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).

  • $\begingroup$ Possibly dumb question: Why is (2) necessary? It can fail in finite characteristic. $\endgroup$ – David E Speyer Jul 1 '10 at 12:22
  • $\begingroup$ In the case $a=2$ and $b=3$, the conditions above allow $d \in \{2,3,6\}$, which isn't correct because it always holds that $d=6$. However, I would also think that the answer to your question depends only on some inequality or divisibility conditions. $\endgroup$ – François Brunault Jul 1 '10 at 14:10
  • $\begingroup$ The new conjecture is wrong. It permits (2,3,3) whereas, in fact, if a=2 and b=3, then d must be 6. $\endgroup$ – David E Speyer Jul 1 '10 at 14:51
  • $\begingroup$ More generally if a < b are coprime, the edited condition means that d >= b divides ab, whereas Isaacs' theorem shows that d = ab. $\endgroup$ – François Brunault Jul 1 '10 at 15:55
  • $\begingroup$ Ewan, why does $k(\alpha,\beta)$ have a special role in your modified conjecture? Sure, the sum is contained in there, but it's in many other fields too and the sum does not "know" about the individual terms $\alpha$ and $\beta$. Generically in a sense the sum has degree $ab$, but I don't know if you can expect some reasonably clean statements about cases of lower degree. Also, is your conjecture meant only for algebraic numbers (and their degree over Q)? This addresses David's comment about characteristic $p$. $\endgroup$ – KConrad Jul 1 '10 at 16:01

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.

  • $\begingroup$ Indeed Gerry. I'll update my conjecture accrodingly $\endgroup$ – Ewan Delanoy Jul 1 '10 at 14:23

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.

  • $\begingroup$ In fact, the question turns out to be group-theoretic rather than (commutative) algebraic. $\endgroup$ – Ewan Delanoy Jul 1 '10 at 17:35

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