# How to show simply that a field is a vector space over any of its subfields?

I know that any field is a vector space over itself, see e.g. here -- Prove that the field F is a vector space over itself.

Is there a correspondingly simple argument showing that this is the case for any subfield?

For example, $\mathbb{C}$ can be thought of as a real vector space. The result in general is true, e.g. here https://cims.nyu.edu/~kiryl/Algebra/Section_5.1--Extension%20Fields.pdf

To me the result is "clear" or "obvious", but when I go about trying to prove it, I cannot think of anything less clunky than to verify all of the scalar multiplication vector space axioms one by one (the additive axioms follow immediately from the fact that any field is an abelian group under addition), which means that I don't really gain anything from the observation.

The example I have in mind is the field of rational functions over a projective curve $V(P) \subset \mathbb{CP}^2$; I want to argue that since all complex scalars $\lambda$ are polynomials hence rational functions on $V(P)$, that the field of rational functions is not only a vector space over itself, but also a complex vector space. Then given any divisor $D$, I can show that $L(D)$ is a complex vector space just by showing that it is closed under addition and scalar multiplication.

Let $$K \subset L$$ be a subfield of $$L$$. Then $$L$$ is a vector space over $$K$$.
Proof: Because $$L$$ is a field, $$(L,+)$$ is an abelian group, thus satisfies the additive axioms for being a vector space over $$K$$.
Because $$L$$ is a field, the laws of scalar multiplication (associativity and distributivity) hold for multiplication within $$L$$ (in particular $$L$$ is a vector space over itself).
Since $$K \subset L$$, associativity and distributivity also hold for scalar multiplication by elements of $$K$$, which means that $$L$$ is a vector space over $$K$$.