# Why do we only work with vector space isomorphisms over a fixed field?

1. Jul 17, 2012

### christoff

I was working on a problem in field extensions (for a 3rd-year ring theory class), and came to a point where I essentially had the following situation...

$F$ is a field isomorphic to $G$, and $G'$ is for all intents and purposes, some set. We can then consider the vector spaces $G'_F$ and $G'_G$.

I wondered to myself if it was true that $G'_F$ was isomorphic to $G'_G$. However, in my mathematical career I've only ever worked with the notion of vector space isomorphisms over a fixed field.

I ended up solving the problem differently, but the question remained... What happens if you don't fix the field?

I suppose the first inherent problem with working with an unfixed field is that the definition of a linear map would have to be changed; it would have to be something like... for $u,v\in V$ and $α$ in the field, a linear map is something which satisfies $L(αv+u)=l(α)L(v)+L(u)$ where $l$ is a field homomorphism specified in the definition of $L$.

Aside from this however, has anybody ever tried doing this, and seeing if properties like dimension are preserved under a "suitable" definition of vector space isomorphism over an unfixed field?

2. Jul 17, 2012

### DonAntonio

...and then $\,G'\,$ is not only "a set": it must be both an abelian group and a module over both fields $\,F\,,\,G\,$ ...

If the module structures over both fields (module over field = vector space, of course) are preserved under the isomorphism

(of rings) $\,F\cong G\,$, then yes: $\,G'_F\cong G'_G$ .

All the algebraic invariants, under the above assumptions I wrote, are preserved: linearly independent

sets, dimensions, etc., and you're right about the definition of linear map but only if we insist in working with both vector

spaces $\,G'_F\,,\,G'_G\,$ , something that seems to me superfluous and confusing.

DonAntonio