Inspired by Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

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.

  • 3
    $\begingroup$ Found the answer, sorry that I didn't look more mathoverflow.net/questions/26832/… $\endgroup$
    – Blah
    Feb 26, 2012 at 10:20
  • 2
    $\begingroup$ 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. $\endgroup$ Feb 26, 2012 at 17:07
  • $\begingroup$ @WillieWong Ah, I see your point now: if it's tagged with galois-theory, then it shouldn't be tagged with field-theory. Sorry! $\endgroup$ Feb 26, 2012 at 17:33
  • $\begingroup$ @Dylan: No problem. Hmmm, let me go and write a stub wiki for galois theory. $\endgroup$ Feb 26, 2012 at 17:39

1 Answer 1


Well, the answer (for $\operatorname{char}k=0$) can be found here


and here (cited there)


The result is quite nice: If $\operatorname{char}k=0$, then $$ [k(a):k][k(b):k] \text{rel. prime} \Rightarrow [k(a+b):k] = [k(a):k][k(b):k] \Rightarrow k(a+b)=k(a,b) $$


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