# Does the sum generate the field compositum when degrees are prime?

Suppose $k$ is a field and $\alpha,\beta$ are algebraic over $k$ such that $[k(\alpha):k]=p$ and $[k(\beta):k]=q$ with distinct primes $p,q$. Then $$1<[k(\alpha,\beta):k(\alpha)]\leq q \quad\text{so}\quad 1<[k(\alpha,\beta):k(\alpha)]\cdot[k(\alpha):k]=[k(\alpha,\beta):k]\leq pq$$ But $[k(\alpha):k]=p$ and $[k(\beta):k]=q$ are divisors of $[k(\alpha,\beta):k]$, so $[k(\alpha,\beta):k]=pq$

Now $k(\alpha+\beta)\subset k(\alpha,\beta)$ is a subfield, so $[k(\alpha+\beta):k]=:n$ divides $pq$.

Is it true that $n\neq p$ and $n\neq q$ and therefore $k(\alpha+\beta)= k(\alpha,\beta)$ or is there a counterexample?

EDIT: The proof for $k=\mathbb Q$ is quite long, maybe in this situation there is a direct proof? And I somehow cannot find a good counterexample for $k$ a finite field.

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Found the answer, sorry that I didn't look more mathoverflow.net/questions/26832/… –  Blah Feb 26 '12 at 10:20
I retagged since the tag wiki for (field-theory) explicitly asks to just use (galois-theory) for these kinds of questions. Also, you should feel free to post either just the link that you found, or a write-up in your own words of the result, as an answer and accept it. –  Willie Wong Feb 26 '12 at 17:07
@WillieWong Ah, I see your point now: if it's tagged with galois-theory, then it shouldn't be tagged with field-theory. Sorry! –  Dylan Moreland Feb 26 '12 at 17:33
@Dylan: No problem. Hmmm, let me go and write a stub wiki for galois theory. –  Willie Wong Feb 26 '12 at 17:39

Well, the answer (for $\operatorname{char}k=0$) can be found here
The result is quite nice: If $\operatorname{char}k=0$, then $$[k(a):k][k(b):k] \text{rel. prime} \Rightarrow [k(a+b):k] = [k(a):k][k(b):k] \Rightarrow k(a+b)=k(a,b)$$