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Jun 11, 2021 at 12:29 comment added David E Speyer Regarding the $i=2$ case: Given a solution to $y-\theta = \theta^2 \alpha^3$, rewrite it as $y - \theta = \theta^{-1} \beta^3$ for $\beta = \theta \alpha$. Then apply the Galois automorphism of $K$ to get $y-(1-\theta) = - \theta \beta^3$ or $(1-y) - \theta = \theta \beta^3$. So, if $y$ solves the $i=2$ case then $1-y$ solves the $i=1$ case (and vice versa). And thanks for this answer!
Jun 11, 2021 at 10:00 history edited Chris Wuthrich CC BY-SA 4.0
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Jun 11, 2021 at 8:52 history answered Chris Wuthrich CC BY-SA 4.0