Why the fraction field of $F[a]$ is isomorphic to $F(a)$ where $F$ is a field? Q1:Why the fraction field of $F[a]$ is isomorphic to $F(a)$ where $F$ is a field? 
Q2:If $\beta$ is algebraic over $F(a_1,a_2,\cdots,a_n)$, then there exists a polynomial $f(x) \in F(a_1,a_2,\cdots,a_n)[x]$ such that $f(\beta) = 0$, then it follows that there exists a polynomial $f(x_1, \cdots, x_{n+1})$ such that $f(a_1,\cdots,a_n,\beta) = 0$. Is this statement right? It seems intuitively right, but I cannot find a way to justify it since I do not even know what does $F(a_1,\cdots,a_n)$ look like.
 A: 1)
The definition of $F(a)$ is that it is the smallest field containing both $F$ and $a$. (This presumes that you are working in some fixed algebraic closure $\Omega$ of $F$.)
Say the fraction field of $F[a]$ is $K$ (also inside $\Omega$). Then $K\subset F(a)$ because $K$ is by definition the smallest field containing $F[a]$ and $F(a)$ does contain $K$.
In the other direction, $K$ contains $F$ and $a$, and since $K$ is a field, it had better contain $F(a)$ the smallest field containing both of those elements. So $K=F(a)$.
In summary the first inclusion is true because of the fraction field property of $K$ and the second is true because of the field property of $K$.

2)
For this we'll use another definition of $F(a_1\cdots a_n)$. 
$$F(a_1\cdots a_n)=\{\frac{f}{g} |\space f,g \text{are relatively prime polynomials in } a_1,\cdots ,a_n \}.$$
So if there is an $f\in F(a_1,\cdots, a_n)[x]=0$ such that $f(\beta)=0$, which in turn means that $f(\beta)=0$.
let's write it out as $$\sum_{i=1}^k\frac{f_i}{g_i}\beta^i=0,$$
where the $f_i,g_i\in F[a_1\cdots a_n]$. Now by clearing denominators we can obtain an expression in $F[a_1\cdots a_n,\beta]$ that vanishes as desired which gives you the desired polynomial in $n+1$ variables.
A: I will answer your first question, since Jake has already answered the second (he was faster ;-)).


*

*Let $a\in K$, $K/k$ a field extension. Use the universal property of the quotient:


\begin{array}{lll}
k[a] & \quad\overset{\varphi}{\hookrightarrow} & k(a)\subseteq K \\
\qquad\underset{\searrow}{i} & & \underset{\nearrow}{\exists!\bar{\varphi}} \\
 & Quot(k[a]) &  & \\
\end{array}
$\varphi$ is an injective homomorphism between a ring and a field. So we have the universal property, that there exists an unique $\bar{\varphi}:Quot(k[a])\to k(a)$, s.t. the diagram above commutes, where $i:k[a]\hookrightarrow Quot(k[a]), i(a)=\frac{a}{1}$ is the natural injection.
Therefore $\bar{\varphi}$ is injective.
The $Im(\bar{\varphi})$ is a subfield of $K$ and $k\subseteq Im(\bar{\varphi}),a\in Im(\bar{\varphi})$, which implies $k(a)\subseteq Im(\bar{\varphi})$, hence $Im(\bar{\varphi})=k(A)$ and $\bar{\varphi}$ is an isomorphism of fields.
