Degree of sum of algebraic numbers This is an elementary question (coming from an undergraduate student) about algebraic numbers, to which I don't have a complete answer.
Let $a$ and $b$ be algebraic numbers, with respective degrees $m$ and $n$. Suppose $m$ and $n$ are coprime. Does the degree of $a+b$ always equal $mn$?
I know that the answer is "yes" in the following particular cases (I can provide details if needed) :
1) The maximum of $m$ and $n$ is a prime number.
2) $(m,n)=(3,4)$.
3) At least one of the fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ is a Galois extension of $\mathbf{Q}$.
4) There exists a prime $p$ which is inert in both fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ (if $a$ and $b$ are algebraic integers, this amounts to say that the minimal polynomials of $a$ and $b$ are still irreducible when reduced modulo $p$).
I can also give the following reformulation of the problem : let $P$ and $Q$ be the respective minimal polynomials of $a$ and $b$, and consider the resultant polynomial $R(X) = \operatorname{Res}_Y (P(Y),Q(X-Y))$, which has degree $mn$. Is it true that $R$ has distinct roots? If so, it should be possible to prove this by reducing modulo some prime, but which one?
Despite the partial results, I am at a loss about the general case and would greatly appreciate any help!
[EDIT : The question is now completely answered (see below, thanks to Keith Conrad for providing the reference). Note that in Isaacs' article there are in fact two proofs of the result, one of which is only sketched but uses group representation theory.]
 A: For those who want to read Isaac's proof --- mentioned in Pete's answer --- in the language of Molière and Bourbaki, there is François Brunault's exposé.
Addendum (30/01/2011).  Today I came across the following related article by Weintraub in the Monatshefte.
A: The following answer was communicated to me by Keith Conrad:
See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25
(1970), 638--641.


http://alpha.math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are
relatively prime, then
$K(a,b)$ = $K(a+b)$, which answers the students question in the
affirmative.  His proof shows the same conclusion holds under the
weaker assumption that
$[K(a,b):K] = [K(a):K][K(b):K]$.
since Isaacs uses the relative primality assumption on the degrees
only to get that degree formula above, which can occur even in cases
where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.
A: What is true for sure is that this statement (in fact, a more general one) holds over a finite field (see Brawley, J. V., Carlitz, L. Irreducibles and the composed product for polynomials over a finite field.  MR0893074 (89g:11118)). In char=0, however, I don't have a good reference, even though I always thought it should be true as well (maybe Brawley and Carlitz have hints on it, but I have no access to their article: Ireland seems to fight recession by discharging online subscriptions to academic journals!)...
A: A counter-example to show that this result does not extend to characteristic $p$: Let $K= \mathbb{F}_p(s,t)$, the rank two transcendental extension of $\mathbb{F}_p$. Let $\alpha$ and $\beta$ be roots of 
$$\alpha^{p-1} - s=0$$
$$\beta^p - s \beta - t=0.$$
Then $\alpha$ and $\beta$ have degrees $p-1$ and $p$ over $K$. The element $\alpha + \beta$ obeys 
$$(\alpha+\beta)^p - s (\alpha+\beta) - t=0.$$
