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The map $|N(\cdot)|^{1/n}$ is a continuous multiplicative extension of $|\cdot|$. By a multiplicative function I mean a function $\chi:L\to [0,\infty)$ such that $\chi(0)=0$, $\chi(1)=1$ and for every $x,y\in L$, $\chi(xy)=\chi(x)\chi(y)$. A multiplicative function which satisfies for every $x,y\in L$, $\chi(x+y)\leq \chi(x)+\chi(y)$ is called an absolute value.

Theorem: Let $K$ be a field and $|\cdot|$ an absolute value on $K$. Assume $K$ is complete with respect to the induced metric. Then for every finite filed extension of $K$, every continuous multiplicative extension of $|\cdot|$ is an absolute value. In fact, there exists a unique such continuous multiplicative extension of $|\cdot|$, which is $|N(\cdot)|^{1/n}$.

The uniqueness follows at once from the fact that all norms are equivalent on finite dimensional vector spaces over complete fields: for multiplicative functions $\chi_1,\chi_2$, the function $\chi_1/\chi_2$ (defined on the multiplicative group) is multiplicative too, hence must be unbounded or trivial. The less trivial part of the theorem is its first part.

In my other answer I gave a proof of this fact which indeed holds in the generality of complete fields. This post is to give an easy proof I found, under the assumption that the fields are locally compact.

We consider a local filed $K$, endowed with an absolute value $|\cdot|$, a finite filed extension $L$ of $K$ and a continuous multiplicative extension $\chi$ of $|\cdot|$. We argue to show that $\chi$ is an absolute value on $L$.

Consider $L$ as a $K$-vector space and consider the corresponding space $\Omega$ consisting of $K$-norms on $L$. Consider the multiplicative group $L^*$ as a locally compact group (see below for justification) and let it act on $\Omega$ by $\|\cdot\|\mapsto \chi(x)^{-1} \|x\cdot\|$ for $x\in L^*$. Note that for $x\in K^*$, $$ \chi(x)^{-1} \|x\cdot\|=\chi(x)^{-1} |x|\|\cdot\|=\|\cdot\|,$$ thus the $L^*$-action on $\Omega$ factors via $L^*/K^*$, which is a compact group, as it is homeomorphic to a projective space. This action admits a fixed point. Indeed, for every norm $\|\cdot\|\in \Omega$, the map $$ L \ni v \mapsto \int_{L^*/K^*} \chi(x)^{-1}\|xv\|~ \text{dHaar}_{L^*/K^*}(x) \in [0,\infty)$$ is easily seen to be an $L^*$-fixed norm on $L$. We let $\|\cdot\|$ be such a fixed point which is normalized to satisfy $\|1\|=1$. Then for every $x\in L^*$, $$ \|x\|=\|x\cdot 1\|=\chi(x)\cdot \chi(x)^{-1}\|x\cdot 1\|= \chi(x)\cdot \|1\|=\chi(x). $$ Thus $\|\cdot\|=\chi$ and in particular, we conclude that $\chi$ is indeed a norm. This finishes the proof.

To see that $L^*$ is a topological group we need to verify the continuity of the inversion map $x\mapsto x^{-1}$ on $L^*$. Its continuity on $K^*$ follows by a standard argument from the existence of the absolute value $|\cdot|$. From the continuity of the inversion on $K^*$ we get its continuity on $\text{GL}_n(K)$, as inversion is polynomial in the matrix entries and $\det(\cdot)^{-1}$, thus also its continuity on $L^*$.

The map $|N(\cdot)|^{1/n}$ is a continuous multiplicative extension of $|\cdot|$. By a multiplicative function I mean a function $\chi:L\to [0,\infty)$ such that $\chi(0)=0$, $\chi(1)=1$ and for every $x,y\in L$, $\chi(xy)=\chi(x)\chi(y)$. A multiplicative function which satisfies for every $x,y\in L$, $\chi(x+y)\leq \chi(x)+\chi(y)$ is called an absolute value.

Theorem: Let $K$ be a field and $|\cdot|$ an absolute value on $K$. Assume $K$ is complete with respect to the induced metric. Then for every finite field extension of $K$, every continuous multiplicative extension of $|\cdot|$ is an absolute value. In fact, there exists a unique such continuous multiplicative extension of $|\cdot|$, which is $|N(\cdot)|^{1/n}$.

The uniqueness follows at once from the fact that all norms are equivalent on finite dimensional vector spaces over complete fields: for multiplicative functions $\chi_1,\chi_2$, the function $\chi_1/\chi_2$ (defined on the multiplicative group) is multiplicative too, hence must be unbounded or trivial. The less trivial part of the theorem is its first part.

In my other answer I gave a proof of this fact which indeed holds in the generality of complete fields. This post is to give an easy proof I found, under the assumption that the fields are locally compact.

We consider a local field $K$, endowed with an absolute value $|\cdot|$, a finite field extension $L$ of $K$ and a continuous multiplicative extension $\chi$ of $|\cdot|$. We argue to show that $\chi$ is an absolute value on $L$.

Consider $L$ as a $K$-vector space and consider the corresponding space $\Omega$ consisting of $K$-norms on $L$. Consider the multiplicative group $L^*$ as a locally compact group (see below for justification) and let it act on $\Omega$ by $\|\cdot\|\mapsto \chi(x)^{-1} \|x\cdot\|$ for $x\in L^*$. Note that for $x\in K^*$, $$ \chi(x)^{-1} \|x\cdot\|=\chi(x)^{-1} |x|\|\cdot\|=\|\cdot\|,$$ thus the $L^*$-action on $\Omega$ factors via $L^*/K^*$, which is a compact group, as it is homeomorphic to a projective space. This action admits a fixed point. Indeed, for every norm $\|\cdot\|\in \Omega$, the map $$ L \ni v \mapsto \int_{L^*/K^*} \chi(x)^{-1}\|xv\|~ \text{dHaar}_{L^*/K^*}(x) \in [0,\infty)$$ is easily seen to be an $L^*$-fixed norm on $L$. We let $\|\cdot\|$ be such a fixed point which is normalized to satisfy $\|1\|=1$. Then for every $x\in L^*$, $$ \|x\|=\|x\cdot 1\|=\chi(x)\cdot \chi(x)^{-1}\|x\cdot 1\|= \chi(x)\cdot \|1\|=\chi(x). $$ Thus $\|\cdot\|=\chi$ and in particular, we conclude that $\chi$ is indeed a norm. This finishes the proof.

To see that $L^*$ is a topological group we need to verify the continuity of the inversion map $x\mapsto x^{-1}$ on $L^*$. Its continuity on $K^*$ follows by a standard argument from the existence of the absolute value $|\cdot|$. From the continuity of the inversion on $K^*$ we get its continuity on $\text{GL}_n(K)$, as inversion is polynomial in the matrix entries and $\det(\cdot)^{-1}$, thus also its continuity on $L^*$.

The map $|N(\cdot)|^{1/n}$ is a continuous multiplicative extension of $|\cdot|$. By a multiplicative function I mean a function $\chi:L\to [0,\infty)$ such that $\chi(0)=0$, $\chi(1)=1$ and for every $x,y\in L$, $\chi(xy)=\chi(x)\chi(y)$. A multiplicative function which satisfies for every $x,y\in L$, $\chi(x+y)=\chi(x)+\chi(y)$ is called an abolute value.

Theorem: Let $K$ be a field and $|\cdot|$ an absolute value on $K$. Assume $K$ is complete with respect to the induced metric. Then for every finite filed extension of $K$, every continuous multiplicative extension of $|\cdot|$ is an absolute value. In fact, there exists a unique such continuous multiplicative extension of $|\cdot|$, which is $|N(\cdot)|^{1/n}$.

The uniqueness follows at once from the fact that all norms are equivalent on finite dimensional vector spaces over complete fields: for multiplicative functions $\chi_1,\chi_2$, the function $\chi_1/\chi_2$ (defined on the multiplicative group) is multiplicative too, hence must be unbounded or trivial. The less trivial part of the theorem is its first part.

In my other answer I gave a proof of this fact which indeed holds in the generality of complete fields. This post is to give an easy proof I found, under the assumption that the fields are locally compact.

We consider a local filed $K$, endowed with an absolute value $|\cdot|$, a finite filed extension $L$ of $K$ and a continuous multiplicative extension $\chi$ of $|\cdot|$. We argue to show that $\chi$ is an absolute value on $L$.

Consider $L$ as a $K$-vector space and consider the corresponding space $\Omega$ consisting of $K$-norms on $L$. Consider the multiplicative group $L^*$ as a locally compact group (see Lemma 3 in my other answer for continuity of inversion) and let it act on $\Omega$ by $\|\cdot\|\mapsto \chi(x)^{-1} \|x\cdot\|$ for $x\in L^*$. Note that for $x\in K^*$, $$ \chi(x)^{-1} \|x\cdot\|=\chi(x)^{-1} |x|\|\cdot\|=\|\cdot\|,$$ thus the $L^*$-action on $\Omega$ factors via $L^*/K^*$, which is a compact group, as it is homeomorphic to a projective space. This action admits a fixed point. Indeed, for every norm $\|\cdot\|\in \Omega$, the map $$ L \ni v \mapsto \int_{L^*/K^*} \chi(x)^{-1}\|xv\|~ \text{dHaar}_{L^*/K^*}(x) \in [0,\infty)$$ is easily seen to be an $L^*$-fixed norm on $L$. We let $\|\cdot\|$ be such a fixed point which is normalized to satisfy $\|1\|=1$. Then for every $x\in L^*$, $$ \|x\|=\|x\cdot 1\|=\chi(x)\cdot \chi(x)^{-1}\|x\cdot 1\|= \chi(x)\cdot \|1\|=\chi(x). $$ Thus $\|\cdot\|=\chi$ and in particular, we conclude that $\chi$ is indeed a norm.

The map $|N(\cdot)|^{1/n}$ is a continuous multiplicative extension of $|\cdot|$. By a multiplicative function I mean a function $\chi:L\to [0,\infty)$ such that $\chi(0)=0$, $\chi(1)=1$ and for every $x,y\in L$, $\chi(xy)=\chi(x)\chi(y)$. A multiplicative function which satisfies for every $x,y\in L$, $\chi(x+y)\leq \chi(x)+\chi(y)$ is called an absolute value.

Theorem: Let $K$ be a field and $|\cdot|$ an absolute value on $K$. Assume $K$ is complete with respect to the induced metric. Then for every finite filed extension of $K$, every continuous multiplicative extension of $|\cdot|$ is an absolute value. In fact, there exists a unique such continuous multiplicative extension of $|\cdot|$, which is $|N(\cdot)|^{1/n}$.

The uniqueness follows at once from the fact that all norms are equivalent on finite dimensional vector spaces over complete fields: for multiplicative functions $\chi_1,\chi_2$, the function $\chi_1/\chi_2$ (defined on the multiplicative group) is multiplicative too, hence must be unbounded or trivial. The less trivial part of the theorem is its first part.

In my other answer I gave a proof of this fact which indeed holds in the generality of complete fields. This post is to give an easy proof I found, under the assumption that the fields are locally compact.

We consider a local filed $K$, endowed with an absolute value $|\cdot|$, a finite filed extension $L$ of $K$ and a continuous multiplicative extension $\chi$ of $|\cdot|$. We argue to show that $\chi$ is an absolute value on $L$.

Consider $L$ as a $K$-vector space and consider the corresponding space $\Omega$ consisting of $K$-norms on $L$. Consider the multiplicative group $L^*$ as a locally compact group (see below for justification) and let it act on $\Omega$ by $\|\cdot\|\mapsto \chi(x)^{-1} \|x\cdot\|$ for $x\in L^*$. Note that for $x\in K^*$, $$ \chi(x)^{-1} \|x\cdot\|=\chi(x)^{-1} |x|\|\cdot\|=\|\cdot\|,$$ thus the $L^*$-action on $\Omega$ factors via $L^*/K^*$, which is a compact group, as it is homeomorphic to a projective space. This action admits a fixed point. Indeed, for every norm $\|\cdot\|\in \Omega$, the map $$ L \ni v \mapsto \int_{L^*/K^*} \chi(x)^{-1}\|xv\|~ \text{dHaar}_{L^*/K^*}(x) \in [0,\infty)$$ is easily seen to be an $L^*$-fixed norm on $L$. We let $\|\cdot\|$ be such a fixed point which is normalized to satisfy $\|1\|=1$. Then for every $x\in L^*$, $$ \|x\|=\|x\cdot 1\|=\chi(x)\cdot \chi(x)^{-1}\|x\cdot 1\|= \chi(x)\cdot \|1\|=\chi(x). $$ Thus $\|\cdot\|=\chi$ and in particular, we conclude that $\chi$ is indeed a norm. This finishes the proof.

To see that $L^*$ is a topological group we need to verify the continuity of the inversion map $x\mapsto x^{-1}$ on $L^*$. Its continuity on $K^*$ follows by a standard argument from the existence of the absolute value $|\cdot|$. From the continuity of the inversion on $K^*$ we get its continuity on $\text{GL}_n(K)$, as inversion is polynomial in the matrix entries and $\det(\cdot)^{-1}$, thus also its continuity on $L^*$.

I found a natural proof.

Consider the multiplicative group $L^*$ as a locally compact group (see Lemma 3 in my other answer for continuity of inversion) and let it act on the space of norms on $L$ (considered as a $K$ vector space) by $\|\cdot\|\mapsto |N(x)|^{1/n}\|x\cdot\|$ for $x\in L^*$. I claim that there are $K^*$-fixed norms on $L$. Indeed, for every given norm on $L$, the corresponding operator norm on $L$, considered as an algebra of operators for the regular action, is $K^*$-fixed. The compact group $L^*/K^*$ (homeomorphic to $\mathbb{P}^{n-1}(L)$) acts on the space of $K^*$-fixed norms. By averaging with respect to the Haar measure we find a fixed point. Normalized it to have $\|1\|=1$, this fixed point must be $|N(\cdot)|^{1/n}$. It follows that the latter is a norm.

The map $|N(\cdot)|^{1/n}$ is a continuous multiplicative extension of $|\cdot|$. By a multiplicative function I mean a function $\chi:L\to [0,\infty)$ such that $\chi(0)=0$, $\chi(1)=1$ and for every $x,y\in L$, $\chi(xy)=\chi(x)\chi(y)$. A multiplicative function which satisfies for every $x,y\in L$, $\chi(x+y)=\chi(x)+\chi(y)$ is called an abolute value.

Theorem: Let $K$ be a field and $|\cdot|$ an absolute value on $K$. Assume $K$ is complete with respect to the induced metric. Then for every finite filed extension of $K$, every continuous multiplicative extension of $|\cdot|$ is an absolute value. In fact, there exists a unique such continuous multiplicative extension of $|\cdot|$, which is $|N(\cdot)|^{1/n}$.

The uniqueness follows at once from the fact that all norms are equivalent on finite dimensional vector spaces over complete fields: for multiplicative functions $\chi_1,\chi_2$, the function $\chi_1/\chi_2$ (defined on the multiplicative group) is multiplicative too, hence must be unbounded or trivial. The less trivial part of the theorem is its first part.

In my other answer I gave a proof of this fact which indeed holds in the generality of complete fields. This post is to give an easy proof I found, under the assumption that the fields are locally compact.

We consider a local filed $K$, endowed with an absolute value $|\cdot|$, a finite filed extension $L$ of $K$ and a continuous multiplicative extension $\chi$ of $|\cdot|$. We argue to show that $\chi$ is an absolute value on $L$.

Consider $L$ as a $K$-vector space and consider the corresponding space $\Omega$ consisting of $K$-norms on $L$. Consider the multiplicative group $L^*$ as a locally compact group (see Lemma 3 in my other answer for continuity of inversion) and let it act on $\Omega$ by $\|\cdot\|\mapsto \chi(x)^{-1} \|x\cdot\|$ for $x\in L^*$. Note that for $x\in K^*$, $$ \chi(x)^{-1} \|x\cdot\|=\chi(x)^{-1} |x|\|\cdot\|=\|\cdot\|,$$ thus the $L^*$-action on $\Omega$ factors via $L^*/K^*$, which is a compact group, as it is homeomorphic to a projective space. This action admits a fixed point. Indeed, for every norm $\|\cdot\|\in \Omega$, the map $$ L \ni v \mapsto \int_{L^*/K^*} \chi(x)^{-1}\|xv\|~ \text{dHaar}_{L^*/K^*}(x) \in [0,\infty)$$ is easily seen to be an $L^*$-fixed norm on $L$. We let $\|\cdot\|$ be such a fixed point which is normalized to satisfy $\|1\|=1$. Then for every $x\in L^*$, $$ \|x\|=\|x\cdot 1\|=\chi(x)\cdot \chi(x)^{-1}\|x\cdot 1\|= \chi(x)\cdot \|1\|=\chi(x). $$ Thus $\|\cdot\|=\chi$ and in particular, we conclude that $\chi$ is indeed a norm.