Basis of vector spaces over fields

[magicarpet512]

If you find the exact same basis for two vector spaces, then is it true that the vector spaces are equal to each other?

 

[wisvuze]

yes, if B is a basis for V then span(B) = V, and if C is a basis for W, then W = Span ( C ) . If B = C then span(C) = span ( B ) = V = W

 

[Hurkyl]

There's some relevant context information that has been left out. e.g. the span of a set doesn't make any sense, although the span of a subset of a particular vector space does.

 

[magicarpet512]

to be more specific...
I believe it is correct that Q($$\sqrt{a}$$,$$\sqrt{b}$$) has a basis of {1, $$\sqrt{a}$$, $$\sqrt{b}$$, $$\sqrt{ab}$$}
and Q($$\sqrt{a}$$ + $$\sqrt{b}$$ has a basis of {1, $$\sqrt{a}$$, $$\sqrt{b}$$, $$\sqrt{ab}$$}.
We know that Q($$\sqrt{a}$$,$$\sqrt{b}$$) = Q($$\sqrt{a}$$ + $$\sqrt{b}$$. Is this true because they have the same basis? Would showing that both of these have the same basis be enough to prove the equality? Does this generalize to any other vector spaces that have the same bases?

 

[Hurkyl]

You can't prove two fields equal by proving them equal as vector spaces -- two fields can be equal as vector spaces but have different multiplication operations.

In this case there is a clear overfield, which simplifies things -- e.g. we could use C or the algebraic closure of Q, or even $\mathbf{Q}(\sqrt{a}, \sqrt{b}, \sqrt{a}+\sqrt{b})$.



That said, is there any particular reason why you aren't arguing that each field contains the generators of the other?