Constructive proof of algebraic elements forming a subfield Let $F \leqslant E$ be a field extension.
If $a, b \in E$ are algebraic over $F$ then $a+b$ and $ab$ are algebraic as well. There is an short proof of this using the extension $E(a,b)$:
$[E(a,b):E]$ is finite so all elements are algebraic otherwise powers non-algebraic would form an infinite linearly independent set. In particular $a+b \in E(a,b)$ and $ab \in E(a,b)$ are algebraic.
But this proof doesn't construct the actual polynomial. Is there a constructive proof or any reason for such a proof not to exist?
 A: This question keeps getting asked on forums, e.g.:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=202382 and www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=180471 .
This fact is a particular case of a more general fact, which claims that if $A$ is a subring of a commutative ring $B$, and if $a$ and $b$ are two elements of $B$ integral over $A$, then $a+b$ and $ab$ are integral over $A$, too. This is a standard result in commutative algebra, and usually proven constructively: e.g. Corollary 2.1.11 in Swanson-Huneke yields it.
PS. This might be clear to you, but there is no chance to get the minimal polynomial of either $a+b$ or $ab$ by any algorithm. What you can get are polynomials, both of degree $\left[E\left(a\right):E\right]\cdot\left[E\left(b\right):E\right]$, that have $a+b$ resp. $ab$ as roots. These can be obtained as resultants in a way similar to Felipe's, or by executing any of the above-mentioned constructive proofs as programs.
A: Let me see if I remember this right. Let $f$ be the minimal polynomial of $a$ and $g$ the minimal polynomial of $b$ over $F$. Consider the polynomial $h(x)$ obtained as the resultant with respect to the variable $y$ of the polynomials $f((x+y)/2),g((x-y)/2)$. A root $c$ of $h$ is therefore an element of the algebraic closure of $F$ for which $f((c+y)/2),g((c-y)/2)$ have a common root $d$, so (up to conjugates) $(c+d)/2=a,(c-d)/2=b$, so $c=a+b$. The correct statement is probably that $h$ has $a+b$ as one of its roots, but may not be irreducible.
There is a similar trick for $ab$ that I don't remember. Also, what I did doesn't work in characteristic two but a variant does, but I don't remember that either.
A: There is a different (on some level) very cute constructive proof:
We need:
Lemma 1 The eigenvalues of $A\otimes B$ are the products of eigenvalues of $A$ and $B.$
and 
Lemma 2 The eigenvalues of $A \otimes I + I \otimes B$ are the sums of the eigenvalues of $A$ and $B.$
Proofs of these are left to the interested reader. In any case, now apply Lemmas 1, 2 to the companion matrices of your favorite algebraic numbers.
