If $E$ is an algebraic extension of $F$, why is $F[\alpha]$ a finite extension of $F$ where $\alpha$ is in some extension field of $E$? From what my friend explained, you take a $g(x)$ in $E$ that has $\alpha$ as a zero. Suppose $g$ has coefficients $c_0, c_1, ... , c_k$ with each coefficient a root of some polynomial in $F[x]$. Then $F[c_0,c_1,...,c_k] = F[c_0][c_1]...[c_k]$ is a finite extension of $F$, since adjoining each coefficient is a finite extension. Then $F[c_0,c_1,...,c_k,\alpha]=F[c_0,c_1,...,c_k][\alpha]$ is a finite extension over F too. Why does this imply that $F[\alpha]$ is a finite extension?
 A: Since $E$ is algebraic extension of $F$ therefore, $\forall \alpha \in E \quad \exists g(x) \in F[x] $ such that $g(\alpha)=0$. Similarly let $\beta \in E$ and if we take the set $G$ of all $g(x) \in F[x]$ to which $\beta$ is a root, then it will form a non-zero ideal. 
Every non-zero ideal is generated by a unique monic irreducible polynomial $p(x)$. So $F[x]/(p(x))$ is a field and isomorphic to $F(\beta)$ under the map $f:F[x] \to F(\beta)$. Since $F[\beta]$ is a set of polynomials in $\beta$ with coefficients in F and so as $F(\beta)$ where $F(\beta)$ is a field.
Hence $F[\beta] = F(\beta) \simeq F[x]/(p(x))$ and since $F[x]/(p(x))$ is finite so as $F[\beta]$.
A: Take $E$ and $F$ to be $\mathbb{Q}$ and $\alpha = \pi$, then $\alpha \in \mathbb{R}$, which is a extension field of $\mathbb{Q}$, but of course $\mathbb{Q}[\pi]$ is not a finite extension of $\mathbb{Q}$ since $\pi$ is not algebraic.
So I think you mean $\alpha$ is in some algebraic extension field of $E$. In this case, remark
$$[F(c_0, \cdots, c_k, \alpha) : F(\alpha)][F(\alpha) : F] = [F(c_0, \cdots, c_k ,\alpha) : F]$$
$[F(c_0, \cdots, c_k ,\alpha) : F]$ is finite as your friend said, then $[F(\alpha) : F]$ has to be finite
