# The Stacks Project

## Tag 02Z7

Example 56.14.8. 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 $$\gamma : \mathop{\rm Spec}(k(x)) \longrightarrow X = \mathbf{A}^1_k/\mathbf{Z}$$ of the generic point of the affine line into the quotient. We claim that this morphism does not factor through any monomorphism $\mathop{\rm Spec}(L) \to X$ of the spectrum of a field to $X$. (Contrary to what happens for schemes, see Schemes, Section 25.13.) In fact, since $\mathbf{Z}$ does not have any nontrivial finite subgroups we see from Lemma 56.14.6 that for any such factorization $k(x) = L$. Finally, $\gamma$ is not a monomorphism since $$\mathop{\rm Spec}(k(x)) \times_{\gamma, X, \gamma} \mathop{\rm Spec}(k(x)) \cong \mathop{\rm Spec}(k(x)) \times \mathbf{Z}.$$

The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 2411–2439 (see updates for more information).

\begin{example}
\label{example-affine-line-translation}
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
$$\gamma : \Spec(k(x)) \longrightarrow X = \mathbf{A}^1_k/\mathbf{Z}$$
of the generic point of the affine line into the quotient.
We claim that this morphism does not factor through any
monomorphism $\Spec(L) \to X$ of the spectrum of
a field to $X$. (Contrary to what happens for schemes, see
Schemes, Section \ref{schemes-section-points}.) In fact, since
$\mathbf{Z}$ does not have any nontrivial finite subgroups we see from
Lemma \ref{lemma-quotient-field-map} that for any such
factorization $k(x) = L$. Finally, $\gamma$ is not a monomorphism
since
$$\Spec(k(x)) \times_{\gamma, X, \gamma} \Spec(k(x)) \cong \Spec(k(x)) \times \mathbf{Z}.$$
\end{example}

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