Extensions of Normalized Valuations

Let $ K$ be a complete field with valuation $ \left\vert \cdot \right\vert$. We consider the following three cases:
$ \left\vert \cdot \right\vert$ is discrete non-archimedean and the residue class field is finite.
The completion of $ K$ with respect to $ \left\vert \cdot \right\vert$ is $ \mathbf{R}$.
The completion of $ K$ with respect to $ \left\vert \cdot \right\vert$ is $ \mathbf{C}$.
(Alternatively, these cases can be subsumed by the hypothesis that the completion of $ K$ is locally compact.)

In case (1) we defined the normalized valuation to be the one such that if Haar measure of the ring of integers $ \O$ is $ 1$, then $ \mu(a\O ) = \left\vert a\right\vert$ (see Definition 17.1.11). In case (2i) we say that $ \left\vert \cdot \right\vert$ is normalized if it is the ordinary absolute value, and in (2ii) if it is the square of the ordinary absolute value:

$\displaystyle \left\vert x+iy\right\vert = x^2 + y^2$   (normalized)$\displaystyle .$

In every case, for every $ a\in K$, the map

$\displaystyle a: x \mapsto a x

on $ K^+$ multiplies any choice of Haar measure by $ \left\vert a\right\vert$, and this characterizes the normalized valuations among equivalent ones.

We have already verified the above characterization for non-archimedean valuations, and it is clear for the ordinary absolute value on $ \mathbf{R}$, so it remains to verify it for $ \mathbf{C}$. The additive group $ \mathbf{C}^+$ is topologically isomorphic to $ \mathbf{R}^+ \oplus \mathbf{R}^+$, so a choice of Haar measure of $ \mathbf{C}^+$ is the usual area measure on the Euclidean plane. Multiplication by $ x+iy\in \mathbf{C}$ is the same as rotation followed by scaling by a factor of $ \sqrt{x^2+y^2}$, so if we rescale a region by a factor of $ x+iy$, the area of the region changes by a factor of the square of $ \sqrt{x^2+y^2}$. This explains why the normalized valuation on $ \mathbf{C}$ is the square of the usual absolute value. Note that the normalized valuation on $ \mathbf{C}$ does not satisfy the triangle inequality:

$\displaystyle \left\vert 1 + (1+i)\right\vert = \left\vert 2+i\right\vert = 2^2...
3= 1^2 + (1^2 + 1^2) = \left\vert 1\right\vert + \left\vert 1+i\right\vert.

The constant $ C$ in axiom (3) of a valuation for the ordinary absolute value on $ \mathbf{C}$ is $ 2$, so the constant for the normalized valuation $ \left\vert \cdot \right\vert$ is $ C\leq 4$:

$\displaystyle \left\vert x+iy\right\vert \leq 1 \implies \left\vert x+iy+1\right\vert \leq 4.

Note that $ x^2 +y^2 \leq 1$ implies

$\displaystyle (x+1)^2 + y^2
= x^2 + 2x + 1 + y^2 \leq 1 + 2x + 1 \leq 4$

since $ x\leq 1$.

Lemma 19.2.1   Suppose $ K$ is a field that is complete with respect to a normalized valuation  $ \left\vert \cdot \right\vert$ and let $ L$ be a finite extension of $ K$ of degree $ N=[L:K]$. Then the normalized valuation $ \left\vert \cdot \right\vert$ on $ L$ which is equivalent to the unique extension of $ \left\vert \cdot \right\vert$ to $ L$ is given by the formula

$\displaystyle \left\Vert a\right\Vert = \left\vert\Norm _{L/K}(a)\right\vert$   all $\displaystyle a\in L.$ (19.3)

Proof. Let $ \left\vert \cdot \right\vert$ be the normalized valuation on $ L$ that extends $ \left\vert \cdot \right\vert$. Our goal is to identify $ \left\vert \cdot \right\vert$, and in particular to show that it is given by (19.2.1).

By the preceding section there is a positive real number $ c$ such that for all $ a\in
L$ we have

$\displaystyle \left\Vert a\right\Vert = \left\vert\Norm _{L/K}(a)\right\vert^c.$

Thus all we have to do is prove that $ c=1$. In case 2 the only nontrivial situation is $ L=\mathbf{C}$ and $ K=\mathbf{R}$, in which case $ \left\vert\Norm _{\mathbf{C}/\mathbf{R}}(x+iy)\right\vert = \left\vert x^2+y^2\right\vert$, which is the normalized valuation on $ \mathbf{C}$ defined above.

One can argue in a unified way in all cases as follows. Let $ w_1, \ldots, w_N$ be a basis for $ L/K$. Then the map

$\displaystyle \varphi :L^+ \to \bigoplus_{n=1}^N K^+, \qquad
\sum a_n w_n \mapsto (a_1,\ldots, a_N)

is an isomorphism between the additive group $ L^+$ and the direct sum $ \oplus_{n=1}^N K^+$, and this is a homeomorphism if the right hand side is given the product topology. In particular, the Haar measures on $ L^+$ and on $ \oplus_{n=1}^N K^+$ are the same up to a multiplicative constant in $ \mathbf{Q}^*$.

Let $ b\in K$. Then the left-multiplication-by-$ b$ map

$\displaystyle b : \sum a_n w_n \mapsto \sum b a_n w_n

on $ L^+$ is the same as the map

$\displaystyle (a_1,\ldots, a_N) \mapsto (ba_1,\ldots, ba_N)

on $ \oplus_{n=1}^N K^+$, so it multiplies the Haar measure by $ \left\vert b\right\vert^N$, since $ \left\vert \cdot \right\vert$ on $ K$ is assumed normalized (the measure of each factor is multiplied by $ \left\vert b\right\vert$, so the measure on the product is multiplied by $ \left\vert b\right\vert^N$). Since $ \left\vert \cdot \right\vert$ is assumed normalized, so multiplication by $ b$ rescales by $ \left\vert b\right\vert$, we have

$\displaystyle \left\Vert b\right\Vert = \left\vert b\right\vert^N.

But $ b\in K$, so $ \Norm _{L/K}(b) = b^N$. Since $ \left\vert \cdot \right\vert$ is nontrivial and for $ a\in K$ we have

$\displaystyle \left\Vert a\right\Vert = \left\vert a\right\vert^N = \left\vert a^N\right\vert = \left\vert\Norm _{L/K}(a)\right\vert,$

so we must have $ c=1$ in (19.2.1), as claimed. $ \qedsymbol$

In the case when $ K$ need not be complete with respect to the valuation  $ \left\vert \cdot \right\vert$ on $ K$, we have the following theorem.

Theorem 19.2.2   Suppose $ \left\vert \cdot \right\vert$ is a (nontrivial as always) normalized valuation of a field $ K$ and let $ L$ be a finite extension of $ K$. Then for any $ a\in

$\displaystyle \prod_{1\leq j \leq J} \left\Vert a\right\Vert _j = \left\vert\Norm _{L/K}(a)\right\vert

where the $ \left\Vert \cdot \right\Vert _j$ are the normalized valuations equivalent to the extensions of  $ \left\vert \cdot \right\vert$ to $ K$.

Proof. Let $ K_v$ denote the completion of $ K$ with respect to $ \left\vert \cdot \right\vert$. Write

$\displaystyle K_v\otimes _K L = \bigoplus_{1\leq j \leq J} L_j.

Then Theorem 19.2.2 asserts that

$\displaystyle \Norm _{L/K}(a) = \prod_{1\leq j\leq J} \Norm _{L_j/K_v}(a).$ (19.4)

By Theorem 19.1.8, the $ \left\Vert \cdot \right\Vert _j$ are exactly the normalizations of the extensions of $ \left\vert \cdot \right\vert$ to the $ L_j$ (i.e., the $ L_j$ are in bijection with the extensions of valuations, so there are no other valuations missed). By Lemma 19.1.1, the normalized valuation $ \left\Vert \cdot \right\Vert _j$ on $ L_j$ is $ \left\vert a\right\vert =
\left\vert\Norm _{L_J/K_v}(a)\right\vert$. The theorem now follows by taking absolute values of both sides of (19.2.2). $ \qedsymbol$

What next?! We'll building up to giving a new proof of finiteness of the class group that uses that the class group naturally has the discrete topology and is the continuous image of a compact group.

William Stein 2004-05-06