# Thread: Subfields and Vector Spaces

1. ## Subfields and Vector Spaces

Consider a vector space V over a field F. Let K be a subfield of F.
Prove that V over K is a vector space if we retain the same addition operation and restrict scalar multiplication to scalars from K

Work so far
I figure if we just prove that there exists the 0 element, and that addition and scalar multiplication are closed, the statement.

I know there exists a zero element, as 0x=0 for $x\in{V}$ and $0\in{F}$, and this 0 is also in K

Under scalar multiplication, all scalars from K are also in F, so if it was closed under F it will also be closed under K

But how do I resolve addition if the same operation is retained.

2. Originally Posted by I-Think
Consider a vector space V over a field F. Let K be a subfield of F.
Prove that V over K is a vector space if we retain the same addition operation and restrict scalar multiplication to scalars from K

Work so far
I figure if we just prove that there exists the 0 element, and that addition and scalar multiplication are closed, the statement.

I know there exists a zero element, as 0x=0 for $x\in{V}$ and $0\in{F}$, and this 0 is also in K

Under scalar multiplication, all scalars from K are also in F, so if it was closed under F it will also be closed under K

But how do I resolve addition if the same operation is retained.
What do you mean? Since it's a field the sum of two field elements is in the field?

3. Yes. If the same operation is retained as V over F, then the sums of elements of V over K should also be closed, shouldn't it?

4. Originally Posted by I-Think
Yes. If the same operation is retained as V over F, then the sums of elements of V over K should also be closed, shouldn't it?
Yes, I think I understand where the confusion is coming in. We have in general for a vector space $\left(V,+,\mu,F\right)$ where $\mu$ is scalar multiplication that $\left(V,+\right)$ is an abelian group, this is independent of the other parts of the vector space. Does that help?