# The Stacks Project

## Tag 02L5

Lemma 34.20.18. The property $\mathcal{P}(f) =$''$f$ is affine'' is fpqc local on the base.

Proof. A base change of an affine morphism is affine, see Morphisms, Lemma 28.11.8. Being affine is Zariski local on the base, see Morphisms, Lemma 28.11.3. Finally, let $g : S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is affine. In other words, $X'$ is affine, say $X' = \mathop{\rm Spec}(A')$. Also write $S = \mathop{\rm Spec}(R)$ and $S' = \mathop{\rm Spec}(R')$. We have to show that $X$ is affine.

By Lemmas 34.20.1 and 34.20.6 we see that $X \to S$ is separated and quasi-compact. Thus $f_*\mathcal{O}_X$ is a quasi-coherent sheaf of $\mathcal{O}_S$-algebras, see Schemes, Lemma 25.24.1. Hence $f_*\mathcal{O}_X = \widetilde{A}$ for some $R$-algebra $A$. In fact $A = \Gamma(X, \mathcal{O}_X)$ of course. Also, by flat base change (see for example Cohomology of Schemes, Lemma 29.5.2) we have $g^*f_*\mathcal{O}_X = f'_*\mathcal{O}_{X'}$. In other words, we have $A' = R' \otimes_R A$. Consider the canonical morphism $$X \longrightarrow \mathop{\rm Spec}(A)$$ over $S$ from Schemes, Lemma 25.6.4. By the above the base change of this morphism to $S'$ is an isomorphism. Hence it is an isomorphism by Lemma 34.20.17. Therefore Lemma 34.19.4 applies and we win. $\square$

The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 4992–4996 (see updates for more information).

\begin{lemma}
\label{lemma-descending-property-affine}
The property $\mathcal{P}(f) =$$f$ is affine''
is fpqc local on the base.
\end{lemma}

\begin{proof}
A base change of an affine morphism is affine, see
Morphisms, Lemma \ref{morphisms-lemma-base-change-affine}.
Being affine is Zariski local on the base, see
Morphisms, Lemma \ref{morphisms-lemma-characterize-affine}.
Finally, let
$g : S' \to S$ be a flat surjective morphism of affine schemes,
and let $f : X \to S$ be a morphism. Assume that the base change
$f' : X' \to S'$ is affine. In other words, $X'$ is affine, say
$X' = \Spec(A')$. Also write $S = \Spec(R)$
and $S' = \Spec(R')$. We have to show that $X$ is affine.

\medskip\noindent
By Lemmas \ref{lemma-descending-property-quasi-compact}
and \ref{lemma-descending-property-separated} we see that
$X \to S$ is separated and quasi-compact. Thus
$f_*\mathcal{O}_X$ is a quasi-coherent sheaf of $\mathcal{O}_S$-algebras,
see Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent}.
Hence $f_*\mathcal{O}_X = \widetilde{A}$ for some $R$-algebra $A$.
In fact $A = \Gamma(X, \mathcal{O}_X)$ of course.
Also, by flat base change
(see for example
Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology})
we have $g^*f_*\mathcal{O}_X = f'_*\mathcal{O}_{X'}$.
In other words, we have $A' = R' \otimes_R A$.
Consider the canonical morphism
$$X \longrightarrow \Spec(A)$$
over $S$ from Schemes, Lemma \ref{schemes-lemma-morphism-into-affine}.
By the above the base change of this morphism to $S'$ is an isomorphism.
Hence it is an isomorphism by
Lemma \ref{lemma-descending-property-isomorphism}.
Therefore Lemma \ref{lemma-descending-properties-morphisms} applies and we win.
\end{proof}

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