Let $\mathbb F_{q^{2n}}$ be a finite field, and let $a,b\sim\text{Unif}\lbrace 0,q^{2n}-1\rbrace$. What is the probability that:
$a - a^q\in\mathbb F_{q^2}$
$a - a^q = b - b^q$
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Daniel Fischer
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PiggyJin
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2Do not re-ask your questions if they don't get an answer. Edit to clarify them. Since this version has an answer, and the old one hadn't I have deleted the old version. – Daniel Fischer Dec 19 '16 at 10:53
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To make sure: the elements of $\Bbb{F}{q^{2n}}$ are not naturally in a bijective correspondence with the set of integers in the range $[0,q^{2n}-1]$. Presumably you are expected to use the uniform distribution of $\Bbb{F}{q^{2n}}$ instead of the one on integers. – Jyrki Lahtonen Dec 19 '16 at 16:45
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The additive version of Hilbert's theorem 90 tells you that the kernel of the trace $\mathbb{F}_{q^{2n}} \rightarrow \mathbb{F}_{q}$ consists precisely of the elements of the form $a - a^q$ with $a \in \mathbb{F}_{q^{2n}}.$ Likewise, the elements of $\mathbb{F}_{q^2}$ of the form $a - a^q$ are also precisely the traceless elements.
Since the trace is a nonzero functional, its kernel has order $q^{2n-1}$ (on $\mathbb{F}_{q^{2n}}$) and $q$ (on $\mathbb{F}_{q^2}$). Combined with the linearity of the map $a \mapsto a - a^q$ this is enough to solve both problems.
Mark Schultz-Wu
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user400429
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For those unfamiliar with it, a discussion of the additive version of Hilbert's Theorem 90 – Mark Schultz-Wu Dec 19 '16 at 07:40
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@PiggyJin You can argue directly that the kernel of $a \mapsto a^q$ is exactly $\mathbb{F}_q$ and count the image that way. The point is that $a - a^q$ is not just an arbitrary polynomial but a linear map, so everything reduces to linear algebra – user400429 Dec 19 '16 at 08:36
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What I understood : suppose that $a - a^q in \mathbb F_{q^2}$. then because of the Hilbert 90 Theorem, $a - a^q$ in the kernal of trace map from $\mathbb F_{q^2}$ to $\mathbb F_{q}$. Is that right? I think this is wrong, so I cannot understand the above argument. – PiggyJin Dec 19 '16 at 08:47
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