Why does the field norm on the field extension $ \mathbb C/\mathbb R $ induce a vector space norm?

Question

There is a general result which holds for the rational numbers $ \mathbb Q $ (as well as number fields in general):

For any completion $ K $ of $ \mathbb Q $ and any finite extension $ L/K $ of degree $ n $, the function $ L \to \mathbb R $ defined by $ x \to \sqrt[n]{|N_{L/K}(x)|} $ gives a norm on $ L $.

The nontrivial part is to prove that the norm thus defined satisfies the triangle inequality. $ K $ is either $ \mathbb R $ or $ \mathbb Q_p $ for a prime $ p $, and for the latter case one can argue using Hensel's lemma that there is an equivalence between having norm $ \leq 1 $ and being integral over the $ p $-adic integers, which combined with the ultrametric property of the $ p $-adic norm is sufficient to prove the claim. I'm satisfied with this argument in the sense that it seems to give a "moral" explanation for why we should expect the claim to be true.

However, the argument for $ K = \mathbb R $ is simply to note that the claim is obviously true for the only nontrivial finite extension of $ \mathbb R $, which is $ \mathbb C/\mathbb R $. While this clearly is sufficient for a proof, I don't understand why one should expect this to be true in advance. Is there a better motivation for why we should expect this result to hold, perhaps an argument which treats all completions symmetrically instead of having one argument for the finite primes and another for the prime at infinity?

Answer 1

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.

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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 $L^*$.

Answer 2

In fact, more is true: for any local field $K$, any degree $n$ field extension $L$ of $K$ and any absolute value $|\cdot|$ on $K$, $|N_K^L(\cdot)|^{1/n}$ is the unique absolute value on $L$ which extends $|\cdot|$. In particular, it is a norm on $L$. In this post I intend to give a proof of this fact which does not rely on properties of $K$ other than local compactness.

Since the post became longer than I expected, here is a summery: it is a general fact that an extension of an absolute value which is a $C$-absolute value is an actual absolute value and the intended map is a $C$-absolute value by the properness of the norm map for a local field extension. Those notions will be explained below. The proofs of the technical Lemma 1 and Lemma 2 are postponed to the end of this text, not to interrupt the reading flow.

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Let me give some preliminaries. We regard here a field $F$ and a multiplicative function $|\cdot|:F\to [0,\infty)$, that is a function satisfying $|0|=0$, $|1|=1$ and $|xy|=|x||y|$ for every $x,y\in F$. For $C\geq 1$ we say that $|\cdot |$ is a $C$-absolute value if for every $x,y\in F$, $|x+y|\leq C(|x|+|y|)$ and if $C=1$ we simply say that $|\cdot|$ is an absolute value. The following is well known.

Lemma 1: A $2$-absolute value is an absolute value.

It is an easy exercise to check that if $|\cdot|$ is a $C$-absolute value and $\alpha\in(0,1]$ then $|\cdot |^\alpha$ is a $C^\alpha$-absolute value. However, this does not work in general for $\alpha>1$. To remify this, we study a more homogenous condition. We say that $|\cdot |$ is a $C$-ultra absolute value if for every $x,y\in L$, $|x+y|\leq C\max\{|x|,|y|\}$ and if $C=1$ we say that $|\cdot|$ is an ultra absolute value. Now we indeed have that if $|\cdot |$ is a $C$-ultra absolute value then for every $\alpha>0$, $|\cdot |^\alpha$ is a $C^\alpha$-ultra absolute value. The two definitions relate trivially: a $C$-ultra absolute value is a $C$- absolute value while a $C$-absolute value is a $2C$-ultra absolute value. In particular, every absolute value is a $2$-ultra absolute value. The following, however, is less trivial.

Lemma 2: An absolute value $|\cdot|$ is a $\max\{1,|2|\}$-ultra absolute value.

Corollary A: A $C$-ultra absolute value $|\cdot|$ is a $\max\{1,|2|\}$-ultra absolute value.

Proof: Set $\alpha=\log_C 2$ and consider the $2$-ultra absolute value $|\cdot |^\alpha$. It is trivially a $2$-absolute value, thus an actual absolute value by Lemma 1. By Lemma 2 it is a $\max\{1,|2|^\alpha\}$-ultra absolute value. Taking now the $1/\alpha$-power, we get that $|\cdot|$ is indeed a $\max\{1,|2|\}$-ultra absolute value. $\square$

Corollary B: A $C$-absolute value $|\cdot|$ is an absolute value iff $|2|\leq 2$.

Proof: If $|\cdot|$ is an absolute value then clearly $|2|=|1+1|\leq |1|+|1|=2$. Assume $|\cdot|$ is a $C$-absolute value and $|2|\leq 2$. Then $|\cdot|$ is a $2C$-ultra absolute value, thus by Corollary A, it is a $\max\{1,|2|\}$-ultra absolute value, hence a $2$-ultra absolute value, as $\max\{1,|2|\}\leq 2$. In particular, $|\cdot|$ is a $2$-absolute value, thus it is an actual absolute value by Lemma 1. $\square$

Corollary C: A $C$-absolute value on $F$ which restricts to an absolute value on a subfield is an absolute value on $F$.

Proof: This follows from Corollary B, as 2 belongs to the subfield. $\square$

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We are now back to the original setting, where $L$ is a finite filed extension of the local field $K$ and $|\cdot|$ is an absolute value on $K$. We treat $L$ as a locally compact space by identifying it with $K^n$, noting that the topology is independent of the choice made. We denote by $\hat{K}$ and $\hat{L}$ the corresponding one point compactifications.

Lemma 3: The inversion map $x\mapsto x^{-1}$ is continuous on $\hat{K}$ and $\hat{L}$ and it interchanges $0$ and $\infty$.

Proof: The statement for $K$ 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 on $L^*$. The multiplication by scalar action of $K^*$ on $L^*$ is cocompact, as the quotient is homeomorphic to $\mathbb{P}^{n-1}(K)$, so there exists a compact subset $B\in L^*$ such that $L^*=K^*B$. Fixing a norm on the $K$ vector space $L$, $B$ and its inversion image are both bounded, and the proof follows easily. $\square$

Recall that a proper map is a continuous map for which preimages of compact sets are compact. Equivalently, maps which are continuous at infinity, that is they extend continuously to the corresponding one point compactifications.

Lemma 4: The map $N=N_K^L:L\to K$ is proper.

Proof: In view of Lemma 3, this follows easily from the continuity of $N$ at 0 and the fact that $N(x^{-1})=N(x)^{-1}$. $\square$

Theorem: The map $|N(\cdot)|^{1/n}:L\to [0,\infty)$ is an absolute value.

Proof: The unit ball $B\subset K$ is compact, hence so is $N^{-1}(B)\subset L$ and its shift $1+N^{-1}(B)$. It follows that the image in $[0,\infty)$ under $|N(\cdot)|$ of $1+N^{-1}(B)$ is bounded by some $C$, thus for $z\in L$, $$ |N(z)|\leq 1 \Rightarrow |N(z)+1|\leq C. $$ It follows that $|N(\cdot)|$ is a $C$-ultra absolute value. Indeed, for $x,y\in L$, assuming wlog $|x|\leq |y|$ and setting $z=xy^{-1}$ we have $$ |N(x+y)|=|N(y)||N(z)+1|\leq C|N(y)||N(z)+1|= C(|N(x)|+|N(y)|).$$ It follows that $|N(\cdot)|^{1/n}$ is a $C^{1/n}$-ultra absolute value, thus by Corollary C, it is an actual absolute value. $\square$

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I will now provide the proofs of Lemma 1 and Lemma 2.

Proof of Lemma 1: Assume $|\cdot|$ is a 2-absolute value.

We first observe that for every natural $j$, and every $2^j$ elements $x_1,\ldots, x_{2^j}\in F$, we have $$ |\sum_{i=1}^{2^j} x_i|\leq 2^j \sum_{i=1}^{2^j} |x_i|.$$ Indeed, this follows easily by induction on $j$. Picking any natural $m$ and considering $j$ such that $2^{j-1}<m\leq 2^j$, we get that for every $m$ elements $x_1,\ldots, x_m\in F$, we have $$ |\sum_{i=1}^{m} x_i|\leq 2m \sum_{i=1}^{m} |x_i|.$$ Indeed, this follows by adding $x_i=0$ for $m<i\leq 2^j$ to the list and observing that $2^j\leq 2m$. In particular, by taking $x_i=1$, we now have for every natural $m$, $|m|\leq 2m$.

We now pick arbitrary $x,y\in F$ and a natural $n$ and make the following estimates: $$ |x+y|^n=|(x+y)^n|=\left|\sum_{k=0}^n {n\choose k} x^ky^{n-k}\right| \leq 2(n+1) \sum_{k=0}^n \left|{n\choose k}\right| |x^k||y^{n-k}| \leq $$ $$ 4(n+1) \sum_{k=0}^n {n\choose k}|x^k||y^{n-k}|=4(n+1)(|x|+|y|)^n.$$ Taking $n$th root an letting $n\to \infty$, we get indeed, $$ |x+y|\leq |x|+|y|.$$ $\square$

For the proof of Lemma 2 we will need the following.

Claim: Assume $|\cdot|$ is an absolute value. Then for every pair of naturals $k< m$, we have $|k|\leq \max\{1,|m|\}$.

Proof of the Claim: Fix a natural $n$ and expand $k^n$ on base $m$, $$ k^n=\sum_{i=0}^{n-1} a_i m^i $$ for some integers $0\leq a_i<m$. Note that indeed it is enough to consider indexes bounded by $n-1$, as $k<m$. Thus we have $$ |k|^n=|k^n|=|\sum_{i=0}^{n-1} a_i m^i|\leq \sum_{i=0}^{n-1} a_i |m|^i < m \sum_{i=0}^{n-1} |m|^i. $$ If $|m|\leq 1$ then we get $ k^n < mn $ and taking $n$th root and $n\to \infty$ we conclude that indeed $|k|\leq 1\leq \max\{1,|m|\}$. If $|m|> 1$ then we get $$ |k|^n <m \sum_{i=0}^{n-1} |m|^i = \frac{m(|m|^n-1)}{|m|-1}$$ and taking $n$th root and $n\to \infty$ we conclude that indeed $|k|\leq |m|\leq \max\{1,|m|\}$. This proves the claim. $\square$

Proof of Lemma 2: We assume $|\cdot|$ is an absolute value. We pick arbitrary $x,y\in F$ and a natural $n$. We note that for $0\leq k\leq n$ we have $|x^k||y^{n-k}|\leq \max\{|x|,|y|\}^n$. Also we have ${n\choose k}\leq 2^n$, thus by the claim $|{n\choose k}|\leq \max\{1,|2^n|\}=\max\{1,|2|\}^n$. We get the following estimates: $$ |x+y|^n=|(x+y)^n|=\left|\sum_{k=0}^n {n\choose k} x^ky^{n-k}\right| \leq \sum_{k=0}^n \left|{n\choose k}\right| |x^k||y^{n-k}| \leq $$ $$ \sum_{k=0}^n \max\{1,|2|\}^n\max\{|x|,|y|\}^n = (n+1)\max\{1,|2|\}^n\max\{|x|,|y|\}^n.$$ Taking $n$th root an letting $n\to \infty$, we get indeed, $$ |x+y|\leq \max\{1,|2|\}\max\{|x|,|y|\}.$$ $\square$

Answer 3

$\newcommand\abs[1]{\lvert#1\rvert}\DeclareMathOperator\N{N}\DeclareMathOperator\Gal{Gal}$You are asking for a conceptual reason why the absolute value on $\mathbf C$ should be given by the usual formula involving the field norm from $\mathbf C$ to $\mathbf R$: why it is the only possible formula for an absolute value on $\mathbf C$ extending the usual absolute value on $\mathbf R$. The other answers so far focus on showing the norm formula really is an absolute value, but I don't think that is what you were actually asking. It looks like you were asking for an explanation of a uniqueness property and were shown instead an explanation for an existence property. These are both important, but they are not the same thing.

The reason for the uniqueness is based on the equivalence of all vector space norms on each finite-dimensional vector space over a complete (not just locally compact) valued field. For example, the equivalence of vector space norms is true when $K = \mathbf C_p$, which is complete but not locally compact, so any proof of equivalence of all vector space norms that relies on local compactness of the scalar field would not cover that case.

Suppose $K$ is a field complete with respect to an absolute value $\abs\cdot$ and $V$ is a finite-dimensional $K$-vector space (such as a finite extension field of $K$). That all $K$-vector space norms on $V$ are equivalent is Theorem 3.2 in my notes Equivalence of norms. It implies that $\abs\cdot$ has at most one extension to an absolute value on each finite extension field $L$ of $K$. Indeed, if $\abs\cdot_1$ and $\abs\cdot_2$ are absolute values on $L$ that extend the absolute value $\abs\cdot$ on $K$ then $\abs\cdot_1$ and $\abs\cdot_2$ are also $K$-vector space norms on $L$ and thus are equivalent as norms, so there are positive constants $A$ and $B$ such that $A\abs x_2 \leq \abs x_1 \leq B\abs x_2$ for all $x \in L$. Replacing $x$ with $x^m$ for a positive integer $m$ and using multiplicativity of absolute values, $A^{1/m}\abs x_2 \leq \abs x_1 \leq B^{1/m}\abs x_2$. Now letting $m \to \infty$, we get $\abs x_2 \leq \abs x_1 \leq \abs x_2$ for all $x \in L$, so $\abs\cdot_1 = \abs\cdot_2$ on $L$.

Now I will explain why the unique extension of an absolute value $\abs\cdot$ on $K$ to a finite extension field of $K$, when $K$ is complete with respect to $\abs\cdot$, must be given by the norm formula when $K$ has characteristic $0$. Note I will not prove the norm formula in fact defines an absolute value (it does), but I don't think that is what you're asking anyway: you are looking for an explanation of why we should anticipate (a "moral" reason) that an extension of the absolute value comes from the field norm. (The norm formula is an absolute value on finite extension fields for all complete $K$, even when $K$ has positive characteristic, but the motivation for using the norm formula is all you are seeking and for that I think characteristic $0$ is adequate.) I will treat the case of $\mathbf C/\mathbf R$ first since the argument in that case contains the basic idea and is more concrete.

The key idea is this: if $L/K$ is a finite extension field and $\abs\cdot_L$ is an absolute value on $L$ that extends $\abs\cdot$ on $K$, then for each $K$-automorphism $\sigma$ of $L$, we have $\abs{\sigma(\alpha)}_L = \abs\alpha_L$ for every $\alpha \in L$. Indeed, $\alpha \mapsto \abs{\sigma(\alpha)}_L$ is an absolute value on $L$ that extends $\abs\cdot$ on $K$ (because $\sigma$ fixes $K$ pointwise), so by the uniqueness of extensions of absolute values on complete fields to finite extension fields (not the existence of them!), we have $\abs{\sigma(\cdot)}_L = \abs\cdot_L$ on $L$.

Theorem: An absolute value on $\mathbf C$ that extends the usual absolute value on $\mathbf R$ must be given by the formula $z \mapsto \sqrt{\abs{\N_{\mathbf C/\mathbf R}(z)}}$.

Proof: Let $\abs\cdot_{\mathbf C}$ be a hypothetical absolute value on $\mathbf C$ that extends the usual absolute value on $\mathbf R$: $\abs x_{\mathbf C} = \abs x$ for $x \in \mathbf R$. Since complex conjugation on $\mathbf C$ is an $\mathbf R$-automorphism of $\mathbf C$, $\abs{\overline{z}}_{\mathbf C} = \abs z_{\mathbf C}$ for all $z \in \mathbf C$ by the argument given in the paragraph preceding this proof. Thus $$ \abs z_{\mathbf C}\abs{\overline{z}}_{\mathbf C} = \abs z_{\mathbf C}^2. $$ The left side is $\abs{z\overline{z}}_{\mathbf C} = \abs{\N_{\mathbf C/\mathbf R}(z)}_{\mathbf C}$. Since $\N_{\mathbf C/\mathbf R}(z)$ is a real number, $$ \abs{\N_{\mathbf C/\mathbf R}(z)}_{\mathbf C} = \abs{\N_{\mathbf C/\mathbf R}(z)}. $$ Thus $$ \abs z_{\mathbf C}^2 = \abs{\N_{\mathbf C/\mathbf R}(z)} $$ and taking square roots of both sides shows $\abs z_{\mathbf C} = \sqrt{\abs{\N_{\mathbf C/\mathbf R}(z)}}$. That ends the proof.

Theorem: If $K$ is a field of characteristic $0$ that is complete with respect to an absolute value $\abs\cdot_K$ and every finite extension field $L$ of $K$ has an absolute value $\abs\cdot_L$ that extends $\abs\cdot_K$ then that extended absolute must be given by the formula $\abs x_L = \sqrt[n]{\abs{\N_{L/K}(x)}_K}$ for all $x \in L$, where $n = [L:K]$.

Proof: Since $K$ has characteristic $0$, we can enlarge $L$ to a finite Galois extension $M/K$, so $M \supset L \supset K$. Set $n = [L:K]$ and $d = [M:K]$, so $n \mid d$. By hypothesis, $\abs\cdot_K$ has extensions $\abs\cdot_L$ and $\abs\cdot_M$ to absolute values on $L$ and $M$, and $\abs\alpha_L = \abs\alpha_M$ for all $\alpha \in L$ by the uniqueness of extensions.

Step 1: $\abs\alpha_M = \sqrt[d]{\abs{\N_{M/K}(\alpha)}_K}$ for all $\alpha \in M$.

To prove Step 1, we have $\abs{\sigma(\alpha)}_M = \abs\alpha_M$ for all $\sigma \in \Gal(M/K)$ by the uniqueness of extensions. Since $\N_{M/K}(\alpha) = \prod_{\sigma} \sigma(\alpha)$, where the product runs over all $\sigma \in \Gal(M/K)$, $$ \abs{\N_{M/K}(\alpha)}_M = \prod_{\sigma} \abs{\sigma(\alpha)}_M = \abs\alpha_M^d. $$ Since $\N_{M/K}(\alpha) \in K$, $\abs{\N_{M/K}(\alpha)}_M = \abs{\N_{M/K}(\alpha)}_K$, so $$ \abs\alpha_M^d = \abs{\N_{M/K}(\alpha)}_K. $$ Take $d$th roots of both sides and we're done with Step 1.

Step 2: $\abs\alpha_L = \sqrt[n]{\abs{\N_{L/K}(\alpha)}_K}$ for all $\alpha \in L$.

By Step 1, $\abs\alpha_M = \sqrt[d]{\abs{\N_{M/K}(\alpha)}_K}$. The left side is $\abs\alpha_L$ by the uniqueness of extensions. On the right side, use transitivity of the norm: $$ \N_{M/K}(\alpha) = \N_{L/K}(\N_{M/L}(\alpha)) = \N_{L/K}(\alpha^{[M:L]}) = \N_{L/K}(\alpha)^{[M:L]}. $$ Since $[M:L] = d/n$, $$ \abs\alpha_L = (\abs{\N_{L/K}(\alpha)^{d/n}}_K)^{1/d} = \abs{\N_{L/K}(\alpha)}_K^{(d/n)/d} $$ so $$ \abs\alpha_L = \abs\alpha_M = \abs{\N_{L/K}(\alpha)}_K^{1/n} = \sqrt[n]{\abs{\N_{L/K}(\alpha)}_K} $$ and that finishes Step 2.

Answer 4

In fact if $(K,\lvert-\rvert)$ is a complete valued field and $L$ is a finite extension of $K$ one can always construct a multiplicative extension of $\lvert-\rvert$ to $L$. Then the argument in KConrad's answer will show that it has to coincide with $\lvert N_{L/K}(-)\rvert^{1/[L:K]}$.

This is theorem 26.3 in Warner's Topological fields. Since the argument there uses many results in topological ring theory I will present an abridged version here for the convenience of the reader.

Let us say that a vector space norm $\|-\|$ on $L$ is multiplicative if $\|1\|=1$ and $\|xy\|\le \|x\|\|y\|$. Moreover we say that $\|-\|$ is spectral if it is multiplicative and we have $\|x^n\|=\|x\|^n$ for every $x\in L$ and $n\ge1$.

Lemma Any two spectral norms on $L$ coincide

Proof: This is the usual argument for the uniqueness of the norms: let $\|-\|_1$ and $\|-\|_2$ be two spectral norms. Then, since any two vector space norms on $L$ are equivalent there exist $c,C>0$ such that $$ c \|x\|_1\le \|x\|_2\le C\|x\|_1$$ foer every $x\in L$. Therefore, by plugging in $x^n$ are taking $n$-th roots we have, by spectrality $$ c^{1/n} \|x\|_1\le \|x\|_2\le C^{1/n}\|x\|_1$$ and letting $n\to \infty$ yields the thesis. $\square$

Lemma There exist a (necessarily unique) spectral norm on $L$.

Proof Take $\|-\|$ a multiplicative norm on $L$ (for example the restriction of the $\ell^1$ operatorn norm on $\operatorname{End}_K(L)$). Then for any $x\in L$ the sequence $\|x^n\|^{1/n}$ is $$\lvert x\rvert:=\lim_{n\ge 1}\|x^n\|^{1/n}$$ (the limit exists because of the multiplicativity of $\|-\|$). We need to prove $\lvert x\rvert$ is a spectral norm on $L$. The only tricky part is to show the triangular identity (note that since $L$ is a field, every multiplicative seminorm is automatically a norm, as its kernel is an ideal). Take $x,y\in L$ and fix $\epsilon>0$. Then there is $N>0$ such that for all $n\ge N$ we have $$\|x^n\|^{1/n}\le \lvert x\rvert+\epsilon\textrm{ and }\|y^n\|^{1/n}\le \lvert y\rvert+\epsilon$$ Therefore there is $C>0$ such that for every $n\ge 0$ we have $$\|x^n\|\le C(\lvert x\rvert+\epsilon)^n\textrm{ and }\|y^n\|\le C(\lvert y\rvert+\epsilon)^n$$ Then $$\|(x+y)^n\|=\left\|\sum_{i=0}^n {n\choose i} x^{n-i}y^i\right\|\le C^2\sum_{i=0}^n {n\choose i} (\lvert x\rvert +\epsilon)^{n-i}(\lvert y\rvert+\epsilon)^i=C^2(\lvert x\rvert +\lvert y\rvert +2\epsilon)^n$$ Finally taking $n$-th roots, letting $n\to \infty$ and $\epsilon\to 0$ proves the thesis. $\square$

Proposition Any spectral norm on $L$ is an absolute value (that is $\lvert xy\rvert=\lvert x\rvert \lvert y\rvert$).

Proof Let $\lvert -\rvert$ be a spectral norm on $L$ and fix $y\in L$. We will show that for every $x\in L$ $\lvert xy \rvert =\lvert x\rvert \lvert y\rvert$. If $y=0$, this is clear, so assume $y\neq 0$, so $\lvert y\rvert\neq 0$. Then define $$ \lvert x\rvert_y:=\lim_{n\to \infty} \lvert xy^n\rvert \lvert y\rvert^{-n}$$ It is easy to verify that $\lvert -\rvert_y$ is a spectral norm on $L$, and therefore $\lvert xy\rvert=\lvert xy\rvert_y$ But we have $\lvert xy\rvert_y=\lvert x\rvert_y \lvert y\rvert$. $\square$

Note that the only fact we have used on $K$ is that any two norms on finite dimensional vector spaces on $K$ are equivalent.