Global Fields

Definition 20.1.1 (Global Field)   A is a number field or a finite separable extension of $ \mathbf{F}(t)$, where $ \mathbf{F}$ is a finite field, and $ t$ is transcendental over $ \mathbf{F}$.

Below we will focus attention on number fields leaving the function field case to the reader.

The following lemma essentially says that the denominator of an element of a global field is only ``nontrivial'' at a finite number of valuations.

Lemma 20.1.2   Let $ a\in K$ be a nonzero element of a global field $ K$. Then there are only finitely many inequivalent valuations $ \left\vert \cdot \right\vert$ of $ K$ for which

$\displaystyle \left\vert a\right\vert > 1.

Proof. If $ K=\mathbf{Q}$ or $ \mathbf{F}(t)$ then the lemma follows by Ostrowski's classification of all the valuations on $ K$ (see Theorem 15.3.2). For example, when $ a=\frac{n}{d}\in\mathbf{Q}$, with $ n,d\in \mathbf{Z}$, then the valuations where we could have $ \left\vert a\right\vert>1$ are the archimedean one, or the $ p$-adic valuations $ \left\vert \cdot \right\vert _p$ for which $ p\mid d$.

Suppose now that $ K$ is a finite extension of $ \mathbf{Q}$, so $ a$ satisfies a monic polynomial

$\displaystyle a^n + c_{n-1} a^{n-1} + \cdots + c_0 = 0,

for some $ n$ and $ c_0,\ldots, c_{n-1}\in\mathbf{Q}$. If $ \left\vert \cdot \right\vert$ is a non-archimedean valuation on $ K$, we have

$\displaystyle \left\vert a\right\vert^n$ $\displaystyle = \left\vert-(c_{n-1} a^{n-1} + \cdots + c_0)\right\vert$    
  $\displaystyle \leq \max(1,\left\vert a\right\vert^{n-1})\cdot \max(\left\vert c_0\right\vert,\ldots,\left\vert c_{n-1}\right\vert).$    

Dividing each side by $ \left\vert a\right\vert^{n-1}$, we have that

$\displaystyle \left\vert a\right\vert \leq \max(\left\vert c_0\right\vert,\ldots,\left\vert c_{n-1}\right\vert),

so in all cases we have

$\displaystyle \left\vert a\right\vert \leq \max(1, \left\vert c_0\right\vert, \ldots,\left\vert c_{n-1}\right\vert)^{1/(n-1)}.$ (20.1)

We know the lemma for  $ \mathbf{Q}$, so there are only finitely many valuations  $ \left\vert \cdot \right\vert$ on  $ \mathbf{Q}$ such that the right hand side of (20.1.1) is bigger than $ 1$. Since each valuation of  $ \mathbf{Q}$ has finitely many extensions to $ K$, and there are only finitely many archimedean valuations, it follows that there are only finitely many valuations on $ K$ such that $ \left\vert a\right\vert>1$. $ \qedsymbol$

Any valuation on a global field is either archimedean, or discrete non-archimedean with finite residue class field, since this is true of $ \mathbf{Q}$ and $ \mathbf{F}(t)$ and is a property preserved by extending a valuation to a finite extension of the base field. Hence it makes sense to talk of normalized valuations. Recall that the normalized $ p$-adic valuation on $ \mathbf{Q}$ is $ \left\vert x\right\vert _p = p^{-\ord _p(x)}$, and if $ v$ is a valuation on a number field $ K$ equivalent to an extension of $ \left\vert \cdot \right\vert _p$, then the normalization of $ v$ is the composite of the sequence of maps

$\displaystyle K\hookrightarrow K_v \xrightarrow{\Norm } \mathbf{Q}_p \xrightarrow{\left\vert \cdot \right\vert _p} \mathbf{R},

where $ K_v$ is the completion of $ K$ at $ v$.

Example 20.1.3   Let $ K=\mathbf{Q}(\sqrt{2})$, and let $ p=2$. Because $ \sqrt{2}\not\in\mathbf{Q}_2$, there is exactly one extension of $ \left\vert \cdot \right\vert _2$ to $ K$, and it sends $ a=1/\sqrt{2}$ to

$\displaystyle \left\vert\Norm _{\mathbf{Q}_2(\sqrt{2})/\mathbf{Q}_2}(1/\sqrt{2})\right\vert^{1/2}_2 = \sqrt{2}.

Thus the normalized valuation of $ a$ is $ 2$.

There are two extensions of $ \left\vert \cdot \right\vert _7$ to $ \mathbf{Q}(\sqrt{2})$, since $ \mathbf{Q}(\sqrt{2})\otimes _\mathbf{Q}\mathbf{Q}_7 \cong \mathbf{Q}_7 \oplus \mathbf{Q}_7$, as $ x^2-2 = (x-3)(x-4)\pmod{7}$. The image of $ \sqrt{2}$ under each embedding into $ \mathbf{Q}_7$ is a unit in $ \mathbf{Z}_7$, so the normalized valuation of $ a=1/\sqrt{2}$ is, in both cases, equal to $ 1$. More generally, for any valuation of $ K$ of characteristic an odd prime $ p$, the normalized valuation of $ a$ is $ 1$.

Since $ K=\mathbf{Q}(\sqrt{2})\hookrightarrow \mathbf{R}$ in two ways, there are exactly two normalized archimedean valuations on $ K$, and both of their values on $ a$ equal $ 1/\sqrt{2}$. Notice that the product of the absolute values of $ a$ with respect to all normalized valuations is

$\displaystyle 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cdot 1
\cdot 1 \cdot 1 \cdots = 1.

This ``product formula'' holds in much more generality, as we will now see.

Theorem 20.1.4 (Product Formula)   Let $ a\in K$ be a nonzero element of a global field $ K$. Let $ \left\vert \cdot \right\vert _v$ run through the normalized valuations of $ K$. Then $ \left\vert a\right\vert _v=1$ for almost all $ v$, and

$\displaystyle \prod_{\text{\rm all }v} \left\vert a\right\vert _v = 1\qquad{\text{\rm (the product

We will later give a more conceptual proof of this using Haar measure (see Remark 20.3.9).

Proof. By Lemma 20.1.2, we have $ \left\vert a\right\vert _v\leq 1$ for almost all $ v$. Likewise, $ 1/\left\vert a\right\vert _v = \left\vert 1/a\right\vert _v\leq 1$ for almost all $ v$, so $ \left\vert a\right\vert _v=1$ for almost all $ v$.

Let $ w$ run through all normalized valuations of $ \mathbf{Q}$ (or of $ \mathbf{F}(t)$), and write $ v\mid w$ if the restriction of $ v$ to $ \mathbf{Q}$ is equivalent to $ w$. Then by Theorem 19.2.2,

$\displaystyle \prod_{v} \left\vert a\right\vert _v = \prod_w \left(\prod_{v\mid...
...ght\vert _v\right)
= \prod_w \left\vert\Norm _{K/\mathbf{Q}}(a)\right\vert _w,

so it suffices to prove the theorem for $ K=\mathbf{Q}$.

By multiplicativity of valuations, if the theorem is true for $ b$ and $ c$ then it is true for the product $ b c$ and quotient $ b/c$ (when $ c\neq 0$). The theorem is clearly true for $ -1$, which has valuation $ 1$ at all valuations. Thus to prove the theorem for $ \mathbf{Q}$ it suffices to prove it when $ a=p$ is a prime number. Then we have $ \left\vert p\right\vert _\infty = p$, $ \left\vert p\right\vert _p = 1/p$, and for primes $ q\neq p$ that $ \left\vert p\right\vert _q = 1$. Thus

$\displaystyle \prod_v \left\vert p\right\vert _v = p \cdot \frac{1}{p} \cdot 1 \cdot 1 \cdot 1 \cdots = 1,$

as claimed. $ \qedsymbol$

If $ v$ is a valuation on a field $ K$, recall that we let $ K_v$ denote the completion of $ K$ with respect to $ v$. Also when $ v$ is non-archimedean, let

$\displaystyle \O _v = \O _{K,v} = \{x \in K_v : \left\vert x\right\vert \leq 1\}

be the ring of integers of the completion.

Definition 20.1.5 (Almost All)   We say a condition holds for elements of a set if it holds for all but finitely many elements.

We will use the following lemma later (see Lemma 20.3.3) to prove that formation of the adeles of a global field is compatible with base change.

Lemma 20.1.6   Let $ \omega_1,\ldots, \omega_n$ be a basis for $ L/K$, where $ L$ is a finite separable extension of the global field $ K$ of degree $ n$. Then for almost all normalized non-archimedean valuations $ v$ on $ K$ we have

$\displaystyle \omega_1 \O _{v} \oplus \cdots \oplus \omega_n \O _{v} = \O _{w_1} \oplus \cdots \oplus \O _{w_g} \subset K_v\otimes _K L,$ (20.2)

where $ w_1,\ldots, w_g$ are the extensions of $ v$ to $ L$. Here we have identified $ a\in
L$ with its canonical image in $ K_v\otimes _K L$, and the direct sum on the left is the sum taken inside the tensor product (so directness means that the intersections are trivial).

Proof. The proof proceeds in two steps. First we deduce easily from Lemma 20.1.2 that for almost all $ v$ the left hand side of (20.1.2) is contained in the right hand side. Then we use a trick involving discriminants to show the opposite inclusion for all but finitely many primes.

Since $ \O _v\subset \O _{w_i}$ for all $ i$, the left hand side of (20.1.2) is contained in the right hand side if $ \left\vert\omega_i\right\vert _{w_j}\leq 1$ for $ 1\leq i\leq n$ and $ 1\leq j\leq
g$. Thus by Lemma 20.1.2, for all but finitely many $ v$ the left hand side of (20.1.2) is contained in the right hand side. We have just eliminated the finitely many primes corresponding to ``denominators'' of some $ \omega_i$, and now only consider $ v$ such that $ \omega_1,\ldots, \omega_n \in \O _{w}$ for all $ w\mid v$.

For any elements $ a_1,\ldots, a_n \in K_v\otimes _K L$, consider the discriminant

$\displaystyle D(a_1,\ldots, a_n) = {\mathrm{Det}}(\Tr (a_i a_j)) \in K_v,

where the trace is induced from the $ L/K$ trace. Since each $ \omega_i$ is in each $ \O _w$, for $ w\mid v$, the traces lie in $ \O _v$, so

$\displaystyle d = D(\omega_1,\ldots, \omega_n)\in \O _v.$

Also note that $ d\in
K$ since each $ \omega_i$ is in $ L$. Now suppose that

$\displaystyle \alpha = \sum_{i=1}^n a_i \omega_i \in \O _{w_1} \oplus \cdots
\oplus \O _{w_g},

with $ a_i \in K_v$. Then by properties of determinants for any $ m$ with $ 1\leq m\leq n$, we have

$\displaystyle D(\omega_1,\ldots, \omega_{m-1}, \alpha, \omega_{m+1}, \ldots, \omega_n) = a_m^2 D(\omega_1,\ldots, \omega_n).$ (20.3)

The left hand side of (20.1.3) is in $ \O _v$, so the right hand side is well, i.e.,

$\displaystyle a_m^2 \cdot d \in \O _v,$   (for $\displaystyle m=1,\ldots, n$)$\displaystyle ,

where $ d\in
K$. Since $ \omega_1,\ldots, \omega_n$ are a basis for $ L$ over $ K$ and the trace pairing is nondegenerate, we have $ d\neq
0$, so by Theorem 20.1.4 we have $ \left\vert d\right\vert _v=1$ for all but finitely many $ v$. Then for all but finitely many $ v$ we have that $ a_m^2\in \O _v$. For these $ v$, that $ a_m^2\in \O _v$ implies $ a_m\in \O _v$ since $ a_m\in K_v$, i.e., $ \alpha$ is in the left hand side of (20.1.2). $ \qedsymbol$

Example 20.1.7   Let $ K=\mathbf{Q}$ and $ L=\mathbf{Q}(\sqrt{2})$. Let $ \omega_1 = 1/3$ and $ \omega_2 = 2\sqrt{2}$. In the first stage of the above proof we would eliminate $ \left\vert \cdot \right\vert _3$ because $ \omega_2$ is not integral at $ 3$. The discriminant is

$\displaystyle d = D\left(\frac{1}{3}, 2\sqrt{2}\right)
\begin{matrix}\frac{2}{9}&0\ 0&16
\end{matrix}\right) = \frac{32}{9}.

As explained in the second part of the proof, as long as $ v\neq 2, 3$, we have equality of the left and right hand sides in (20.1.2).

William Stein 2004-05-06