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Let $k$ be a field of char $0$ and let $\mathbb{Z}$ act on $\mathbb{A}^1_k$ by the action induced by $G\to\mathrm{Aut}_k(k[X]), n\mapsto X+n$. It is rather easy to show that the orbit space $\mathbb{A}^1_k/\mathbb{Z}$ is just $\mathrm{Spec}(k)$. At least to me this is surprising at the moment since dividing out the analogue action of $\mathbb{Z}$ on the topological spaces $\mathbb{R}$ resp. $\mathbb{C}$ gives the sphere $S^1$ (up to homotopy).

Q.: Is that simply a pathological example of a categorical quotient or is there a geometric explanation why this is the right orbit space?

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The first and second Z-actions are rather different. Why would you expect them to give similar results? – Qiaochu Yuan Sep 17 '10 at 22:14
You're using the "wrong" notion of "orbit space". The quotient $\mathbf{A}^1/\mathbf{Z}$ makes good sense as a 1-dimensional smooth algebraic space over $k$, far bigger than ${\rm{Spec}}(k)$. Its geometric points are the $\mathbf{Z}$-orbits of the geometric points of the affine line, so one of your objections is handled well. However, it is not quasi-separated, so in that sense it is not such a good space for the purposes of algebraic geometry. – BCnrd Sep 18 '10 at 5:12
up vote 2 down vote accepted

Your observation $\mathbb{A}^1/\mathbb{Z}=Spec(k)$ refers to the fact that there are no non-constant, $\mathbb{Z}$-invariant polynomial functions. On the other hand there are plenty $\mathbb{Z}$-invariant continues functions on $\mathbb{A}^1(\mathbb{R})$, hence you get a non-trivial quoteint.

Also note that $\mathbb{Z}$ is no group-scheme on the nose. Taking the group ring yields $k[\mathbb{Z}] = k[t,t^{-1}]$ which is a torus. So to get a group-scheme $G$ with $G(k)=\mathbb{Z}$ you need to consider an infinite union of copies of $Spec(k)$, which would be not finitely generated.

There is no good way to take quotients of schemes by infinite, discrete groups in general. For example there is no way to produce curves as quoteints of the upper half plane $\mathbb{H}/\Gamma$, or abelian varietis as quotients $\mathbb{C}^n/ \Lambda$, in the realm of scheme theory.

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