I have trouble following the logic in this proof. In particular, why is the following equality is true: $$\displaystyle\sum_{x \in K^\times} x^u = \displaystyle\sum_{x \in K^\times} y^u x^u$$
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Similar to this question. +1 to y'all – Jyrki Lahtonen Jun 23 '15 at 10:41
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As $y \in K^{\times}$, the map $\varphi_y : K^{\times} \to K^{\times}$, $x \mapsto yx$ is a bijection (with inverse $\varphi_{y^{-1}}$). So the only difference between the two summations is the order of the summands, so they are equal.
For example, if $u = 1$, $K = \mathbb{Z}/5\mathbb{Z}$, and $y = 2$, then we have
\begin{align*} \sum_{x\in K^{\times}}x &= 1 + 2 + 3 + 4\\ \sum_{x\in K^{\times}}2x &= 2 + 4 + 1 + 3. \end{align*}
Michael Albanese
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So is there any explicit reason the blogger mentions we can let $y$ equal a generator? – user240033 Jun 22 '15 at 21:34
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I'm not sure I understand your question. Are you asking why the author chooses $y$ such that $y^u \neq 1$? – Michael Albanese Jun 22 '15 at 21:36
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The author says "We can let $y$ be a generator of $K^\times$". But as you say, $\varphi_y$ is a bijection as $y \in K^\times$. – user240033 Jun 22 '15 at 21:38
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2Choosing $y$ to be a generator guarantees that $y^u \neq 1$ as $u$ is not divisible by $q - 1$. – Michael Albanese Jun 22 '15 at 21:42
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@orangeskid: Of course. The proof that the OP refers to shows that they must be zero. – Michael Albanese Jun 22 '15 at 23:34