How do I know that $q(\alpha)\neq 0$ in this proof? And one more question. 
Theorem If $[F(\alpha):F]=n$ is finite, then $\alpha$ is the root of some irreducible polynomial $p(x) \in F[x]$ of degree $n$. Furthermore, $F(\alpha)=F[\alpha]$.
Proof. If $[F(\alpha):F]=1$ then $F(\alpha)=F$ so $\alpha \in F$ and the corresponding polynomial $p(x)$ is $p(x)=x-a$. We assume from now on that $[F(\alpha):F]>1$.
The field $F(\alpha)$ includes, among other elements, linear combinations of the form
$a_0+a_1\alpha+a_2\alpha^2+...+a_k\alpha^k$
where $a_i \in F$. All these linear combinations are polynomials in $F[x]$ evaluated at $\alpha$. Since $F(\alpha)$ has dimension $n$ as a vector space over $F$, the set $\{1, \alpha, \alpha^2, ... \alpha^n \}$ is linearly dependent. Thus, there exists some nontrivial polynomial $q(x)$ of degree at most $n$ such that $q(\alpha)=0$. Let $p(x)$ be a polynomial of least degree such that $p(\alpha)=0$ and let us call $\operatorname{deg}p(x)=d$. (The fact that such a polynomial exists follows from the well-ordering principle of the integers and the fact that $[F(\alpha):F]$ is finite.)
We claim that $p(x)$ is irreducible in $F[x]$. Suppose that $p(x)$ is not irreducible. Then $p(x)=p_1(x)p_2(x)$ with $p_1(x), p_2(x) \in F[x]$ with positive degrees. Thus, evaluating $p(x)$ at the root $\alpha$ gives
$0=p(\alpha)=p_1(\alpha)p_2(\alpha)$.
Since there are no zero divisors in $F$, $p_1(\alpha)=0$ or $p_2(\alpha)=0$. This contradicts the fact that $p(x)$ is a polynomial in $F[x]$ of minimal degree for which $\alpha$ is a root. Hence, we conclude that $p(x)$ is irreducible.
Writing $p(x)=c_0+c_1x+...+c_dx^d$, we note that $c_d \neq 0$, so
  $\alpha^d=-\frac{c_0}{c_d}-\frac{c_1}{c_d}\alpha-...-\frac{c_{d-1}}{c_d}\alpha^{d-1}$.
This expresses $\alpha^d$ as a linear combination of $\{1,\alpha,...,\alpha^{d-1}\}$. By a recursion argument, we can see that for all $m \geq 0$, the element $\alpha^m$ can also be written as a linear combination of $\{1,\alpha,...,\alpha^{d-1}\}$. Hence, the set of powers of $\alpha$ spans $F[\alpha]$ as a vector field over $F$. However, a priori $F[\alpha]$ is only a subset of $F(\alpha)$.
By definition, every element in $F(\alpha)$ can be written as a rational expression of $\alpha$, namely
$\gamma=\frac{a(\alpha)}{b(\alpha)}$ where $a(x),b(x) \in F[x]$ and $b(\alpha) \neq 0$.
Suppose also that $a(x)$ and $b(x)$ are chosen such that $b(x)$ has minimal degree and $\gamma=\frac{a(\alpha)}{b(\alpha)}$. Performing the Euclidean division of $p(x)$ by $b(x)$ we get $p(x)=b(x)q(x)+r(x)$, where $\operatorname{deg}r(x)\lt\operatorname{deg}b(x)$ or $r(x)=0$. Assume that $r(x) \neq 0$. Then
$\gamma=\frac{a(\alpha)}{b(\alpha)}=\frac{a(\alpha)q(\alpha)}{b(\alpha)q(\alpha)}=\frac{a(\alpha)q(\alpha)}{p(\alpha)-r(\alpha)}=-\frac{a(\alpha)q(\alpha)}{r(\alpha)}$.
Hence, $a(\alpha)/b(\alpha)$ can be written as $a_2(\alpha)/b_2(\alpha)$ where $\operatorname{deg}b_2(x)\lt\operatorname{deg}b(x)$. This contradicts the choice that $b(x)$ has minimal degree. Consequently, $r(x)=0$ and hence $b(x)$ divides $a(x)$. Then the expression $\gamma=a(\alpha)/b(\alpha)$ can only be such that $b(x)$ has minimal degree among such rational expressions if $b(x)$ is a constant. Consequently, $F(\alpha) \subseteq F[\alpha]$ and therefore $F(\alpha)=F[\alpha]$.
Consequently, it also follows that $d=dim_FF(\alpha)=n$ and $p(x)$ is an irreducible polynomial such that $p(\alpha)=0$.

The above are a theorem and a proof from Lovett. "Abstract Algebra." pp. 324-325. (Section 7.1. Introduction to Field Extensions.)
I don't understand two parts in this proof:
1) "$\frac{a(\alpha)}{b(\alpha)}=\frac{a(\alpha)q(\alpha)}{b(\alpha)q(\alpha)}$". Question: How do I know that $q(\alpha)\neq 0$?
2) "Consequently, $r(x)=0$ and hence $b(x)$ divides $a(x)$." Question: If $r(x)=0$, I see that $b(x)$ divides $p(x)$, but why does the author say that $b(x)$ divides $a(x)$?
To help people avoid the unnecessary reading, I'll summarize the proof from the start to the place I didn't understand.
Summary of the Proof We proved that the theorem holds if $[F(\alpha):F]=1$. So we assume from now on that $[F(\alpha):F] \gt 1$. We proved that there exists a nonzero polynomial $q(x) \in F[x]$ such that $q(\alpha)=0$. We let $p(x)$ be such polynomial having smallest degree. We proved that $p(x)$ is irreducible. Let $\operatorname{deg}p(x)=d$. We proved that $\{1,\alpha,...,\alpha^{d-1}\}$ spans $F[\alpha]$. It is known that $F(\alpha)$ is isomorphic to the field of fractions of $F[\alpha]$ so for each $\gamma \in F(\alpha)$, we can write $\gamma=\frac{a(\alpha)}{b(\alpha)}$ where $a(x),b(x) \in F[x]$ and $b(\alpha) \neq 0$. Suppose also that $a(x)$ and $b(x)$ are chosen such that $b(x)$ has minimal degree. Performing the Euclidean division of $p(x)$ by $b(x)$ we get $p(x)=b(x)q(x)+r(x)$, where $\operatorname{deg}r(x) \lt\operatorname{deg}b(x)$ or $r(x)=0$. Assume that $r(x) \neq 0$. Then $\gamma=\frac{a(\alpha)}{b(\alpha)}=\frac{a(\alpha)q(\alpha)}{b(\alpha)q(\alpha)}=\frac{a(\alpha)q(\alpha)}{p(\alpha)-r(\alpha)}=-\frac{a(\alpha)q(\alpha)}{r(\alpha)}$. (The rest omitted.)
 A: I will slightly change the argument in the proof to come with something that seems a little more natural to me. Nonetheless, I will still provide an answer to both of your questions.
First, let me justify why $q$ is not zero. Note that $b$ is chosen so that it has minimal degree. Hence, we must have $\operatorname{deg}(b)<\operatorname{deg}(p)$, because otherwise dividing $b$ by $p$ produces a remainder $b'$ whose degree is less that the degree of $p$ and whose value at $\alpha$ is the same as $b$. This would contradict the minimality of $b$.
Then, we know that $q$ is not zero. 
Now, I claim that we have $\operatorname{deg}(p)=\operatorname{deg}(q)+\operatorname{deg}(b)$. This is just because $\operatorname{deg}(r)<\operatorname{deg}(b)$, so the polynomial $r$ is not relevant to the computation of the degree of the sum $bq+r$ (given that $q$ is not zero).  
Now, if $b$ is constant, then we are done (there is nothing to do).
Otherwise, we have $\operatorname{deg}(p)>\operatorname{deg}(q)$, so that $q(\alpha)$ can not be zero (it would otherwise contradict the minimality of $p$).  
Now, let us look back at $p=qb+r$. Evaluating this relation at $\alpha$ gives us $r(\alpha)=-q(\alpha)b(\alpha) \not = 0$. Hence, $r$ is not the zero polynomial, and in particular its value at $\alpha$ is non zero. Now, the computation done in the proof of the book, that is 
$$\gamma=\frac{a(\alpha)}{b(\alpha)}=\frac{a(\alpha)q(\alpha)}{b(\alpha)q(\alpha)}=\frac{a(\alpha)q(\alpha)}{p(\alpha)-r(\alpha)}=-\frac{a(\alpha)q(\alpha)}{r(\alpha)}$$
proves that $\gamma$ may be written as a quotient whose denominator is the image at $\alpha$ of a polynomial of degree less than the degree of $b$, which comes in contradiction with the minimality of $b$.  
The hypothesis that is contradicted is the assumption I did at "Otherwise", ie the assumption that $b$ is not constant. So actually, we deduce that $b$ must be constant, so that indeed $b$ divides $a(x)$, and this concludes the proof.
NB: Thank you very much for your summary, it is helpful. You should precise in it that $b$ is chosen to have minimal degree.
