My question is the following: Are there algebraic norms on the fields $\mathbb{R}, \mathbf{Q_p}$ ''other'' than the absolute value, respectively $|\cdot|_p$?

Now phrasing more precisely: If generally $F$ is a field, an algebraic norm is a map $|\cdot| : K \to [0, \infty)$ such that

1) $|x| = 0 \iff x = 0$

2) $|xy| = |x||y| \forall x,y \in F$

3) $|x + y| \leq |x| + |y| \forall x,y \in F$

Two such norms $|\cdot|, ||\cdot||$ are called equivalent if and only if they generate the same topology. One can show that this is the case iff. there is an $s > 0$ such that $|x| = ||x||^s$ for all $x$. The question is: On the fields $\mathbb{R}, \mathbf{Q_p}$ where the latter one means the p-adic numbers, are there algebraic norms that are not equivalent to the usual absolute value, respectively the p-adic norm?

Of course, every such norm induces a norm on $\mathbb{Q}$, so restricted to $\mathbb{Q}$ it must be either trivial or equivalent to one of the norms mentioned above by the Thm. of Ostrowski. The problem is that i was unable to extend that to the whole field.

Thanks in advance

Fabian Werner


2 Answers 2


It is known that $\mathbb C$ is isomorphic to $\mathbb C_p$ (completion of $({\mathbb Q}_p)^{\rm{alg}}$) as fields. See e.g. Theorem 5 in this paper. So $\mathbb R$ inherts an ultrametric absolute value from that of $\mathbb C_p$. Conversely, $\mathbb Q_p$ inherts from $\mathbb C$ an archimedian absolute value.

  • $\begingroup$ Interesting. Thanks. $\endgroup$ Dec 30, 2012 at 9:37
  • $\begingroup$ the link is broken. Could you update it please? $\endgroup$
    – Chilote
    Nov 2, 2016 at 18:36

The $p$-adic norm on $\mathbb Q$ can be extended to $\mathbb R$.

We can consider pairs $(F,|\cdot|_F)$ where $F$ is a field with $\mathbb Q\subseteq F\subseteq \mathbb R$ and $|\cdot|_F$ is a norm on $F$ that extends $|\cdot |_p$. The set $\mathcal F$ of such pairs is inductively ordered: We can define $(F,|\cdot|_F)\le (E,|\cdot|_E)$ if $F\subseteq E$ and $|\cdot|_F$ is the restriction of $|\cdot|_E$ to $F$. If a subset $\mathcal T\subseteq \mathcal F$ is totally ordered, then we can let $L=\bigcup_{(F,|\cdot|_F)\in\mathcal T}F$ and define $|\cdot|_L\colon L\to[0,\infty)$ by letting $|x|_L:=|x|_F$ where $(F,|\cdot|_F)\in \mathcal T$ with $x\in F$. Then $(F,|\cdot|_F)\le (L,|\cdot|_L)$ for all $(F,|\cdot|_F)\in\mathcal T$, that is every totally ordered subset of $\mathcal F$ has an upper bound in $\mathcal F$. By Zorn's lemma, $\mathcal F$ contains a maximal element $(M,|\cdot|_M)$.

Assume there exists $\alpha\in\mathbb R\setminus M$. Then we can define $|\cdot|$ on $M(\alpha)$:

Case (i): $\alpha$ is transcendental. Every nonzero element $\beta\in M(\alpha)$ can be written as $\beta=\alpha^k\frac{f(\alpha)}{g(\alpha)}$ with $k\in\mathbb Z$ and $f,g\in M[X]$ with $f(0)\ne 0, g(0)=1$. Set $|\beta|:= |f(0)|_M$. One quickly verifies that this is a norm on $M(\alpha)$.

Case (ii): $\alpha$ is algebraic with minimal polynomial $p\in M[X]$. If $p(X)=a_0+a_1X+\cdots + a_{n-1} X^{n-1}+X^n$, select $v>0$ such that $v^n=\max_{k<n}\{v^k|a_k|_M\}$. Then define $|\sum_{k=0}^{n-1} c_k\alpha^k| = \max_{k<n}\{v^k|c_k|_M\}$. If I'm not mistaken, this is a norm on $M(\alpha)$.

In both cases we can extend $(M,|\cdot|_M)$, contradicting the maximality of $(M,|\cdot|_M)$. We conclude that $M=\mathbb R$.

  • $\begingroup$ I am having trouble verifying the triangle inequality for the simple case "$\alpha$ transcendental": Choose $q,w \in \mathbb{Q}$ arbitrary with $q \neq -w$ and $\beta = \alpha^0 \frac{q\alpha + 1}{1}$ and $\beta' = \alpha^0 \frac{w\alpha - 1}{1}$ then $\beta + \beta' = \alpha^1 \frac{q - w}{1}$ so their norm ist $|\beta + \beta'| = |q+w|_p$ which can be an arbitrary value but $|\beta| + |\beta'| = 1+1 = 2$. Could you check? $\endgroup$ Nov 29, 2012 at 12:39
  • $\begingroup$ @FabianWerner Hm, I think you are right. So can we extend $|\cdot|_p$ actually only to $\overline{\mathbb Q}$? Maybe one rather needs to assign values to $f(\alpha)$ for all irrducible polynomials $f(X)$, but how to ensure triangle inequality? If $M$ is algebraically closed this might be easy. Thus maybe one should argue that according to case ii, $M$ is algebraically closed, hence the only irreducibles are linear; set $|a+b\alpha|_p=???\max\{|a|_p,|b|_p\}???$ and everything else follows by multiplication. But again the triangle inequality looks hard. $\endgroup$ Nov 29, 2012 at 22:14
  • $\begingroup$ I was very impressed by your proof, because it seems really strange to me that there is more than just the ''usual'' absolute value on $\mathbb{R}$... everything that one learned in analysis should then be doubted, because this awkward absolute value might be the ''better'' one to do analysis... $\endgroup$ Nov 30, 2012 at 11:45
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    $\begingroup$ @HagenvonEitzen: I think your construction is correct if you take an ultrametric absolue value on $M$ and define (for $\alpha$ transcendental) $|\sum_n a_n\alpha^n|=\max_n\{ |a_n|_M \}$. $\endgroup$
    – user18119
    Dec 22, 2012 at 21:45

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