Normed vector spaces are typically defined over the reals or complex numbers.

Is there any "standard," well-behaved construction that generalizes this to a vector space over a finite field, such as $\Bbb F_2$?

I'm looking for something kind of like the class of $\ell_p$ norms, except designed with finite fields in mind.

Ideally, something that has deep fundamental properties making it well-behaved in the same way that the Euclidean norm is.

  • 2
    $\begingroup$ This question discusses inner product spaces over finite fields. $\endgroup$ Commented Feb 19, 2017 at 18:24

1 Answer 1


There is a "standard" way to consider normed spaces over arbitrary fields but these are not well-behaved in the case of scalars in finite fields. If you want to work with norms on vector spaces over fields in general, then you have to use the concept of valuation.

Valued field: Let $K$ be a field with valuation $|\cdot|:K\to\mathbb{R}$. This is, for all $x,y\in K$, $|\cdot|$ satisfies:

  1. $|x|\geq0$,
  2. $|x|=0$ iff $x=0$,
  3. $|x+y|\leq|x|+|y|$,
  4. $|xy|=|x||y|$.

The set $|K|:=\{|x|:x\in K-\{0\}\}$ is a multiplicative subgroup of $(0,+\infty)$ called the value group of $|\cdot|$. The valuation is called trivial, discrete or dense accordingly as its value group is $\{1\}$, a discrete subset of $(0,+\infty)$ or a dense subset of $(0,+\infty)$. For example, the usual valuations in $\mathbb{R}$ and $\mathbb{C}$ are dense valuations. The valuation is said to be non-Archimedean when it satisfies the strong triangle inequality $|x+y|\leq\max\{|x|,|y|\}$ for all $x,y\in K$. In this case, $(K,|\cdot|)$ is called a non-Archimedean valued field and $|n1_K|\leq1$ for all $n\in\mathbb{Z}$. Common examples of non-Archimedean valuations are the $p$-adic valuations in $\mathbb{Q}$ or the valuations of a field that is not isomorphic to a subfield of $\mathbb{C}$.

Norm: Let $(K,|\cdot|)$ be a valued field and $X$ be a vector space over $(K,|\cdot|)$. A function $p:X\to \mathbb{R}$ is a norm iff for each $a,b\in X$ and each $k\in K$, it satisfies:

  1. $p(a)\geq0$ and $p(a)=0$ iff $a=0_X$,
  2. $p(ka)=|k|p(a)$,
  3. $p(a+b)\leq p(a)+p(b)$

In the case of a finite field, the valuation $|\cdot|$ must be the trivial one. In fact, if there is nonzero scalar $x\in K$ such that $|x|\neq1$, then $\{|x^n|:n\in\mathbb{Z}\}=\{|x|^n:n\in\mathbb{Z}\}$ is an infinite subset of $\mathbb{R}$, so $|K|$ is infinite as well which is a contradiction.

Example of Normed space over a finite field: Let $K$ be any field with the trivial valuation (e.g. a finite field) and let $X$ be an infinite-dimensional vector space with Hamel basis $B$. We can define a norm $p$ by saying $p(e)$ is the number of nonzero coefficients there are when we write $e$ as a linear combination of elements of $B$.

But in this context, we have unexpected situations. For example, two norms may induce the same topology without being equivalent. In fact, consider the trivial norm $q$ on $X$ defined by $q(e)=1$ for all nonzero $e\in X$. Then both norms, $p$ and $q$, induce the discrete topology, but $p/q$ is unbounded. So there are no constant $C$ such that $ p\leq Cq$.

A comprehensive starting point to read about normed spaces in this context is the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - Dekker New York (1978).

For more information on finite fields, I recommend the paper: Non-archimedean Banach spaces over trivially valued fields, Borrey, S., P-adic functional analysis, Editorial Universidad de Santiago, Chile, 17 - 31. (1994). There, the norm is assumed to satisfy the strong triangle inequality.

For the study of more advanced stuff, like locally convex spaces over valued fields I recommend the book: Locally Convex Spaces over non-Arquimedean Valued Fields - [C.Perez-Garcia,W.H.Schikhof] - Cambridge Studies in Advanced Mathematics (2010).

  • $\begingroup$ Shunning $\mathbb R$, what if e.g. we replaced each inequality $a\leq b$ by the assertion that $b-a$ is a square in $\mathbb K$? Can this make an interesting different kind of "absolute value" or "norm"? $\endgroup$
    – mr_e_man
    Commented Sep 27, 2023 at 1:57
  • $\begingroup$ For example $|x|=x^2$ when $\text{char}(\mathbb K)=2$, or any group homomorphism from $\mathbb K^*$ to itself when $\mathbb K$ has no non-squares. $\endgroup$
    – mr_e_man
    Commented Sep 27, 2023 at 2:10

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