Q:

What is the set of possible values of the degree of the sum of two algebraic numbers with fixed degrees?

This question is related to Degree of sum of algebraic numbers and algebraic numbers of degree 3 and 6, whose sum has degree 12. 
  In this last question I asked a very special case of the following problem : given two
algebraic numbers $\alpha$ and $\beta$ with degrees $a$ and $b$ respectively,
what can the degree of $\alpha+\beta$ be ?
I believe the answer is as follows : the degree of $\alpha+\beta$ can equal some
value $d$ iff
(1) $d \leq ab$ and $a \leq db$ and $b \leq da$. (this condition is obviously
necessary)
(2) $d$ divides $ab$, or $a$ divides $db$, or $b$ divides $da$.
The "if" part probably involves Galois theory as in Gerry's answer to the special case.  
EDIT 07/01/2010 : As Gerry noted, the conjecture above is grossly false. Below is a "corrected version"
of my conjecture.
I believe the answer is as follows : the degree of $\alpha+\beta$ can equal some
value $d$ iff
(*)  There is some integer $e$ divisible by all of $a,b,d$, and lower than or equal to
all of $ab,ad,bd$ (this is a necessary condition, as is seen by taking $e$ to be the degree
of the extension $k(\alpha,\beta)/k$, where $k$ is the base field). 

A:

I'm not sure what the answer is, but it can't be what you suggest. Your conditions allow $a=3$, $b=6$, 
$d=7$. But ${\bf Q}(\alpha+\beta)$ is contained in $K={\bf Q}(\alpha,\beta)$, so $d$ must divide the degree of $K$, which must be a multiple of 6 (since it contains ${\bf Q}(\beta)$) but no greater than 18. 

A:

This isn't a solution, just a comment that got too long for the comment box: Assuming that all finite groups occur as Galois groups over $k$, the answer to this question should only depend on the characteristic of $k$.
Consider the more detailed question: 

For a finite group $G$, and subgroups
  $H_1$, $H_2$ and $H_3$, is there a
  Galois extension $L$ of $k$ with
  Galois group $G$, and elements $v_i
> \in L$ such that the stabilizer of
  $v_i$ is $H_i$ and $v_1+v_2+v_3=0$.

I claim that, for given $(G, H_1, H_2, H_3)$, assuming that there is some $G$-extension of $k$, the answer to this question only depends on the characteristic of $k$. Proof: As a $G$-representation, $L$ is the permutation representation on $X:=G/(H_1 \cap H_2 \cap H_3)$. The question, then, is whether we can find $L$-valued functions, $f_i$ on $X$, such that $f_i$ is constant on $H_i$ orbits (but not for any larger subgroup) and $f_1+f_2+f_3=0$. This is a collection of linear equalities and inequalities in $3 |X|$ variables, with integer coefficients. So whether or not they have a solution depends only on the characteristic of $k$ (we are using that $k$ is infinite). To see that the characteristic can matter, take $G=S_3$, $H_1$ and $H_2$ two different subgroups of order $2$ and $H_3$ the subgroup of order $3$. You should get a solution in characteristic $3$, and not otherwise.
Of course, answering the original question just means answering this question for all $(G, H_1, H_2, H_3)$ with $|G/H_i| = d_i$. (One can immediately make two reductions. First, a necessary condition is that $H_1 \cap H_2 = H_1 \cap H_3= H_2 \cap H_3$. Second, one can immediately reduce to the case that $H_1 \cap H_2$ contains no nontrivial normal subgroup. The latter means that $|G| \leq (d_1 d_2)!$, so the problem is finite.) 
I see no reason to believe that you will get a nicer answer by forgetting the groups and only remembering the degrees, but of course I haven't thought very hard about the problem.

