Let $F$ be a field, and let $F(X)$ be the fraction field of the polynomial ring $F[X]$. Let $p(X)\in F[X]$ be a non-constant polynomial. Then the question is : What is the dimension of $F(X)$ over $F(p(X))$ ?
My guess is : $\left\lbrack F(X):F(p(X)) \right\rbrack =\text{ deg }p(X)$.
For example, take $p(X)$ as some monic polynomial of degree $1$, i.e., $p(X)=X+c$ in $F[X]$.
Now notice that $\left(F(X+c)\right)(X)=F(X)$. Also the polynomial $q(Y)=Y+c-(X+c)\in F(X+c)[Y]$ is the minimal polynomial having root as $X$. Therefore $\left\lbrack F(X):F(X+c) \right\rbrack =\text{ deg }q(Y)=1$.
So the main question is : Is the above guess true for degrees of all positive integers ? If yes, then how to prove it ? Any help will be appreciated.