Degree of field extension using minimum polynomial Let $K \subset L$ an algebraic extension. I want to prove for $a,b \in L$ that 
$$[K(b): K] \geq [K(a,b): K(a)],$$
Where for example $[K(b):K]$ is the degree of the field extension $K \subset K(b)$. My ideas so far are:


*

*The degree of $[K(a):K]$ is equal to $\deg(f_K^a)$, where $f_K^a$ is the minimum polynomial. So
$$K[X]/(f_K^a) \stackrel{\sim}{\to} K(a). $$
and I want to find a minimum polynomial with root $a$.


My question is: how do I find the minimum polynomials?
 A: You don't need to find the minimal polynomial.  That's because whatever the polynomial is for $b$ over $K$, call it $p$, it is still a polynomial for $b$ over $K(a)$.  Therefore, the minimal polynomial for $b$ over $K(a)$ divides $p$ and therefore has degree no greater than $\deg(p)=[K(b):K]$.
A: You can't really find minimal polynomials in a constructive way, unless you know more about the fields themselves and the elements $a,b$.  Most of abstract field theory is very nonconstructive stuff, unfortunately.  However, you can argue as follows:
Let $E \subseteq E'$ be fields, and assume that $b$ is algebraic over $E$.  Let $f(X) \in E[X]$ be the minimal polynomial of $b$ over $E$.  Let $g(X) \in E'[X]$ be the minimal polynomial of $b$ over $E'$.  Then $$[E(b): E] = \textrm{deg } f$$ $$[E'(b) : E'] = \textrm{deg } g$$ Now $g$ has the following property: if $h$ is a polynomial in $E'[X]$, and $h(b) = 0$, then the degree of $h$ is $\geq$ the degree of $g$.  But $f$ is a polynomial in $E'[X]$ with $f(b) = 0$.  So $$[E(b) : E] = \textrm{deg } f \geq \textrm{deg } g = [E'(b) : E']$$
Now apply what I just said with $E = K$ and $E' = K(a)$.  Remember that $E'(b) = K(a)(b) = K(a,b)$.
