$V$ is finite dimentional over field $K\iff$ field extension $L/K$ is finite Let $L/K$ be a field extension and $V$ a non-zero vector space over $L$. Prove that:
$V$ is finite dimensional over $K\iff V$ is finite dimensional over $L$ and $[L:K]<\infty$
for the first direction: $V$ is finite dimensional over $K\implies[L:K]<\infty$ and $V$ is finite dimensional over $L$
If $V$ is finite dimentional over $K$, then $\exists\alpha_i\in V$ for $1\leq i\leq n$, s.t $\forall v\in V$, $v=c_1\alpha_1 + ... + c_n\alpha_n$. Where $c_i\in K$. Since $K\subset L$, $c_i\in L$, hence $V$ is finite dimensional over $L$. How can I use the fact that $V$ is finite dimensional over both $K$ and $L$ to show that $[L:K]$ is finite.
I think I've managed to prove the second half, $[L:K]<\infty$ and $V$ is finite dimensional over $L\implies V$ is finite dimensional over $K$.
Since $[L:K]$ is finite $\exists \beta_j\in L$ for $1\leq j\leq m$ s.t $\forall l_i\in L$, $l_i=d_{i1}\beta_1 + ... + d_{im}\beta_m$ for some $d_{ij}\in K$. Also since we know that $V$ is finite dimensional over $L$, we can write $v=l_1 v_1 + ... + l_n v_n$, substituting each $l_i$ for their linear combination of the $K$-basis we get
$v=(d_{11}\beta_1 + ... + d_{1m}\beta_m)v_1+...+(d_{n1}\beta_1 + ... + d_{nm}\beta_m)v_n$
$=((v_1\beta_1)d_{11}+...+(v_i\beta_j)d_{ij}+...+(v_n\beta_m)d_{nm})$.
Clearly $V$ is finite dimensional over $K$ as we can express every element in $V$ as a linear combination of the basis $d_{ij}$. Have I missed anything important in my proof
 A: This is a variant of the dimension theorem about field extensions.
Suppose $V$ is finite dimensional over $L$ and $L$ is finite dimensional over $K$. Let $\{v_1,\dots,v_m\}$ be a basis of $V$ over $L$ and $\{l_1,\dots,l_n\}$ be a basis of $L$ over $K$. Then, if $v\in V$, we have
$$
v=\sum_{i=1}^m\alpha_iv_i
$$
with $\alpha_i\in L$. Therefore we have
$$
\alpha_i=\sum_{j=1}^n \beta_{ij}l_j \qquad (i=1,\dots,m)
$$
with $\beta_{ij}\in K$, and so
$$
v=\sum_{i,j}\beta_{ij}(l_jv_i)
$$
so the set $\{l_jv_i:1\le i\le m,1\le j\le n\}$ is a finite spanning set of $V$ over $K$ (it's also a basis, by the way, but it's not needed).
Conversely, suppose $V$ is finite dimensional over $K$. Then, choose $v\in V$, $v\ne0$. The map of $K$-vector spaces $\alpha\mapsto \alpha v$ is an isomorphism between $L$ and $Lv$; in particular $L$ is finite dimensional over $K$. A finite spanning set of $V$ over $K$ is obviously a spanning set of $V$ as $L$-vector space.
A: For the direction:
$V$ is finite over $K\implies V$ is finite over $L$ and $[L:K]<\infty$
Since $V$ is finite over $K$, we can write any $v\in V$ as $v=\alpha_1 k_1 +...+\alpha_n k_n$, where $\alpha_i\in V$ is a $K-$basis for $V$, and $k_i\in K$, since $K\subseteq L$, each $k_i\in L$. Therefore, $V$ is finite over $L$ and also $dim_L(V)\leq dim_K(V)$. So now each $v\in V$ may be written in the form $v=\beta_1 l_1 +...+\beta_m l_m$, where $m\leq n$.
We know $L/K$ is a field extension, assume that $L/K$ is infinite, i.e $\exists l_i=k_1 l_{i1}+k_2l_{i2}+...$. let this $l_i$ be in the $L-$basis for $V$. Substituting $l_i$ in the linear expression of $v$ gives us, $v=\beta_1 l_1+...+\beta_i(k_1 l_{i1}+k_2l_{i2}+...)+...+\beta_m l_m$. $k_i$ are linearly independent in $L$, hence must be linearly independent in $V$. So $V$ has an infinite $K-$basis, contradiction since we assumed that $V$ is finite over $K$, hence $L/K$ is finite
The reverse direction is in the question
