Extensions of Valuations

In this section we continue to tacitly assume that all valuations are nontrivial. We do not assume all our valuations satisfy the triangle

Suppose $K\subset L$ is a finite extension of fields, and that $\left\vert \cdot \right\vert{}$ and $\left\vert \cdot \right\vert{}$ are valuations on $K$ and $L$, respectively.

Definition 19.1.1 (Extends)   We say that $\left\vert \cdot \right\vert{}$ $\left\vert \cdot \right\vert$ if $\left\vert a\right\vert = \left\Vert a\right\Vert$ for all $a\in K$.

Theorem 19.1.2   Suppose that $K$ is a field that is complete with respect to $\left\vert \cdot \right\vert$ and that $L$ is a finite extension of $K$ of degree $N=[L:K]$. Then there is precisely one extension of $\left\vert \cdot \right\vert{}$ to $K$, namely

$$\left\Vert a\right\Vert = \left\vert\Norm _{L/K}(a)\right\vert^{1/N},$$ (19.1)

where the $N$th root is the non-negative real $N$th root of the nonnegative real number $\left\vert\Norm _{L/K}(a)\right\vert$.

Proof. We may assume that $\left\vert \cdot \right\vert$ is normalized so as to satisfy the triangle inequality. Otherwise, normalize $\left\vert \cdot \right\vert$ so that it does, prove the theorem for the normalized valuation $\left\vert \cdot \right\vert^c$, then raise both sides of (19.1.1) to the power $1/c$. In the uniqueness proof, by the same argument we may assume that $\left\vert \cdot \right\vert$ also satisfies the triangle inequality.

Uniqueness. View $L$ as a finite-dimensional vector space over $K$. Then $\left\vert \cdot \right\vert$ is a norm in the sense defined earlier (Definition 18.1.1). Hence any two extensions $\left\vert \cdot \right\vert _1$ and $\left\vert \cdot \right\vert _2$ of $\left\vert \cdot \right\vert$ are equivalent as norms, so induce the same topology on $K$. But as we have seen (Proposition 16.1.4), two valuations which induce the same topology are equivalent valuations, i.e., $\left\Vert \cdot \right\Vert _1 = \left\Vert \cdot \right\Vert _2^c$, for some positive real $c$. Finally $c=1$ since $\left\Vert a\right\Vert _1 = \left\vert a\right\vert = \left\Vert a\right\Vert _2$ for all $a\in K$.

Existence. We do not give a proof of existence in the general case. Instead we give a proof, which was suggested by Dr. Geyer at the conference out of which [Cas67] arose. It is valid when $K$ is locally compact, which is the only case we will use later.

We see at once that the function defined in (19.1.1) satisfies the condition (i) that $\left\Vert a\right\Vert\geq 0$ with equality only for $a = 0$, and (ii) $\left\Vert ab\right\Vert=\left\Vert a\right\Vert\cdot \left\Vert b\right\Vert$ for all $a,b\in L$. The difficult part of the proof is to show that there is a constant $C>0$ such that

$$\left\Vert a\right\Vert\leq 1 \implies \left\Vert 1+a\right\Vert\leq C.$$

Note that we do not know (and will not show) that $\left\vert \cdot \right\vert$ as defined by (19.1.1) is a norm as in Definition 18.1.1, since showing that $\left\vert \cdot \right\vert$ is a norm would entail showing that it satisfies the triangle inequality, which is not obvious.

Choose a basis $b_1,\ldots, b_N$ for $L$ over $K$. Let $\left\Vert \cdot \right\Vert _0$ be the max norm on $L$, so for $a=\sum_{i=1}^N c_i b_i$ with $c_i\in K$ we have

$$\left\Vert a\right\Vert _0 = \left\Vert\sum_{i=1}^N c_i b_i\right\Vert _0 = \max \{\left\vert c_i\right\vert : i=1,\ldots, N\}.$$

(Note: in Cassels's original article he let $\left\Vert \cdot \right\Vert _0$ be any norm, but we don't because the rest of the proof does not work, since we can't use homogeneity as he claims to do. This is because it need not be possible to find, for any nonzero $a\in L$ some element $c\in K$ such that $\left\Vert ac\right\Vert _0=1$. This would fail, e.g., if $\left\Vert a\right\Vert _0\neq \left\vert c\right\vert$ for any $c\in K$.) The rest of the argument is very similar to our proof from Lemma 18.1.3 of uniqueness of norms on vector spaces over complete fields.

With respect to the $\left\Vert \cdot \right\Vert _0$-topology, $L$ has the product topology as a product of copies of $K$. The function $a\mapsto \left\Vert a\right\Vert$ is a composition of continuous functions on $L$ with respect to this topology (e.g., $\Norm _{L/K}$ is the determinant, hence polynomial), hence $\left\vert \cdot \right\vert$ defines nonzero continuous function on the compact set

$$S = \{a \in L : \left\Vert a\right\Vert _0 = 1\}.$$

By compactness, there are real numbers $\delta,\Delta\in\mathbf{R}_{>0}$ such that

$$0 < \delta \leq \left\Vert a\right\Vert \leq \Delta$$   for all $a&isin#in;S$$$.$$

For any nonzero $a\in L$ there exists $c\in K$ such that $\left\Vert a\right\Vert _0 = \left\vert c\right\vert$; to see this take $c$ to be a $c_i$ in the expression $a=\sum_{i=1}^N c_i b_i$ with $\left\vert c_i\right\vert\geq \left\vert c_j\right\vert$ for any $j$. Hence $\left\Vert a/c\right\Vert _0 = 1$, so $a/c\in S$ and

$$0 \leq \delta \leq \frac{\left\Vert a/c\right\Vert}{\left\Vert a/c\right\Vert _0} \leq \Delta.$$

Then by homogeneity

$$0 \leq \delta \leq \frac{\left\Vert a\right\Vert}{\left\Vert a\right\Vert _0} \leq \Delta.$$

Suppose now that $\left\vert a\right\vert\leq 1$. Then $\left\Vert a\right\Vert _0\leq \delta^{-1}$, so

$$\left\Vert 1+a\right\Vert \leq \Delta\cdot \left\Vert 1+a\right\Vert _0$$

$$\leq \Delta\cdot \left( \left\Vert 1\right\Vert _0 + \left\Vert a\right\Vert _0\right)$$

$$\leq \Delta\cdot \left( \left\Vert 1\right\Vert _0 + \delta^{-1}\right)$$

$$=C ,$$

as required. $\qedsymbol$

Example 19.1.3   Consider the extension $\mathbf{C}$ of $\mathbf{R}$ equipped with the archimedean valuation. The unique extension is the ordinary absolute value on $\mathbf{C}$:

$$\left\Vert x+iy\right\Vert = \left(x^2 + y^2\right)^{1/2}.$$

Example 19.1.4   Consider the extension $\mathbf{Q}_2(\sqrt{2})$ of $\mathbf{Q}_2$ equipped with the $2$-adic absolute value. Since $x^2-2$ is irreducible over $\mathbf{Q}_2$ we can do some computations by working in the subfield $\mathbf{Q}(\sqrt{2})$ of $\mathbf{Q}_2(\sqrt{2})$.

> K<a> := NumberField(x^2-2);
> K;
Number Field with defining polynomial x^2 - 2 over the Rational Field
> function norm(x) return Sqrt(2^(-Valuation(Norm(x),2))); end function;
> norm(1+a);
1.0000000000000000000000000000
> norm(1+a+1);
0.70710678118654752440084436209
> z := 3+2*a;
> norm(z);
1.0000000000000000000000000000
> norm(z+1);
0.353553390593273762200422181049

Remark 19.1.5   Geyer's existence proof gives (19.1.1). But it is perhaps worth noting that in any case (19.1.1) is a consequence of unique existence, as follows. Suppose $L/K$ is as above. Suppose $M$ is a finite Galois extension of $K$ that contains $L$. Then by assumption there is a unique extension of $\left\vert \cdot \right\vert{}$ to $M$, which we shall also denote by $\left\vert \cdot \right\vert$. If $\sigma\in\Gal (M/K)$, then

$$\left\Vert a\right\Vert _\sigma := \left\Vert\sigma(a)\right\Vert$$

is also an extension of $\left\vert \cdot \right\vert{}$ to $M$, so $\left\Vert \cdot \right\Vert _\sigma = \left\Vert \cdot \right\Vert$, i.e.,

$$\left\Vert\sigma(a)\right\Vert = \left\Vert a\right\Vert$$   for all $a&isin#in;M$$$.$$

But now

$$\Norm _{L/K}(a) = \sigma_1(a) \cdot \sigma_2(a) \cdots \sigma_N(a)$$

for $a\in K$, where $\sigma_1,\ldots, \sigma_N\in \Gal (M/K)$ extend the embeddings of $L$ into $M$. Hence

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

$$= \prod_{1\leq n \leq N} \left\Vert\sigma_n(a)\right\Vert$$

$$= \left\Vert a\right\Vert^N,$$

as required.

Corollary 19.1.6   Let $w_1, \ldots, w_N$ be a basis for $L$ over $K$. Then there are positive constants $c_1$ and $c_2$ such that

$$c_1 \leq \frac{\left\Vert\displaystyle \sum_{n=1}^{N} b_n w_n\right\Vert}{\max \{ \left\vert b_n\right\vert : n = 1,\ldots, N\}} \leq c_2$$

for any $b_1,\ldots, b_N\in K$ not all 0.

Proof. For $\left\vert\sum_{n=1}^N b_n w_n\right\vert$ and $\max{\left\vert b_n\right\vert}$ are two norms on $L$ considered as a vector space over $K$.

I don't believe this proof, which I copied from Cassels's article. My problem with it is that the proof of Theorem 19.1.2 does not give that $C\leq 2$, i.e., that the triangle inequality holds for $\left\vert \cdot \right\vert$. By changing the basis for $L/K$ one can make any nonzero vector $a\in L$ have $\left\Vert a\right\Vert _0=1$, so if we choose $a$ such that $\left\vert a\right\vert$ is very large, then the $\Delta$ in the proof will also be very large. One way to fix the corollary is to only claim that there are positive constants $c_1, c_2,c_3, c_4$ such that

$$c_1 \leq \frac{\left\Vert\displaystyle \sum_{n=1}^{N} b_n w_n\rig... ...^{c_3}}{\max \{ \left\vert b_n\right\vert^{c_4} : n = 1,\ldots, N\}} \leq c_2.$$

Then choose $c_3, c_4$ such that $\left\Vert \cdot \right\Vert^{c_3}$ and $\left\vert \cdot \right\vert^{c_4}$ satisfies the triangle inequality, and prove the modified corollary using the proof suggested by Cassels. $\qedsymbol$

Corollary 19.1.7   A finite extension of a completely valued field $K$ is complete with respect to the extended valuation.

Proof. By the proceeding corollary it has the topology of a finite-dimensional vector space over $K$. (The problem with the proof of the previous corollary is not an issue, because we can replace the extended valuation by an inequivalent one that satisfies the triangle inequality and induces the same topology.) $\qedsymbol$

When $K$ is no longer complete under $\left\vert \cdot \right\vert$ the position is more complicated:

Theorem 19.1.8   Let $L$ be a separable extension of $K$ of finite degree $N=[L:K]$. Then there are at most $N$ extensions of a valuation $\left\vert \cdot \right\vert$ on $K$ to $L$, say $\left\Vert \cdot \right\Vert _j$, for $1\leq j \leq J$. Let $K_v$ be the completion of $K$ with respect to $\left\vert \cdot \right\vert$, and for each $j$ let $L_j$ be the completion of $L$ with respect to $\left\Vert \cdot \right\Vert _j$. Then

$$K_v \otimes _K L \cong \bigoplus_{1\leq j\leq J} L_j$$ (19.2)

algebraically and topologically, where the right hand side is given the product topology.

Proof. We already know (Lemma 18.2.1) that $K_v\otimes _K L$ is of the shape (19.1.2), where the $L_j$ are finite extensions of $K_v$. Hence there is a unique extension $\left\vert \cdot \right\vert _j^*$ of $\left\vert \cdot \right\vert$ to the $L_j$, and by Corollary 19.1.7 the $L_j$ are complete with respect to the extended valuation. Further, the ring homomorphisms

$$\lambda_j : L \to K_v\otimes _K L \to L_j$$

are injections. Hence we get an extension $\left\Vert \cdot \right\Vert _j$ of $\left\vert \cdot \right\vert$ to $L$ by putting

$$\left\Vert b\right\Vert _j = \left\vert\lambda_j(b)\right\vert _j^*.$$

Further, $L\cong \lambda_j(L)$ is dense in $L_j$ with respect to $\left\Vert \cdot \right\Vert _j$ because $L = K\otimes _K L$ is dense in $K_v\otimes _K L$ (since $K$ is dense in $K_v$). Hence $L_j$ is exactly the completion of $L$.

It remains to show that the $\left\Vert \cdot \right\Vert _j$ are distinct and that they are the only extensions of $\left\vert \cdot \right\vert{}$ to $L$.

Suppose $\left\vert \cdot \right\vert$ is any valuation of $L$ that extends $\left\vert \cdot \right\vert$. Then $\left\vert \cdot \right\vert$ extends by continuity to a real-valued function on $K_v\otimes _K L$, which we also denote by $\left\vert \cdot \right\vert$. (We are again using that $L$ is dense in $K_v\otimes _K L$.) By continuity we have for all $a,b\in K_v\otimes _K L$,

$$\left\Vert ab\right\Vert = \left\Vert a\right\Vert\cdot \left\Vert b\right\Vert$$

and if $C$ is the constant in axiom (iii) for $L$ and $\left\vert \cdot \right\vert$, then

$$\left\Vert a\right\Vert\leq 1 \implies \left\Vert 1+a\right\Vert\leq C.$$

(In Cassels, he inexplicable assume that $C=1$ at this point in the proof.)

We consider the restriction of $\left\vert \cdot \right\vert$ to one of the $L_j$. If $\left\Vert a\right\Vert\neq 0$ for some $a\in L_j$, then $\left\Vert a\right\Vert = \left\Vert b\right\Vert\cdot \left\Vert ab^{-1}\right\Vert$ for every $b\neq 0$ in $L_j$ so $\left\Vert b\right\Vert \neq 0$. Hence either $\left\vert \cdot \right\vert$ is identically 0 on $L_j$ or it induces a valuation on $L_j$.

Further, $\left\vert \cdot \right\vert$ cannot induce a valuation on two of the $L_j$. For

$$(a_1,0,\ldots, 0)\cdot (0,a_2,0,\ldots,0) = (0,0,0,\ldots,0),$$

so for any $a_1\in L_1$, $a_2\in L_2$,

$$\left\Vert a_1\right\Vert\cdot \left\Vert a_2\right\Vert = 0.$$

Hence $\left\vert \cdot \right\vert{}$ induces a valuation in precisely one of the $L_j$, and it extends the given valuation $\left\vert \cdot \right\vert$ of $K_v$. Hence $\left\Vert \cdot \right\Vert= \left\Vert \cdot \right\Vert _j$ for precisely one $j$.

It remains only to show that (19.1.2) is a topological homomorphism. For

$$(b_1,\ldots, b_J)\in L_1\oplus \cdots \oplus L_J$$

put

$$\left\Vert(b_1,\ldots, b_J)\right\Vert _0 = \max_{1\leq j \leq J} \left\Vert b_j\right\Vert _j.$$

Then $\left\Vert \cdot \right\Vert _0$ is a norm on the right hand side of (19.1.2), considered as a vector space over $K_v$ and it induces the product topology. On the other hand, any two norms are equivalent, since $K_v$ is complete, so $\left\Vert \cdot \right\Vert _0$ induces the tensor product topology on the left hand side of (19.1.2). $\qedsymbol$

Corollary 19.1.9   Suppose $L=K(a)$, and let $f(x)\in K[x]$ be the minimal polynomial of $a$. Suppose that

$$f(x) = \prod_{1\leq j\leq J} g_j(x)$$

in $K_v[x]$, where the $g_j$ are irreducible. Then $L_j = K_v(b_j)$, where $b_j$ is a root of $g_j$.