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Last year, in a talk of Michel Waldschmidt's, I remember hearing a statement along the lines of the title of this question, that is, "The Galois group of $\pi$ is $\mathbb{Z}$.". In what sense/framework is this true? What was meant exactly - and can this notion be made precise?

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Crossposted from math.SE: math.stackexchange.com/q/94994 –  Theo Buehler Dec 30 '11 at 2:52
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See this nice survey by Andre: arxiv.org/abs/0805.2569 –  user18237 Dec 30 '11 at 3:04
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@JSeaton: When you crosspost to/from math.SE, please inform readers (in both sites) of the fact; this helps prevent duplication of efforts. If you get a good answer in one site, please be sure to note it in the other. –  Arturo Magidin Dec 30 '11 at 4:29
    
@Arturo: Okay, noted. –  Joshua Seaton Dec 30 '11 at 16:31

1 Answer 1

I gather that the idea behind $\mathrm{Gal}(\pi)=\mathbb{Z}\backslash\{0\}$ (not $\mathbb{Z}$, $0$ is not a conjugate of $\pi$!) comes from Euler's formula:

$$\prod_{n\in \mathbb{Z\backslash\{0\}}}\bigg(1-\frac{x}{n\pi}\bigg)=\frac{\mathrm{sin}(x)}{x} \in\mathbb{Q}\{x\} $$

which can make you think of those $n\pi$ as the conjugates of $\pi$.

But in order for the Galois groups to act transitively on the conjugates you need all the non-zero rationals, so that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$.

For some actual evidence that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$ is the right answer (this is, consistent with the conjectural picture of periods and motives), see sections 3 and 5 of Galois theory, motives and transcendental numbers by Yves Andre, alredy mentioned in the comments.

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You mean $\mathbb{Z}\setminus\{0\}$ and not $\mathbb{Z}^\times=\{\pm 1\}$. –  Johannes Hahn Feb 14 at 15:37
    
@JohannesHahn. Yes, by $K^{\times}$ I mean the nonzero elements of $K$. –  Myshkin Feb 14 at 15:47
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Which is nonstandard notation. $k^\times$ is standard notation for the unit group of $k$. If $k$ is a field, then of course this coincides with $k\setminus\{0\}$. But $\mathbb{Z}$ isn't a field. –  Johannes Hahn Feb 14 at 16:57

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