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AlgebraZahlentheorie
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Fix typo

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Stefan Kebekus 2024-02-02 13:51:13 +01:00
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commit 07461e0969
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@ -93,7 +93,7 @@ französischer Mathematiker.}-Symbol ein.
Nichtreste gibt, ist $\sum_{k=1}^{p-1} \left(\frac{k}{p}\right) = 0$ und Nichtreste gibt, ist $\sum_{k=1}^{p-1} \left(\frac{k}{p}\right) = 0$ und
deshalb deshalb
\begin{equation}\label{eq:g4.2} \begin{equation}\label{eq:g4.2}
\sum_{k=1}^{p-2} \left(\frac{k}{p}\right) = -\left(\frac{p-1}{p}\right) = -\left(\frac{-1}{p}\right) ∈ F. \sum_{k=1}^{p-2} \left(\frac{k}{p}\right) = -\left(\frac{p-1}{p}\right) = -\left(\frac{-1}{p}\right) ∈ \bN.
\end{equation} \end{equation}
\end{beobachtung} \end{beobachtung}