Algebraic Number Theory

The valuation that $w$ induces on the completion $L_w$ can only be the unique valuation extending the valuation that $v$ induces on $K_v$. If $\tau :L_w\to \overline{K_v}$ is any field embedding fixing $K_v$, then $\overline{v}\circ \tau$ is a valuation on $L_w$ that extends $v$ on $K_v$, and hence it must be $w$.

For the second statement, suppose first that ${\tau }^{\prime }=\sigma \circ \tau$ with $\sigma \in {\mathrm{Aut}}_{K_v}(\overline{K_v})$. Again, $\overline{v}$ is the unique valuation on $\overline{K_v}$ extending the valuation on $K_v$, so $\overline{v}\circ \sigma =\overline{v}$. But then we have $\overline{v}\circ {\tau }^{\prime }=\overline{v}\circ \tau$, and restricting to $L$, this means that $w^{\prime }=w$.

Conversely, suppose that $w^{\prime }=w$. Note that ${\tau }^{\prime }\circ {\tau }^{-1}:\tau (L)\to {\tau }^{\prime }(L)$ is an isomorphism fixing $K$. Suppose that $L∕K$ is finite. As $K$ is dense in $K_v$ and $\tau (L)$ is also a finite extension of $K$, we have that $\tau (L)$ is dense in $\tau (L)\cdot K_v$ (and similarly for ${\tau }^{\prime }$). Define

by

for $\alpha \in \tau (L)\cdot K_v$, choosing a sequence ${({\alpha }_n)}_n$ of elements of $L$ such that ${(\tau ({\alpha }_n))}_n$ converges to $\alpha$. Note that this is independent of choice, as if ${\alpha }_n^{\prime }$ is another such sequence, then the limits of ${({\tau }^{\prime }({\alpha }_n))}_n$ and ${({\tau }^{\prime }({\alpha }_n^{\prime }))}_n$ are the same by continuity of ${\tau }^{\prime }$. But then ${\tau }^{\prime }=\sigma \circ \tau$, and $\sigma$ is the unique isomorphism fixing $K_v$ with this property. We then extend $\sigma$ to an element of ${\mathrm{Aut}}_{K_v}(\overline{K_v})$, obtaining the desired map.

In general, $L$ is a union of finite extensions $E$ of $K$. For each such $E$, we have defined a unique isomorphism ${\sigma }_E:\tau (E)K_v\to {\tau }^{\prime }(E)K_v$ such that ${\tau }^{\prime }{|}_E={\sigma }_E\circ \tau {|}_E$. If $E$ and $E^{\prime }$ are two finite extensions of $K$ in $L$, then ${\sigma }_E$ and ${\sigma }_{E^{\prime }}$ agree on $\tau (E\cap E^{\prime })K_v\subseteq \tau (E)K_v\cap \tau (E^{\prime })K_v$ by uniqueness. Together, the collection of maps ${\sigma }_E$ defines an embedding of the compositum of the fields $\tau (E)K_v$ into $\overline{K_v}$. We then extend this embedding to an automorphism of $\overline{K_v}$ fixing $K_v$, and by definition, it has the property that ${\tau }^{\prime }=\sigma \circ \tau$. □

As $L∕K$ is finite and separable, there exists an element $𝜃\in L$ such that $L=K(𝜃)$. Let $f$ be the minimal polynomial of $𝜃$ over $K$, and let $f={\prod }_{i=1}^m{f}_i$ over $K_v$, where the $f_i$ are irreducible (and necessarily distinct by separability of $f$). Choose a root ${𝜃}_i$ of $f_i(x)$ for each $i$ inside a fixed algebraic closure $\overline{K_v}$ of $K$. Proposition 1.1.1 provides an isomorphism

such that $𝜃\otimes 1$ is sent to ${𝜃}_i$ in the $i$th coordinate. By Theorem 5.4.7, the field $K_v({𝜃}_i)$ is necessarily complete with respect to a valuation $w_i$ extending $v$. The embedding ${\tau }_i:L\to K_v({𝜃}_i)$ sending $𝜃$ to ${𝜃}_i$ and fixing $K$ has dense image, so $K_v({𝜃}_i)$ is isomorphic to the completion of $L$ with respect to $w_i$.

If $w$ is any valuation on $L$ extending $v$, then Theorem 6.1.4 yields an embedding $\tau$ of $L$ into $\overline{K_v}$ such that $w=\overline{v}\circ \tau$, where $\overline{v}$ is the unique extension of $v$ from $K_v$ to $\overline{K_v}$. Then $\tau (𝜃)$ is the root of $f_i$ for some $i$, so there exists an automorphism $\sigma$ over $\overline{K_v}$ fixing $K_v$ such that $\tau =\sigma \circ {\tau }_i$ and therefore $w=\overline{v}\circ \sigma \circ {\tau }_i=\overline{v}\circ {\tau }_i=w_i$. □

Let $m_{\beta }:L\to L$ be left multiplication. This indues a $K_v$-linear transformation $m_{\beta }\otimes {\mathrm{id}}_{K_v}$ on $K_v$, and the characteristic polynomials of $m_{\beta }$ and $m_{\beta }\otimes {\mathrm{id}}_{K_v}$ agree. Noting that the isomorphism of Proposition 6.1.6 is one of $L$-vector spaces, the characteristic polynomial of $m_{\beta }\otimes {\mathrm{id}}_{K_v}$ coincides with the product of the characteristic polynomials of multiplication by $\beta$ on the $L_w$ for $w\mid v$. The result is then a consequence of Proposition 1.3.3. □

First, we note that both the domain and codomain of ${\kappa }^{\prime }$ are free ${𝒪}_v$-modules of rank $n=[L:K]$. Moreover, $\kappa$ is injective, so it suffices to show that ${\kappa }^{\prime }$ is surjective. For this, note that the trace pairing $\psi :L\times L\to K$ given by

extends to a unique $K_v$-bilinear pairing

which is given on simple tensors by the equation

Let $w_1,\dots ,w_g$ be the places of $L$ lying over $v$. Since ${𝒪}_v$ is a PID, each valuation ring ${𝒪}_{w_i}$ is a free ${𝒪}_v$-module of rank $n_i=[L_{w_i}:K_v]$. So, for $1\le i\le g$, let

and let $({\lambda }_{m_i+1},\dots ,{\lambda }_{m_{i+1}})$ be an ordered basis of ${𝒪}_{w_i}$ as a free ${𝒪}_v$-module. We view each ${\lambda }_i$ as sitting in the product ${\prod }_{i=1}^g{𝒪}_{w_i}$ by taking the other coordinates to be zero. Let $A=(a_{i,j})\in {\mathrm{GL}}_n(K_v)$ be the matrix such that

for each $1\le i\le n$. We must show that $A$ has entries in ${𝒪}_v$.

Via $\kappa$, the pairing ${\psi }_v$ gives rise to a pairing

that is given on the basis $({\lambda }_1,\dots ,{\lambda }_n)$ by

Here, of course, we have applied Proposition 6.1.9.

Note that ${\lambda }_i{\lambda }_j$ has nontrivial component in $L_{w_h}$ if and only if $m_h+1\le i,j\le m_{h+1}$. If this is the case, then

and otherwise $\tilde{\psi }({\lambda }_i,{\lambda }_j)=0$. It follows that the matrix $M=(\tilde{\psi }({\lambda }_i,{\lambda }_j))$ is block-diagonal with determinant

and this is exactly the discriminant of $\tilde{\psi }$ relative to the basis $({\lambda }_1,\dots ,{\lambda }_n)$.

On the other hand, the discriminant of $\tilde{\psi }$ relative to the basis $({\beta }_1,\dots ,{\beta }_n)$ is $D({\beta }_1,\dots ,{\beta }_n)$ by definition, and we have by (6.1.1) and Lemma 1.4.6 that

Since $D({\lambda }_{m_i+1},\dots ,{\lambda }_{m_{i+1}})\in {𝒪}_v^{\times }$for each $i$ and $D({\beta }_1,\dots ,{\beta }_n)\in {𝒪}_v^{\times }$ as well, we therefore have $\det (A)\in \underset{v}{\overset{\times }{𝒪}}$. Since the inverse of $A$ has coefficients in ${𝒪}_v$, we therefore have that $A$ does as well. □

Remark 5.4.2 and the definition of $\parallel {\parallel }_v$ tell us that for a place $w$ of $L$ over $v$, we have

for all $\alpha \in K$. Noting Proposition 6.1.9, we then have

□

By Propositions 5.2.34 and 5.2.35, we have the result for $\mathbb{Q}$ and ${𝔽}_p(t)$ for all primes $p$. The field $K$ is in the general case a finite extension of exactly one of these fields, which we denote by $F$. By Lemma 6.1.12, we have

□

There is only one nonzero prime ideal in the valuation ring of $L$, and it lies over the maximal ideal of the valuation ring of $K$. Theorem 2.5.11 then yields the result. □

We know that the residue field of $K$ for $𝔭$ is isomorphic to the residue field of $K_{𝔭}$, and similarly for $L$ and $L_{𝔓}$. Therefore, the second equality holds. As for the first, note that the valuation on $K_{𝔭}$ (resp., $L_{𝔓}$) is just the unique extension of that on $K$ (resp., $L$). If ${\pi }_K$ (resp., ${\pi }_L$) is a uniformizer of $K_{𝔭}$ (resp., $L_{𝔓}$), and $𝒪$ is the valuation ring of $L_{𝔓}$ then we have

so the ramification index of $L_{𝔓}∕K_{𝔭}$ is $v_{𝔓}({\pi }_K)=e_{𝔓∕𝔭}$. □

Note that any $\sigma \in \mathrm{Gal}(L_w∕K_v)$ acts continuously on $L_w$, since $w=\sigma (w)$, as there is a unique valuation on $L_w$ extending the restriction of $w$ to $K_v$. If $\sigma (\beta )=\beta$ for all $\beta \in L$, then continuity forces $\sigma (\beta )=\beta$ for all $\beta \in L_w$, since $L$ is dense in $L_w$. So, the restriction map is injective, and its image is by definition contained in the decomposition group $G_w$ of $G=\mathrm{Gal}(L∕K)$. Any $\tau$ in the decomposition group of $w$ in $G$ satisfies $w=\tau (w)$ on $L$, so on $L_w$ as well by continuity. Therefore, $\tau :L\to L$ is continuous, and its composite with the natural inclusion $L\hookrightarrow L_w$ extends to a unique element of $\mathrm{Gal}(L_w∕K_v)$ by continuity, as in Proposition 5.3.12. That is, the image of restriction is $G_w$.

For $w$ nonarchimedean, it remains to show that the image of the inertia group in $\mathrm{Gal}(L_w∕K_v)$ is the inertia group $I_w$ in $G$. By definition, the image is contained in this group. Let $\tau \in I_w$, and take $\beta \in {𝒪}_w$. As $L_w$ is the completion of $L$, we have $|\beta -b{|}_w<1$ for some $b\in L$ with $|b{|}_w\le 1$. Set ${\beta }^{\prime }=\beta -b$. Since $w$ is nonarchimedean and $|\tau (b)-b{|}_w<1$, we have $|\tau (\beta )-\beta {|}_w<1$ if and only if $|\tau ({\beta }^{\prime })-{\beta }^{\prime }{|}_w<1$. But $|\tau ({\beta }^{\prime })-{\beta }^{\prime }{|}_w\le |{\beta }^{\prime }{|}_w<1$ as $|\tau ({\beta }^{\prime }){|}_w=|{\beta }^{\prime }{|}_w$. Therefore, the extension of $\tau$ to an element of $\mathrm{Gal}(L_w∕K_v)$ lies in the inertia subgroup. □

Let ${\alpha }_1,\dots ,{\alpha }_n$ be a basis of $L$ as a $K$-vector space that consists of elements of $B$. Let $d=D({\alpha }_1,\dots ,{\alpha }_n)$. By Lemma 1.4.19, we have that $\mathit{𝑑ℭ}\subseteq B$, so $ℭ$ is a fractional ideal of $B$. □

We have

which implies that ${𝔇}_{C∕A}\subseteq {𝔇}_{C∕B}{𝔇}_{B∕A}$. On the other hand, we may compute

so ${\mathrm{Tr}}_{M∕L}({𝔇}_{C∕A}^{-1})\subset {𝔇}_{B∕A}^{-1}$. We therefore have that

so ${𝔇}_{C∕B}\subseteq {𝔇}_{B∕A}^{-1}{𝔇}_{C∕A}$, which is to say that ${𝔇}_{C∕B}{𝔇}_{B∕A}\subseteq {𝔇}_{C∕A}$. □

Note that

so ${𝔇}_{S^{-1}B∕S^{-1}A}\subseteq S^{-1}{𝔇}_{B∕A}$. On the other hand, we have

so for each $\alpha \in {𝔇}_{S^{-1}B∕S^{-1}A}^{-1}$, there exists $s\in S$ such that ${\mathrm{Tr}}_{L∕K}(\mathit{𝑠𝛼})=s{\mathrm{Tr}}_{L∕K}(\alpha )\in B$, so $\mathit{𝑠𝛼}\in {𝔇}_{B∕A}^{-1}$. Therefore, we have the other containment. □

Let $\alpha \in {𝔇}_{B∕A}^{-1}$, and let $\beta \in {𝒪}_{𝔓}$. We claim that ${\mathrm{Tr}}_{L_{𝔓}∕K_{𝔭}}(\mathit{𝛼𝛽})\in {𝒪}_{𝔭}$. Let ${𝔓}_1,\dots ,{𝔓}_g$ be the prime ideals of $B$ lying over $𝔭$, taking ${𝔓}_1=𝔓$. For this, let ${({\beta }_n)}_n$ be a sequence in $B$ with limit $\beta$ in the $𝔓$-adic topology and with limit $0$ in the ${𝔓}_i$-adic topology for $i\ge 2$, which exists by the Chinese remainder theorem. Since the trace map is continuous, we have

Moreover, we have by Proposition 6.1.9 that

Note that ${\mathrm{Tr}}_{L∕K}(\alpha {\beta }_n)\in A$ and that, for $i\ge 2$, the sequence of elements ${\mathrm{Tr}}_{L_{{𝔓}_i}∕K_{𝔭}}(\alpha {\beta }_n)\in K_{𝔭}$ tends to $0$ in the $𝔭$-adic topology, again by continuity of the trace map. Therefore, for sufficiently large $n$, the element ${\mathrm{Tr}}_{L_{𝔓}∕K_{𝔭}}(\alpha {\beta }_n)$ lies in ${𝒪}_{𝔭}$, proving the claim. In particular, we have ${𝔇}_{B∕A}^{-1}\subseteq {𝔇}_{{𝒪}_{𝔓}∕{𝒪}_{𝔭}}^{-1}$.

On the other hand, if $\alpha \in {𝔇}_{{𝒪}_{𝔓}∕{𝒪}_{𝔭}}^{-1}$, then we may write $\alpha$ as the limit of a sequence ${({\alpha }_n)}_n$ in $L$ that has limit $0$ in $L_{{𝔓}_i}$ for $i\ge 2$. For $\beta \in B$, we have

Since $K\cap {𝒪}_{𝔭}=A_{𝔭}$, we have that ${\mathrm{Tr}}_{L∕K}({\alpha }_n\beta )\in A_{𝔭}$ for sufficiently large $n$. By Lemma 6.2.6, we therefore have ${\alpha }_n\in S^{-1}{𝔇}_{B∕A}^{-1}$ for all such $n$, where $S$ is the complement of $𝔭$ in $A$. But then $\alpha \in {𝔇}_{B∕A}^{-1}{𝒪}_{𝔓}$ as the limit of these elements. We therefore have the reverse containment ${𝔇}_{{𝒪}_{𝔓}∕{𝒪}_{𝔭}}^{-1}\subseteq {𝔇}_{B∕A}^{-1}{𝒪}_{𝔓}$ and therefore equality. Since the these fractional ideals in ${𝒪}_{𝔓}$ agree, so do their inverses, proving the lemma. □

Write $f={\sum }_{i=0}^n{a}_i{x}^i$ for some $a_i\in A$ with $a_n=1$. Let ${\beta }_1,\dots ,{\beta }_n$ be the roots of $f$ in an algebraic closure of $L$. We claim that for any nonnegative integer $k<n$, we have

To see this, note that the two sides of (6.2.1) are equal upon evaluation at each ${\beta }_i$, yet both sides are polynomials of degree less than $n$, so their difference is identically zero.

We have

for some $b_j\in A[\beta ]$, so

and therefore