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Is there a map $\sigma$ such that at some $\tau\in\Bbb F_p^\times\backslash\{1\}$ we have at every $\alpha,\beta\in\Bbb F_p$ $$\sigma(\alpha+\tau\cdot\beta)=\alpha-\tau\cdot\beta$$ holding true?

Turbo
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1 Answers1

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Setting $\beta=0$ gives $\sigma(\alpha)=\alpha$ for all $\alpha$. Setting $\alpha=0$ gives $\sigma(\tau\beta)=-\tau\beta$ so letting $\gamma=\tau\beta$, $\sigma(\gamma)=-\gamma$ for all $\gamma$ ($\gamma$ can be anything since $\tau\neq 0$). So this is only possible for $p=2$, but then $\mathbb{F}_p^\times\setminus\{1\}$ is empty so it is trivially impossible.

Eric Wofsey
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  • what about if $\tau$ alone is in $\Bbb F_p^2$ and $\alpha,\beta\in\Bbb F_p$? – Turbo Sep 06 '16 at 03:30
  • What does $\tau\cdot \beta$ mean then? – Eric Wofsey Sep 06 '16 at 03:34
  • $\beta\in\Bbb F_p\subseteq\Bbb F_{p^2}$ and so it should be defined. – Turbo Sep 06 '16 at 03:35
  • Oh, you meant $\mathbb{F}_{p^2}$. – Eric Wofsey Sep 06 '16 at 03:36
  • yes correct it is. – Turbo Sep 06 '16 at 03:37
  • Then such a map $\sigma$ trivially exists for any $\tau\not\in\mathbb{F}_p$: just define $\sigma(\alpha+\tau\beta)=\alpha-\tau\beta$ for any $\alpha,\beta\in\mathbb{F}_p$, and define $\sigma$ arbitrarily elsewhere. This is well defined since $1$ and $\tau$ are linearly independent over $\mathbb{F}_p$. – Eric Wofsey Sep 06 '16 at 03:38
  • I only get to modify $\alpha+\tau\cdot\beta$ in a blak box manner and I do not know $\alpha$ and $\beta$ individually (or equivalently $\alpha+\tau\cdot\beta$) and that is why I want an expicit map that operates on black box manner. – Turbo Sep 06 '16 at 03:39
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    @Turbo: If we are allowed to pick $\tau\in\Bbb{F}{p^2}$, then such mappings do exist as Eric (+1) explained. But when $1,\tau$ are linearly independent over $\Bbb{F}_p$, the value of $z=\alpha+\beta\tau$ does determine both $\alpha$ and $\beta$ uniquely. So I don't know about the black box... Anyway, if $n$ is a quadratic non-residue modulo $p$, then you can use $\tau=\sqrt{n}\in\Bbb{F}{p^2}$. In that case the Frobenius mapping $\sigma(z)=z^p$ is even an automorphism of fields, and does behave like $\sigma(\alpha+\tau\beta)=\alpha-\tau\beta$. – Jyrki Lahtonen Sep 06 '16 at 05:01