Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 (Local Fields)

The inverse different isIt is an ${\mathcal{O}}_L$-submodule of $L$. Choose an integral basis $e_1,\dots ,e_n$ of a finite-index free submodule of ${\mathcal{O}}_L$. Nondegeneracy of the trace pairing gives a dual $K$-basis $e_1^{\ast },\dots ,e_n^{\ast }$, and the codifferent lies between two finitely generated full ${\mathcal{O}}_K$-lattices obtained from these bases. It is therefore a fractional ${\mathcal{O}}_L$-ideal.

Every algebraic integer has integral trace, so ${\mathcal{O}}_L\subseteq {\mathfrak{D}}_{L/K}^{-1}$. Consequently its inverseis contained in ${\mathcal{O}}_L$. It is thus an integral ${\mathcal{O}}_L$-ideal, called the different ideal.

Let $n=[L:K]$ and write $g(X)={\prod }_{j=1}^n(X-{\alpha }_j)$. Lagrange interpolation, followed by summing over the conjugates, shows that the trace-dual of the power basis $1,\alpha ,\dots ,{\alpha }^{n-1}$ is contained in $g^{\prime }(\alpha {)}^{-1}{\mathcal{O}}_K[\alpha ]$ and has the same determinant. Hence

For $L=\mathbb{Q}(\sqrt{3})$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{3}]$. Taking $g=X^2-3$ givesFor $L=\mathbb{Q}(\sqrt{5})$, use $\alpha =(1+\sqrt{5})/2$ and $g=X^2-X-1$. Then

The ring $\mathbb{Z}[T]$ is Noetherian and integrally closed, but it is not a Dedekind domain because it has Krull dimension two. Concretely, the nonzero prime ideal $(2)$ is properly contained in the prime ideal $(2,T)$ and is therefore not maximal.

The element $T$ belongs to the fraction field of $R=\mathbb{C}[T^2,T^3]$ and is integral over $R$, since it satisfies the monic polynomial $X^2-T^2$. But $T\notin R$. Thus $R$ is not integrally closed domain and hence is not a Dedekind domain.

The ring $\mathbb{Z}[{\zeta }_3]$ is the full ring of integers of the quadratic number field $\mathbb{Q}(\sqrt{-3})$. Every ring of integers of a number field is a Dedekind domain, so this ring is Dedekind.

Suppose first that the valuation is discrete and the residue field $k$ is finite. For a uniformizer $\pi$, every quotient ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$ is finite, and completeness givesThis inverse limit is compact. Since ${\mathcal{O}}_K$ is a compact neighborhood of zero, $K$ is locally compact.

Conversely, local compactness gives a compact ball about zero, which can be rescaled to make ${\mathcal{O}}_K$ compact. Its distinct residue classes are disjoint open balls of radius below one, so compactness forces the residue field to be finite. Cover ${\mathcal{O}}_K$ by finitely many balls of some radius $r<1$. Applying the ultrametric inequality to centers lying in the maximal ideal produces $\rho <1$ such that every nonunit has absolute value at most $\rho$. Hence the value group has a largest value below one, and the valuation is discrete. This proves the local compactness criterion for a complete non-Archimedean field.

An algebraically closed valued field has an $n$th root of every element. If its valuation were discrete and $\pi$ were a uniformizer, then $v(\sqrt[n]{\pi })=1/n$ would contradict discreteness. Therefore an algebraically closed non-Archimedean field cannot be locally compact.

If $|\cdot |$ is non-Archimedean, then $|n|=|1+\cdots +1|\le 1$. Conversely suppose $|n|\le M$ for every integer $n$. The binomial theorem and the ordinary triangle inequality giveTaking $N$th roots and letting $N\to \infty$ proves $|x+y|\le \max (|x|,|y|)$. This is the bounded-integer criterion for a non-Archimedean absolute value.

In characteristic $p$, the image of $\mathbb{Z}$ is the finite prime field ${\mathbb{F}}_p$, so every absolute value is bounded on it. Thus every absolute value on ${\mathbb{F}}_p(t)$ is non-Archimedean.

The polynomialhas no root in ${\mathbb{F}}_3$ and is therefore irreducible. Definefor any fixed $c>1$. Its valuation ring has residue fieldThe completion at an irreducible polynomial over a finite field identifies the completion with ${\mathbb{F}}_{27}((T))$, where $T$ corresponds to the uniformizer $h(t)$.

Let ${\mathfrak{m}}_K$ be the maximal ideal. For all sufficiently large $r$, the convergent p-adic logarithm and p-adic exponential give the principal-unit logarithm isomorphismMultiplication by a power of a uniformizer identifies the additive group ${\mathfrak{m}}_K^r$ with $({\mathcal{O}}_K,+)$. Since $1+{\mathfrak{m}}_K^r$ has finite index in ${\mathcal{O}}_K^{\times }$, it is the required subgroup.

Put $\alpha =\sqrt[p]{p}$. The polynomial $X^p-p$ is Eisenstein, so ${\mathbb{Q}}_p(\alpha )/{\mathbb{Q}}_p$ is totally ramified of degree $p$. The cyclotomic extension ${\mathbb{Q}}_p({\zeta }_p)/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$. Their coprime degrees make their intersection trivial, sohas degree $p(p-1)$ and is totally ramified. It is the splitting field of $X^p-p$, hence Galois.

Normalize $v_K$ by $v_K(K^{\times })=\mathbb{Z}$. Thensois a uniformizer. Write an automorphism aswhere $a\in {\mathbb{F}}_p^{\times }$ and $b\in {\mathbb{F}}_p$. Sincethe uniformizer criterion for lower ramification groups gives valuation one for $\sigma (\varpi )-\varpi$ when $a\ne 1$, and valuation $p+1$ when $a=1$, $b\ne 0$. Thereforeand $G_i=1$ for $i\ge p+1$. These are the ramification groups of the splitting field of Xp minus p over the p-adic numbers.

One strong form of the Hensel lemma is this: if a complete discretely valued field $K$, a polynomial $f\in {\mathcal{O}}_K[X]$, and $a\in {\mathcal{O}}_K$ satisfythen there is a unique root $\alpha$ in the ball $v(\alpha -a)>v(f^{\prime }(a))$.

Set $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Taylor expansion shows that the valuation of the error at least doubles at each step, while $v(f^{\prime }(a_n))$ remains constant. Thus the corrections tend to zero geometrically, so completeness gives a limit $\alpha$. Continuity gives $f(\alpha )=0$. Applying the same Taylor estimate to two roots in the stated ball proves uniqueness. This is Newton iteration over a valued field.

Write $x={\pi }^{r}u$ with $r\in \mathbb{Z}$ and $u\in {\mathcal{O}}_K^{\times }$. If $x$ is an $m$th power, then $m$ divides $r$. Divisibility by infinitely many $m$ forces $r=0$, so $x$ is a unit.

Conversely, if $m$ is coprime to both the residue characteristic $p$ and $q-1$, exponentiation by $m$ is an automorphism on the finite residue-unit group and on every principal unit quotient; equivalently, use Hensel's lemma on $Y^m-u$. Hence every unit is an $m$th power for infinitely many such $m$. Thereforeas recorded by elements that are powers of infinitely many degrees in a local field.

Put $Y=X^2$. Modulo two, $Y^2+9Y-2$ has the two simple roots zero and one, so Hensel lemma lifts them to roots $y_0,y_1\in {\mathbb{Z}}_2$, with $v_2(y_0)=1$ and $y_1\equiv 1(\mathrm{mod}8)$. The 2-adic unit criterion makes $y_1$ a square, so $X^2-y_1$ splits into two linear factors. The polynomial $X^2-y_0$ is Eisenstein and remains irreducible. Thus $X^4+9X^2-2$ has three irreducible factors, of degrees $1,1,2$, as in factorization of X4 plus 9X2 minus 2 over the 2-adic numbers.

Seek $F=L+F_2+F_3+\cdots$, with each $F_d$ homogeneous of degree $d$. Suppose the terms below degree $d$ have been chosen. Comparing degree $d$ ingives a linear equation for $F_d$ whose coefficient is $\pi -{\pi }^d=\pi (1-{\pi }^{d-1})$. The already known term is divisible by $\pi$ because $f(X),g(X)\equiv X^q(\mathrm{mod}\pi )$ and every residue ${\bar{a}}_i\in {\mathbb{F}}_q$ satisfies ${\bar{a}}_i^q={\bar{a}}_i$. Since $1-{\pi }^{d-1}$ is a unit, this determines a unique integral $F_d$. Induction constructs a unique $F\in {\mathcal{O}}_K[[X_1,\dots ,X_n]]$. This is the Lubin–Tate functional equation lemma.

Take $f=g$ and $L=X_1+\cdots +X_n$. Permuting the variables produces another solution with the same linear term, so uniqueness givesFor $a\in {\mathcal{O}}_K$, the one-variable case of part i gives a unique ${\theta }_a(X)\equiv aX(\mathrm{mod}{X}^2)$ commuting with $f$. Applying uniqueness once more to the two ways of composing ${\theta }_a$ with $F$ givesThus $F$ is the addition law and the ${\theta }_a$ are scalar endomorphisms of the Lubin–Tate formal group.

Since $f(X)=X(\pi +X^{q-1})$, its iterates satisfyThe polynomial ${\Phi }_m$ is Eisenstein: it is monic, every nonleading coefficient is divisible by $\pi$, and its constant term is $\pi$. It is also separable, since $q-1$ is prime to the residue characteristic and the iterates have nonzero derivative.

If $\alpha \ne 0$ and $m$ is least with $f_m(\alpha )=0$, then ${\Phi }_m(\alpha )=0$. Eisenstein irreducibility makes ${\Phi }_m$ its minimal polynomial, so $K(\alpha )/K$ is totally ramified and separable. For $\alpha =0$ the extension is trivial and has the same properties. This is the Eisenstein layers of Lubin–Tate torsion argument.