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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:

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?

I imagine that this would draw on the fact that the $\sin x$ is an 'infinite polynomial' over $\mathbb{Q}$ whose roots are the integer multiples of $\pi$. However, if this is the case, then how can we possibly view this function as a sort of generalized 'minimal polynomial' while we have factorizations like $\sin x = 2 \cos (x/2) \cdot \sin (x/2)$?

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You could consider the Galois group of $\mathbb{Q}(\pi)$ over $\mathbb{Q}$. Since $\mathbb{Q}(\pi)\cong \mathbb{Q}(x)$, this would be equivalent to asking for the set of all $\mathbb{Q}$-automorphism of the function field $\mathbb{Q}(x)$. But then you would get the group of invertible $2\times 2$ matrices with coefficients in $\mathbb{Q}$, rather than $\mathbb{Z}$. E.g., here. –  Arturo Magidin Dec 29 '11 at 18:27
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My guess is that $\pi$ is (related to) the period attached to the motive attached to the multiplicative group scheme $\mathbb{G}_m$, which over the complex numbers is $\mathbb{C}^{\times}$, which has fundamental group $\mathbb{Z}$, so should have etale fundamental group something like $\mathbb{Z}$ or its profinite completion, and etale fundamental groups generalize Galois groups. (Actually from mathoverflow.net/questions/15153/periods-and-l-values it looks like the right period is $2\pi i$.) –  Qiaochu Yuan Dec 29 '11 at 18:28
    
@Arturo: I appreciate the comment, however, this is clearly not what the speaker had in mind. –  Joshua Seaton Dec 29 '11 at 18:30
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@Arturo: don't you mean $\text{PGL}_2(\mathbb{Q})$? –  Qiaochu Yuan Dec 29 '11 at 18:33
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Crossposted to MO –  t.b. Dec 30 '11 at 2:53
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