Ttyjfyk

I'm doing an introductory module in number theory, and came across the definition of a norm on a field. It seems to agree with the definition of a norm on a vector space over a field (just view the field as a vector space in the former definition). Then I read Ostrowski's Theorem that says 'every non trivial norm on $mathbbQ$ is equivalent either to the standard absoltute value or to the p-adic norm for some prime p. All these norms are inequivalent.'

I don't see how this theorem doesn't contradict the theorem that all norms on a finite dimensional vector space are equivalent.....However I see a difference in the way 'equivalence' is being defined between norms in the case where the norms are on a field as opposed to a general vector space. Everything seems a bit all over the place in my head and I don't know why there is a difference in definition of norms being equivalent on fields and norms on vector spaces when surely a field is a vector space...

If someone could clear things up I'd appreciate it.

The difference is how multiplication by scalars is handled.

When considering a norm $VertcdotVert$ on a vector space $V$ over some field $F$, you need to fix the notion of absolute value on the scalar-field $F$, so that the axiom
$$Vert ccdot xVert = vert cvert cdotVert xVert,quad cin F, xin V$$
makes sense.

The statement that 'all norms on a finite dimensional vector space are equivalent' makes use of the assumption that the above axiom is satisfied for the same notion of absolute value. This axiom fails if we try to view the $p$-adic norms as norms on vector spaces over $mathbbQ$ with the usual notion of absolute value. E.g. for the $2$-adic norm, we have
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$

I think that everything that @Eero has said is true, but there’s more wrong in the situation than that.

You have not quoted the theorem about norms on a finite-dimensional vector space fully. First, you have to start with a normed field $k$, and ask about the norms on a finite $k$-space that in some sense extend that norm. Second, the theorem does not hold if $k$ is not complete under the given norm.

So there’s no contradiction, because in talking about $Bbb Q$, you have not started with a pre-given norm on a field. Rather, you are asking whether there may be more than one norm on a field that has been given abstractly. Since $Bbb Q$ has many topologies on it, each of them making it a topological field (with continuity of $+$, $times$, and $div$, for instance) and all of them different as topologies, it should not be surprising that the completions are different.