Why is F($\beta$) a subfield of F($\alpha$)? There is a corollary in my book that says:

If E is an extension field of F, $\alpha \in E$ is algebraic over $F$,
  and $\beta \in F(\alpha)$, then $\deg(\beta,F)$ divides
  $\deg(\alpha,F)$.

In the proof they use that $F(\beta)\leq F(\alpha)$. But why do we have this? I mean since $\alpha$ is algebraic over F, and $\beta \in F(\alpha)$. We know that $\beta=c_0+\ldots+c_{n-1}\alpha^{n-1}.$
Now let $\gamma \in F(\beta)$. If $\beta$ is algebraic over F. Then $\gamma=d_0+\ldots +d_{m-1}\beta^{m-1}$. Now we can substitute $\beta$ with $\beta=c_0+\ldots+c_{n-1}\alpha^{n-1}$ and use that $\alpha^n=-a_{n-1}-\ldots-a_0$.(This comes from irr($\alpha$,F)).
But what if $\beta$ is transcendental over F? How do I show it then?
 A: The field $F(\beta)$ can be viewed/defined as the intersection of all subfields of $K$ containing $\beta$ for some ambient field $K$ containing $F$ and $\beta$.  Thus $F(\beta) \subset F(\alpha)$ follows directly from $\beta \in F(\alpha)$.  You can also see this directly: if $\beta$ is in a field containing $F$, then so too is any rational function in $\beta$ with $F$ coefficients.
It is a also a fact (although the argument might not be immediately obvious) that if $\alpha$ is algebraic over $F$, then any polynomial in $\alpha$ with coefficients in $F$ is also algebraic over $F$.  This means that $\beta $ cannot be transcendental as long as $\alpha$ is algebraic.
A: Edited: for clarity, I'll try to reason constructively (avoid things like intersecting all subfields containing $\beta$) and in very small steps.
If $\alpha$ is algebraic over $F$, then $\dim_F(F(\alpha))$ is finite (namely the degree$~n$ of the minimal polynomial of$~\alpha$). This was given, so we are working in an $n$-dimensional space over$~F$.
Consider the successive powers $\beta^0=1,\beta^1=\beta,\beta^2,\beta^3,\ldots$, of which initial portions will be linearly independent up to some point, but must then become linearly dependent (certainly $[1,\beta,\ldots,\beta^n]$ is linearly dependent, being $n+1$ vectors in an $n$-dimensional space). Let $d>0$ be minimal such that $[1,\beta,\ldots,\beta^d]$ is linearly dependent. By minimality, the linear dependence relation involves $\beta^d$ with a nonzero coefficient, and dividing by that coefficient we obtain a relation that involves $\beta^n$ with coefficient$~d$, in other words, a monic polynomial $P$ of degree$~d$ such that $P[\beta]=0$. This shows that $\beta$ is certainly algebraic.
Now the $F$-subspace generated by $1,\beta,\ldots,\beta^{d-1}$ is closed under multiplication by $\beta$ (check this on the generating vectors), and therefore contains all positive powers of $\beta$, so it is equal to $F[\beta]$. This is a priori a subring of $F(\alpha)$ (its elements are all polynomials in$~\beta$, and as such it is closed under addition and multiplication). However one sees that $F[\beta]$ is a field by a general result that says that if a ring$~R$ is both (1) an integral domain, and (2) finite dimensional over some sub-field $F$ of $R$, then $R$ is itself a field. The following proof of this fact is simple linear algebra: consider multiplication by a nonzero element $r\in R$ as $F$-linear operator $m_r:R\to R$. By hypothesis (1) $m_r$ is injective, so that $\det_F(m_r)\neq0$, but then $m_r$ is also surjective, in particular there is $s\in R$ such that $m_r(s)=1\in R$, which means that $s$ is the inverse of$~r$.
There is a somewhat more constructive version of the argument that $F[\beta]$ is a field when $\beta$ is algebraic. The fact that $F[\beta]$ is an integral domain (as before because we are inside the field $F(\alpha)$) implies that the minimal polynomial $P\in F[X]$ of$~\beta$ is irreducible: if one had $P=Q_1Q_2$ with $Q_1,Q_2$ both of degree strictly less than $\deg(P)$, then by minimality of$~P$ neither $Q_1[\beta]$ nor $Q_2[\beta]$ would be zero, but $ Q_1[\beta].Q_2[\beta]=P[\beta]=0$, contradicting the integral domain condition. Now any nonzero element of $F[\beta]$ can be written uniquely as $Q[\beta]$ with $Q$ nonzero and $\deg(Q)<\deg(P)$. Then $\gcd(P,Q)=1$, whence there exist Bézout coefficients $S,T\in F[X]$ such that $SP+TQ=1$, and evaluating this at $X=\beta$ gives $T[\beta]Q[\beta]=1$, showing that our nonzero element $ Q[\beta]\in F[\beta]$ is invertible.
