_2,\dots$. Induction givesThe remainder tends to zero, proving the Teichmuller expansionReduction after subtracting successive partial sums also proves uniqueness of the digits.

For sufficiently large $r$, the series for the p-adic logarithm and p-adic exponential converge on $1+{\pi }^r{\mathcal{O}}_K$ and ${\pi }^r{\mathcal{O}}_K$ and are inverse homomorphisms. Hence the principal-unit logarithm giveswhere the last map is division by ${\pi }^r$.

Now take $K={\mathbb{Q}}_3({\zeta }_3)$ and $\pi ={\zeta }_3-1$. This is a uniformizer, the residue field is ${\mathbb{F}}_3$, and its Teichmuller units are ${\mu }_2=\{\pm 1\}$. ThusThe image of ${\zeta }_3$ generates $U_1/U_2$, and ${\mu }_3\cap U_2=1$, so $U_1={\mu }_3\times U_2$. The logarithm and exponential already converge inversely on $U_2$, giving $U_2\cong ({\pi }^2{\mathcal{O}}_K,+)\cong ({\mathcal{O}}_K,+)$. Since ${\mu }_2\times {\mu }_3\cong {\mu }_6$, this proves the unit group of Q3 zeta3 decomposition

The Ostrowski theorem says that every nontrivial absolute value on a field defined on $\mathbb{Q}$ is equivalent either to the usual absolute value or to $|\cdot {|}_p$ for a unique prime $p$.

Let an absolute value on the number field $K$ extend $|\cdot {|}_p$. Its valuation ring determinesa prime ideal satisfying $\mathfrak{p}\cap \mathbb{Z}=(p)$. Conversely, each prime $\mathfrak{p}$ above $p$ defines the normalized absolute valuewhere $e_{\mathfrak{p}}=v_{\mathfrak{p}}(p)$. It restricts to $|\cdot {|}_p$. The correspondence between extensions and primes follows either from the valuation ring or from local factorization and extended absolute values; distinct primes give inequivalent valuations. Thus these $|\cdot {|}_{\mathfrak{p}}$ are exactly the extensions, up to equivalence.

For the tensor-product assertion, choose a primitive element of a field extension $\alpha$ for $K/\mathbb{Q}$, with minimal polynomial $f$. Because number fields are separable, over ${\mathbb{Q}}_p$ it factors into distinct irreduciblesindexed by the primes $\mathfrak{p}\mid p$. The Chinese remainder theorem givesThe $i$th factor is the completion of a number field at a prime ideal $K_{{\mathfrak{p}}_i}$. Under these identifications the isomorphism is the natural diagonal map $x\otimes a\mapsto (ax{)}_{\mathfrak{p}}$, proving the p-adic tensor decomposition of a number field

Let $\alpha =\sqrt[3]{2}$. Its minimal polynomial is $f(X)=X^3-2$, whose discriminant is $-108$. Since $5\nmid 108$, the prime $5$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$, so the Dedekind factorization theorem applies at $5$. In ${\mathbb{F}}_5[X]$,The quadratic factor has discriminant $3$, which is a quadratic nonresidue modulo $5$, so it is irreducible. ThereforewhereTheir residue-field degrees are one and two. Both factors of $f$ occur with multiplicity one, so both prime-ideal exponents are one. Hence neither ${\mathfrak{p}}_1$ nor ${\mathfrak{p}}_2$ is ramified.