$[K(a):F(a)]=[K:F]$ if $a$ is transcendental over $K$. Let $F$ and $K$ be subfields of the complex number $\mathbb{C}$ such that $K$ is a finite extension of $F$. Let $a\in \mathbb{C}$. If $a$ is not algebraic over $K$, prove that $[K(a):F(a)]=[K:F]$. 
I want to ask is there any short solution (uses well known theorem)? I have no idea where to begin, thanks in advance.
 A: It is actually not too dificult to prove if one treats the element $a$ as an indeterminate (which makes no difference). It works in general like this:

Let $F\subseteq K$ be a finite field extension. Then
  $[K(x):F(x)]=[K:F]$.

Proof: 
Take $b_1, \dots, b_n$ a basis of $K$ over $F$. First claim is that $b_1, \dots, b_n$ are linearly independent as elements of $K(x)$ over $F(x)$. 
So suppose $$ \sum_{i=1}^{n}\frac{f_i(x)}{g_i(x)}b_i=0 , \;\; f_i(x), g_i(x) \in F[x], g_i(x)\neq 0. $$
Then, multiplying both sides by the common denominator of the fractions, one obtains
$$\sum_{i=1}^{n}f_i(x)h_i(x)b_i=0 $$
for suitable nonzero polynomials $h_i(x)\in F(x)$. Denote by $a_{ij}$ the coefficient by the monomial $x^j$ in $f_i(x)h_i(x)$.
Then the equation above implies series of equations
$$\sum_{i=1}^{n}a_{ij}b_i=0,  \;\; j \in \mathbb{N},$$
 which, by linear independence of $b_i$'s over $F$, implies 
$$a_{ij}=0, \;\; j \in \mathbb{N}, i=1, \dots, n.$$
Hence all the polynomials $f_i(x)h_i(x)$ are zero (all their coefficients are), hence (since $h_i$'s were nonzero), $f_i(x)=0$ for every $i$. Hence $f_1(x)/g_i(x)=0, \;\; i=1, \dots, n$.
The more interesting (less computational) part is to show that $b_1, \dots, b_n$ generate $K(x)$ over $F(x)$. Since $F \subseteq K$ is algebraic, it follows that $F(x)[K]$ (the smallest subring of $K(x)$ containing both $F(x)$ and $K$) is a field. Hence $K(x)=F(x)[K]$ (since the RHS contains both $x$ and $K$ and $K(x)$ is smallest such).
Hence every element $g(x)$ of $K(x)$ can be written in the form
$$g(x)=\sum_{i=1}^m r_i(x)k_i, \;\;\; r_i(x) \in F(x), k_i \in K$$
(observe that this is indeed a general form of an element of $F(x)[K]$, since $K$ is multiplicatively closed).
But then $$k_i=\sum_{j=1}^{m_i}f_{ij}b_j, i=1, \dots, n$$
for suitable coefficients $f_{ij} \in F$, hence
$$g(x)=\sum_{i=1}^m \sum_{j=1}^{m_i} r_i(x)f_{ij}b_j$$
and we are done.
