Degree of sum of algebraic numbers

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://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 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 $$(\alpha+\beta)^p - s (\alpha+\beta) - t=0.$$

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.