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It is widely known that on a finite-dimensional vector space over a complete field, every norm is equivalent. However, I'm looking for a counterexample over a field which is not complete.

I have found no such counterexample neither by myself nor in the internet, but it is frequently stated that such result does not hold for non-complete fields, without a proof.

Edit: The stackexchange app notified me of a comment which I believe is now deleted. Still, I will address its concern in case somebody has the same idea.

I already tried to treat $\mathbb{Q}$ as a vector space over itself and find two non-equivalent norms. As for Ostrowski's theorem we know that, for example, the absolute value and the 2-adic norm are not equivalent. However, I noticed that the $p$-adic norm, though being a norm on the field $\mathbb{Q}$, it is not a norm on the vector space $(\mathbb{Q},|\cdot|)$, and thus is not a valid counterexample.

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  • $\begingroup$ In looking for an example over $\mathbb Q$, it seems that the crucial question is, can there be a seminorm on $\mathbb R^n$ that takes rational values on $\mathbb Q^n$ but is not a norm? $\endgroup$
    – Tom Goodwillie
    Commented Mar 21, 2017 at 17:41
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    $\begingroup$ When considering a norm on a vector space over a field we must fix first an absolute value on the field itself. Take for example $\mathbb{Q}$ endowed with its usual (real) absolute value. Fix a number field $k$. Consider the various embedding of $k$ into $\mathbb{C}$. For each embedding pull back the usual absolute value of $\mathbb{C}$. You will get inequivalent norms on the $\mathbb{Q}$- vector space $k$. $\endgroup$
    – Uri Bader
    Commented Mar 21, 2017 at 18:01
  • $\begingroup$ Explicitly, and related to the answer be Gerald Edgar, when $k=\mathbb{Q}(\sqrt{2})$ you will get the two norms on $\mathbb{Q}^2$: $(x,y)\mapsto |x+\sqrt{2}y|$ and $(x,y)\mapsto |x-\sqrt{2}y|$. $\endgroup$
    – Uri Bader
    Commented Mar 21, 2017 at 18:04
  • $\begingroup$ I misinterpreted the question, I'm sure. $\endgroup$
    – Tom Goodwillie
    Commented Mar 21, 2017 at 18:05
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    $\begingroup$ @TomGoodwillie (assuming I did not misinterpreted the question myself) you probably misinterpreted the question because the notion of norm on a vector space was not defined in it. There is a standard definition (which is easy to guess), but norms are not defined merely on vectors spaces over fields but rather on vector spaces over fields endowed with an absolute value. This should be clarified by the OP. In this case, I don't think this is a "research level question". $\endgroup$
    – Uri Bader
    Commented Mar 21, 2017 at 18:20

2 Answers 2

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Field $\mathbb Q$ with the usual absolute value $|\cdot|$ from the real numbers.

Two norms on $\mathbb Q^2$ ... $$ \|(x,y)\|_1 = |x|+|y| $$ and $$ \|(x,y)\|_2 = \left|\,x+\sqrt{2}\;y\,\right| $$ In both of these, $|\cdot|$ is still the usual absolute value for the real numbers.

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  • $\begingroup$ But I think the norm was meant to take values in the given field. $\endgroup$
    – Tom Goodwillie
    Commented Mar 21, 2017 at 17:42
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    $\begingroup$ @TomGoodwillie ... The OP should tell us if Tom is right. If so, please provide a definition of "norm" for us. (The usual definition says real values, even for a non-real field.) $\endgroup$
    – Gerald Edgar
    Commented Mar 21, 2017 at 17:47
  • $\begingroup$ Yes, I'm probably wrong. $\endgroup$
    – Tom Goodwillie
    Commented Mar 21, 2017 at 18:06
  • $\begingroup$ Do you know any example in the case of Non-archimedean norms? @GeraldEdgar $\endgroup$
    – FNH
    Commented Nov 15, 2022 at 0:46
  • $\begingroup$ @FawzyHegab All norms on the two-dimensional vector space are equivalent, provided the field is complete (as the OP noted). So counterexamples must use an incomplete field. Non-archimedean? Try to do it where the field is $\mathbb Q$ with the $p$-adic absolute value (do not complete it to the $p$-adic numbers). Imitate the proof above, but in place of $\sqrt{2}$ use an irrational $p$-adic number. $\endgroup$
    – Gerald Edgar
    Commented Nov 15, 2022 at 17:23
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A variant on other comments: as a fairly cliched number-theoretic riff, let $\mathbb Q$ have the "usual" absolute value $|\cdot|$. Let $V=\mathbb Q[x]/\langle x^2-2\rangle$, and $\alpha$ the image of $x$ in that quotient. Let $\sqrt{2}$ be the usual real number $>0$ whose square is $2$. Then there are two distinct absolute values $\|\cdot\|_i$ (with $i=1,2$) such that $\|t\cdot v\|_i=|t|\cdot \|v\|_i$ for $i=1,2$, $t\in \mathbb Q$, and $v\in V$: $\|a+b\alpha\|_1=|a+b\sqrt{2}|$ and $\|a+b\alpha\|_2=|a-b\sqrt{2}|$, where the single-bar absolute values are the usual absolute value on $\mathbb R$.

(No, these $\|\cdot\|$ do not take values in $\mathbb Q$.)

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