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I asked a similar auxiliary question but have since realized it was erroneous. The setup is that I have an irreducible separable polynomial $f(x) \in F[x]$ of degree $n$ with $\alpha$ and $\beta$ distinct roots. I want to show that $[F(\alpha + \beta):F] \leq \frac{n(n-1)}{2}$. (This problem comes from a past qualifying exam).

I've already shown that $[F(\alpha, \beta):F] \leq n(n-1)$. My thought was that we can write $$ [F(\alpha + \beta): F] = \frac{[F(\alpha, \beta): F]}{[F(\alpha, \beta): F(\alpha + \beta)]} \leq \frac{n(n-1)}{[F(\alpha, \beta): F(\alpha + \beta)]}$$ and if we can show that $[F(\alpha, \beta): F(\alpha + \beta)] \geq 2$, then we would be done. My impression from the answers in the auxiliary question I linked is that this latter inequality is not true in general. So I'm wondering if I'm missing something here, or there's a missing assumption in the problem as stated.

Dalop
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1 Answers1

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The $F$-conjugates of $\alpha+\beta$ must lie among all the sums of two different roots of $f(x)$. And we know $\alpha+\beta$ is separable over $F$ since $\alpha$ and $\beta$ are, so $[F(\alpha+\beta):F]$ is bounded above by the number of distinct sums of two different roots of $f(x)$. At most how many such sums are there?

KCd
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