Extensions of Normalized Valuations

Let $K$ be a complete field with valuation $\left\vert \cdot \right\vert$. We consider the following three cases:

(1)

$\left\vert \cdot \right\vert$ is discrete non-archimedean and the residue class field is finite.

(2i)

The completion of $K$ with respect to $\left\vert \cdot \right\vert$ is $\mathbf{R}$.

(2ii)

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:

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

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

$$a: x \mapsto a x$$

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:

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

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 $C\leq 4$:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $a\in L$,

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

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

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

Then Theorem 19.2.2 asserts that

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