Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 136 (Local Fields)

One form of the Hensel lemma is the following. Let $R$ be a complete discrete valuation ring with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $a_0\in R$ satisfiesthen there is a unique $a\in R$ with $f(a)=0$ and $a\equiv a_0(\mathrm{mod}\mathfrak{m})$.

Define $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Since $f^{\prime }(a_n)$ remains a unit, Taylor expansion givesso the valuations of the errors at least double. The corrections tend to zero, making $(a_n)$ a Cauchy sequence; completeness gives a limit $a$, and continuity gives $f(a)=0$. If $a,b$ are two such roots, thenwith $u\in \mathfrak{m}$, so the second factor is a unit and $a=b$.

Put $f(X)=X^3+5X-12$. Any root in ${\mathbb{Q}}_p$ is a p-adic integer: if its valuation were negative, $X^3$ would be the unique term of least valuation.

For $p=2$, reduction modulo two has roots zero and one. The root zero is simple because $f^{\prime }(0)=5$ is odd, so it lifts uniquely. An odd integer $x$ satisfies $x^2\equiv 1(\mathrm{mod}8)$, and henceThus there is no odd $2$-adic root and the number of roots is one.

For $p=3$,All three roots are simple because $f^{\prime }(X)=3X^2+5\equiv 2(\mathrm{mod}3)$. Each lifts uniquely, giving three roots in ${\mathbb{Q}}_3$.

For $p=5$, reduction gives $X^3-2$, whose unique root is $X=3$; it is simple because $3X^2\not\equiv 0(\mathrm{mod}5)$. It lifts uniquely, so there is one root in ${\mathbb{Q}}_5$.

For odd $p$ there is a decompositioninto the valuation factor, the Teichmuller representative factor, and the group of principal units. If $x$ is a $((p-1)m)$th power for every $m$ coprime to $p$, its valuation is divisible by every such $(p-1)m$, and is therefore zero. Its residue in ${\mathbb{F}}_p^{\times }$ is a $(p-1)$st power, hence is one. Thus $x\in 1+p{\mathbb{Z}}_p$.

Conversely, exponentiation by any integer $n$ coprime to $p$ is an automorphism of $1+p{\mathbb{Z}}_p$. This follows either from the principal-unit logarithm, under which it becomes multiplication by $n$, or by applying the Hensel lemma to $Y^n-x$. Since $(p-1)m$ is coprime to $p$, every $x\in 1+p{\mathbb{Z}}_p$ has a $((p-1)m)$th root for every allowed $m$.

The non-Archimedean part of the Ostrowski theorem says that every nontrivial Non-Archimedean absolute value on $\mathbb{Q}$ is equivalent to the p-adic absolute value for a unique prime $p$.

Indeed, $|n|\le 1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$, and there cannot be two such primes because the Bezout identity would make $1$ a sum of two terms of absolute value below one. If $a$ is coprime to $p$, another Bezout identity shows $|a|=1$. Consequentlyfor every $x\in {\mathbb{Q}}^{\times }$, which is a positive real power of $|x{|}_p$.

More generally, the Non-Archimedean absolute values on a number field $K$ are indexed, up to equivalence, by the nonzero prime ideals $\mathfrak{p}\subset {\mathcal{O}}_K$. The value attached to $\mathfrak{p}$ isTo prove completeness of the list, restrict an absolute value to $\mathbb{Q}$ and obtain a rational prime $p$. Its valuation ring contains ${\mathcal{O}}_K$ away from a unique prime above $p$; equivalently, its centeris a nonzero prime ideal. Since ${\mathcal{O}}_{K,\mathfrak{p}}$ is a discrete valuation ring, every $x\in K^{\times }$ is a unit times a power of a uniformizer, so the given value is equivalent to the displayed $\mathfrak{p}$-adic value. An absolute value trivial on $\mathbb{Q}$ is trivial on the algebraic extension $K$, so no further cases occur.

Choose a $K$-basis of $L$. Any extension $|\cdot {|}_L$ defines a norm on the finite-dimensional $K$-vector space $L$, and all norms on such a space over a complete valued field induce the same topology. Hence any two extensions induce the same topology and are equivalent absolute values. The coordinate sup norm is complete because $K$ is complete, so the equivalent topology defined by $|\cdot {|}_L$ is complete as well. This proves the unique extension of an absolute value to a finite extension and the completeness of $L$.

Completeness is essential. Give $K=\mathbb{Q}$ its $5$-adic value and take $L=\mathbb{Q}(i)$. Sincein $\mathbb{Z}[i]$, the two primes $(2+i)$ and $(2-i)$ define two inequivalent extensions. For example, $(2+i)/(2-i)$ has positive valuation for one and negative valuation for the other. This is the valuation ring need not equal the integral closure over a noncomplete field example.

For $a\in k$ and each $n$, choose a lift $x_n\in {\mathcal{O}}_K$ of the unique $p^n$th root $a^{p^{-n}}$, which exists because $k$ is a perfect field. DefineChanging $x_n$ by an element of the maximal ideal changes its $p^n$th power by an element whose valuation tends to infinity, so the limit exists and is independent of all choices. In characteristic $p$, the Frobenius endomorphism satisfies $(x+y{)}^{p^n}=x^{p^n}+y^{p^n}$, making $[\cdot ]$ a ring homomorphism lifting the identity on $k$. If $s:k\to {\mathcal{O}}_K$ is any other such section, thenfor every $n$, and the same limiting construction forces $s(a)=[a]$. This proves uniqueness of the Teichmuller lift.

Choose a uniformizer $t$. Repeatedly subtracting the lift of the residue and dividing by $t$ gives every $x\in {\mathcal{O}}_K$ a unique convergent Teichmuller expansionBecause the lift is a ring map, this identifies ${\mathcal{O}}_K$ with $k[[t]]$ and its fraction field with the Laurent series field $k((t))$. This is the equal-characteristic complete discretely valued field classification.

If $K$ is locally compact, its compact valuation ring has only finitely many disjoint residue-class balls. Thus $k$ is finite, as also follows from the local compactness criterion for a complete non-Archimedean field.

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. In particular $G_0$ is the inertia group, and $G_1$ is the wild inertia group. When the extension is totally ramified, the uniformizer criterion for lower ramification groups permits the equivalent test on one uniformizer.

LetThe extension $K_2/K$ is the unramified quadratic extension. Since ${\mathbb{F}}_{p^2}^{\times }$ contains all $(p^2-1)$st roots of unity, $L/K_2$ is a cyclic, tamely and totally ramified extension of degree $p^2-1$, with automorphisms $u\mapsto \zeta u$.

The Frobenius automorphism of ${\mathbb{F}}_{p^2}/{\mathbb{F}}_p$ extends by fixing $u$ and conjugates $\zeta$ to ${\zeta }^p$. Hence $L/K$ is Galois, its inertia group isand its residue-field Galois group isThe tame ramification also gives $G_1=1$.

Suppose first that $L=K(\alpha )$ and the degree-$n$ minimal polynomial of $\alpha$ is an Eisenstein polynomial. It is irreducible, $\alpha$ is a uniformizer of $L$, and its valuation shows that $e(L/K)\ge n=[L:K]$. Equality follows, so $L/K$ is a totally ramified extension.

Conversely, suppose $L/K$ is totally ramified of degree $n$ and choose a uniformizer $\alpha$ of $L$. The field $K(\alpha )$ already has ramification index at least $n$, so it equals $L$. Letbe the minimal polynomial. All conjugates of $\alpha$ have $L$-valuation one. Each $a_i$ with $i<n$ is an elementary symmetric polynomial in products of at least one conjugate and therefore has positive $L$-valuation. Since valuations of elements of $K$ are multiples of $n$, every $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverso $a_0$ is not divisible by the square of that ideal. Thus $f$ is Eisenstein, proving the Eisenstein generator of a totally ramified extension criterion.

Put ${\pi }_n={\zeta }_{p^n}-1$. The shifted cyclotomic polynomialis Eisenstein at $p$. Therefore it is irreducible, ${\pi }_n$ is a uniformizer, andso the extension is totally ramified. It is the splitting field of ${\Phi }_{p^n}$, and every automorphism is uniquelyThis proves the cyclotomic extension of a p-adic field isomorphismRestriction in the cyclotomic tower corresponds to reduction of $a$, so taking the inverse limit gives

The Local Kronecker-Weber theorem says that every finite abelian extension of ${\mathbb{Q}}_p$ is contained in a cyclotomic extension obtained by adjoining roots of unity. It identifies the totally ramified cyclotomic part throughand the maximal unramified part through its Frobenius generator.

Choose the normalization in which a uniformizer maps to arithmetic Frobenius. For $x=p^{r}u$ with $r\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_p^{\times }$, define ${\mathrm{Art}}_{{\mathbb{Q}}_p}(x)$ to act by ${\mathrm{Frob}}^r$ on the maximal unramified extension and byon every $p$-power root of unity. These compatible actions define an element of the abelian Weil group, because its residue action is an integral power of Frobenius. The resulting continuous homomorphismis the Local Artin map; reversing both Frobenius conventions replaces the displayed inverse by the corresponding opposite normalization.

A one-dimensional commutative formal group law over a ring $R$ is a series $F(X,Y)\in R[[X,Y]]$ satisfyingThe identity axiom gives $F(X,Y)=X+Y+$ terms of total degree at least two. SeekAfter $a_2,\dots ,a_{n-1}$ have been chosen, the coefficient of $X^n$ in $F(X,i(X))$ is $a_n$ plus a known expression in the earlier coefficients. There is therefore a unique choice of $a_n$ making it zero. Recursion constructs the formal inverse $i(X)$ with $F(X,i(X))=0$ and $i(X)\equiv -X(\mathrm{mod}{X}^2)$.

The formal additive group and formal multiplicative group have lawsIf $R$ is an algebra over a field whose scalar field is $\mathbb{Q}$, the serieshas linear coefficient one and satisfiesIt is therefore the exponential isomorphism between the additive and multiplicative formal groups, with inverse $\log (1+X)$.

Repeated formal addition givesfor positive $n$, and the same formulas extend to all $n\in \mathbb{Z}$ using the formal inverse.

Now let $R$ be a field of characteristic $p>0$ and let $h:{\hat{\mathbb{G}}}_a\to {\hat{\mathbb{G}}}_m$ be a homomorphism. Compatibility with multiplication by $p$ givesThe left side is $h(0)=0$, while the Frobenius identity gives the right side asThe formal power series ring over a field is an integral domain, so $h(X{)}^p=0$ implies $h=0$. Hence there are no nonzero homomorphisms from the formal additive group to the formal multiplicative group.