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There is a point I don't understand in : Edward, Fermat's Last Theorem. More precisely in the last paragraph of page 173.

Let $\alpha =\exp(2i\pi/p)$, and $u=F(\alpha) \in \mathbb{Z}[\alpha]^\times$ such that $u=-\overline{u}$, where $F(X) = \sum_{i=0}^{p-1} b_i X^i$. Then this implies $F(\alpha)=-F(\alpha^{-1})$. Assume without loss of generality $b_0=0$.

Why does it implies the following : $F(\alpha) = \sum_{i=1}^{p-1} b_i(\alpha^i-\alpha^{-i})$ ?

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FTL = Faster Than Light. FLT: Fermat's Last Theorem. – Arturo Magidin Dec 5 '11 at 13:46
up vote 3 down vote accepted

Since $\overline{\alpha^i} = \alpha^{-i} = \alpha^{p-i}$, it follows from $u = -\overline{u}$ that $b_i = -b_{p-i}$. Thus each $b_i$ pairs up with $b_{p-i}$. So $$F(\alpha) = \sum_{i=1}^{p-1} b_i \alpha^i = \sum_{i=1}^{(p-1)/2} (b_i \alpha^i - b_i \alpha^{p-i}) = \sum_{i=1}^{(p-1)/2} b_i (\alpha^i - \alpha^{-i})$$ (the last sum goes from 1 to $(p-1)/2$, not $p-1$, because we've paired up the elements).

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Ok, I was confused because in the book he wrote : $F(\alpha) = b_1(\alpha-\alpha^{-1}) + b_2(\alpha^2-\alpha^{-2}) + \cdots$. Since $(\alpha^i)_{i=1..p-1}$ is a basis of $Z[\alpha]$, we can identify the terms. – user10676 Dec 5 '11 at 9:33

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