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Let $F$ be a field. I have a separable, irreducible polynomial $f(x) \in F[x]$ and $\alpha$ and $\beta$ are two distinct roots of $f$. I want to show that $[F(\alpha, \beta): F(\alpha + \beta)] \geq 2$.

A hint I was given was to consider the action of a suitable Galois group. Let $K$ be the splitting field of $f$ over $F$. It follows that $$ [F(\alpha, \beta): F(\alpha + \beta)] = \frac{[K: F(\alpha + \beta)]}{[K: F(\alpha, \beta)]}$$ So I'm guessing that a potential solution would be to prove there exists automorphism in $\text{Gal}(K/F)$ that fixes $\alpha + \beta$ but neither $\alpha$ nor $\beta$ individually. I have not been successful in this approach because while the automorphisms are certainly transitive (as $f$ is irreducible), one cannot specify where too many roots go at the risk of specifying a map that is not actually an automorphism. If this is in fact the correct direction, I would appreciate any hints. If not, I'd appreciate hints in a different direction.

Dalop
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  • Where did you find this problem? – ShyamalSayak Dec 27 '23 at 23:58
  • For more counterexamples consider the irreducible polynomial $x^p-x-1\in\Bbb{F}_p[x]$, $p>2$ a prime. If $\alpha$ is one of the zeros then $\alpha+1,\alpha+2,\ldots,\alpha+p-1$ are the others. So irrespective of choice of $\beta$ we have $\alpha+\beta=2\alpha+k$, $k\in\Bbb{F}_p$. Hence $\Bbb{F}_p(\alpha+\beta)=\Bbb{F}_p(\alpha,\beta)=\Bbb{F}_p(\alpha)$. – Jyrki Lahtonen Dec 28 '23 at 05:34
  • Having said that, I upvoted the question because you identified a key problem in your argument in that you cannot control the automorphic images of both $\alpha$ and $\beta$ simultaneously. If you know that the Galois group acts doubly transitively, then you would have that kind of control, and the argument would work. – Jyrki Lahtonen Dec 28 '23 at 05:39
  • @ShyamalSayak Even though your first attempt at a counterexample failed, I think you can find a biquadratic polynomial that does work. If you pick one that has a cyclic Galois group, $\simeq C_4$, then it should work for appropriate choices of $\alpha$ and $\beta$ as the Galois group is generated by the 4-cycle $\alpha\mapsto \beta\mapsto -\alpha\mapsto-\beta\mapsto \alpha$. – Jyrki Lahtonen Dec 28 '23 at 05:50
  • Indeed. Well spotted, @user297024. – Jyrki Lahtonen Dec 28 '23 at 20:02
  • In many cases, in zero characteristic, $m\alpha+n\beta$ with $m.n$ rational will be a primitive element of $F(\alpha, \beta)$, in particular when $m=n=1$. In other words, $[F(\alpha, \beta): F(\alpha + \beta)]=1$ and $F(\alpha, \beta)=F(\alpha + \beta)$ – Piquito Jan 02 '24 at 18:07

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