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

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

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