Sums and products of algebraic numbers 
How does one go about proving that the sums and products of two algebraic numbers over a field $F$ (say $a,b\in K$, where $K/F$ is a field extension) is also algebraic?

If we call $f_a$ and $f_b$ the min. poly's of $a$ and $b$, then I'm assuming the answer involves such polynomials. Perhaps looking at their roots in splitting fields for both of them? And finding a "big" splitting field, constructed from those two other ones? 

In particular, I'd like a way of explicitly constructing the minimal polynomials $\ f_{ab}$ of $ab$ and $f_{a+b}$ of $a+b$.  

I read somewhere that $g(x)=\Pi_j\Pi_i (x-\alpha_i\beta_j)$ works for $ab$, where the $\alpha_i$ and $\beta_j$ are the roots of $f_a$ and $f_b$, respectively, but I do not know why $g(x)\in F[x]$.  Similar remarks for $a+b$
 A: Lemma. If $F$ is a finite extension of $E$, and $a\in F$, then $a$ is algebraic over $E$.
Proof. Let $[F:E]=n$. Then $1,a,a^2,\ldots,a^n$ are linearly dependent over $F$, so there exists $c_0,c_1,\ldots,c_n\in E$, not all zero, such that
$$c_01 + c_1a + \cdots + c_na^n = 0.$$
Set $f(x) = c_0 + c_1x+\cdots+c_nx^n\in F[x]$. Then $f(x)\neq 0$, and $f(a) = 0$, so $a$ is algebraic over $E$. $\Box$
Lemma. Let $K$ be an extension of $E$, and let $a,b\in K$ be algebraic over $F$. Then $[F(a,b):F(a)]\leq [F(b):F]$.
Proof. $[F(a,b):F(a)]$ is the degree of the minimal polynomial of $b$ over $F(a)$. Since the minimal polynomial of $b$ over $F$ is a multiple of the minimal polynomial of $b$ over $F(a)$, the latter has degree less than or equal to the former; the degree of the former equals $[F(b):F]$, so we are done. $\Box$
Corollary. If $a$ and $b$ are algebraic over $F$, then so are $a+b$ and $ab$.
Proof. Fix an algebraic closure of $F$; $[F(a,b):F] = [F(a,b):F(a)][F(a):F]\leq [F(b):F][F(a):F] \lt\infty$. Therefore, $F(a,b)$ is a finite extension of $F$, so each element of $F(a,b)$ is algebraic over $F$. In particular, $a+b,ab\in F(a,b)$, so they are algebraic over $F$. $\Box$
