The Stacks project

Example 65.14.8. Let $k$ be a field of characteristic zero. Let $U = \mathbf{A}^1_ k$ and let $G = \mathbf{Z}$. As action we take $n(x) = x + n$, i.e., the action of $\mathbf{Z}$ on the affine line by translation. The only fixed point is the generic point and it is clearly the case that $\mathbf{Z}$ injects into the automorphism group of the field $k(x)$. (This is where we use the characteristic zero assumption.) Consider the morphism

\[ \gamma : \mathop{\mathrm{Spec}}(k(x)) \longrightarrow X = \mathbf{A}^1_ k/\mathbf{Z} \]

of the generic point of the affine line into the quotient. We claim that this morphism does not factor through any monomorphism $\mathop{\mathrm{Spec}}(L) \to X$ of the spectrum of a field to $X$. (Contrary to what happens for schemes, see Schemes, Section 26.13.) In fact, since $\mathbf{Z}$ does not have any nontrivial finite subgroups we see from Lemma 65.14.6 that for any such factorization $k(x) = L$. Finally, $\gamma $ is not a monomorphism since

\[ \mathop{\mathrm{Spec}}(k(x)) \times _{\gamma , X, \gamma } \mathop{\mathrm{Spec}}(k(x)) \cong \mathop{\mathrm{Spec}}(k(x)) \times \mathbf{Z}. \]

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View Example 65.14.8 as pdf