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

[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?

 

[DonAntonio]

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.

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

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 .

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?

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