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As the title says:

Given $\alpha, \beta \in \bar{F}$ are separable over $F$, prove that $\alpha + \beta$ is also separable over $F$.

I'd like a push in the right direction, not a complete answer. Thanks!

GMB
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1 Answers1

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Separable extensions are a distinguished class. Furthermore, if $\alpha$ is separable over $F$, then it is separable over any algebraic extension of $F$. Is that enough of a hint?

Little further hint: $E/F$ is separable if and only if every $\alpha\in E$ is separable as an element over $F$.

Ian Coley
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  • Yep! I had to stare at it for awhile but I think I've got it now. Thanks. – GMB Mar 27 '13 at 04:47
  • Cool. Want to post your proof? I'll double-check it. – Ian Coley Mar 27 '13 at 04:49
  • Can I check if I've understood correctly? $\alpha$ and $\beta$ separable $\implies $ $F(\alpha, \beta)$ is separable $\implies$ any $\gamma \in F(\alpha, \beta)$ is separable. So in particular $\alpha + \beta$ is separable. – eatfood Sep 24 '19 at 09:12