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Let $ X $, $ Y $ and $ Z $ be $ S $-schemes. Suppose that we have a morphism $ \varphi : X \times_{S} Z \to Y \times_{S} Z $. Let $ V $ be an open affine subset of $ Y $ and let $ U' = \varphi ^ { -1 } ( V \times _{S} Z ) $. Is it true that there is an open affine cover of $ U ' $ by opens of the form $ U _{i} \times _{S} Z $ where $ U_{i} $ are open affine in $ X $?

P.S. The situation where I need this is the following: $ S $ is locally Noetherian, $ Z = \text{Spec} O_{S,s} $ for some $ s \in S $ and $ Y $ is finite type over $ S $.

Eric Wofsey
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Hodge-Tate
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1 Answers1

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This is not true in general. For instance, let $S=\operatorname{Spec} k$ for some field $k$ and let $X=Y=Z=\mathbb{A}^1$. Let $\varphi:\mathbb{A}^2\to\mathbb{A}^2$ be defined by $\varphi(x,y)=(x-y,0)$, and let $V=\mathbb{A}^1\setminus\{0\}$. Then $U'$ is the complement of the diagonal in $\mathbb{A}^2$, which does not contain $U\times\mathbb{A}^1$ for any nonempty open $U\subseteq\mathbb{A}^1$.

I don't know whether your additional assumptions make it true. In particular, the assumption that $Z$ has the form $Z = \operatorname{Spec} O_{S,s}$ is extremely restrictive and may make it true, but at the moment I don't see a way to prove it.

Eric Wofsey
  • 330,363