Proving $\dim_K V = \dim_E V \cdot [E:K]$ for a field extension $K \subseteq E$

Question

Let $K \subseteq E$ be a field extension. Then according to Wikipedia, for a vector space $V$ over $E$ we have that $$\dim_K V = \dim_E V \cdot [E:K]$$ where $[E:K]$ denotes the degree of the field extension $K \subseteq E$. This formula makes sense to me. However, since I am not really familiar with field extensions, I tried looking for resources, i.e. a proof of this. I checked my algebra books, but did not find something. So my question is: Has anyone a good resource of this or even a proof? I think it should be not that hard. Since $E$ can be seen as a vector space over $K$, we could write elements out in a basis.

Answer 1

HINT: Let $\{v_1,...,v_s\}$ be a basis of $V$ over $E$ and let $\{\lambda_1,...,\lambda_r\}$ be a basis of $E$ over $K$.

Then the set $\{\lambda_iv_j\}_{1\leq i\leq r,1\leq j\leq s}$ is a basis of $V$ over $K$.

Answer 2

I'll assume $V$ finite dimensional over $E$ and $E$ a finite field extension of $K$.

If $\dim_E V=0$, the statement is clear. Suppose $\dim_E V\ne0$. Then there is $v\in V$, $v\ne0$. Let $U$ be a complement of $Ev$ (the $E$-subspace generated by $v$). Then $$ V=Ev\oplus U $$ Since $Ev\cong E$ as vector spaces over $E$ (and over $K$), $\dim_K Ev=[E:K]$. Since $\dim_E U=\dim_E V-1$, we can apply induction.