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?



Constructive proof of algebraic elements forming a subfield
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.


