TITLE: What is the set of possible values of the degree of the sum of two algebraic numbers with fixed degrees?
QUESTION [6 upvotes]: 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).

REPLY [3 votes]: 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.