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I'm reading a proof that starts:

Proof: Assume $f$ separated. Suppose $(U,V)$ is a pair as in (1). Let $W=\operatorname{Spec}(R)$ be an affine open {subset ?} of $S$ containing both $f(U)$ and $f(V)$. Write $U=\operatorname{Spec}(A)$ and $V=\operatorname{Spec}(B)$ for $R$-algebras $A$ and $B$.


But anyhow, I don't see how $A$ is an $R$-algebra which leads to my more general question:

If $f: X \to \mathrm{Spec}(R)$ is a morphism of schemes, and $U \cong \mathrm{Spec}(A)$ is an open affine of $X$, how is $A$ an $R$-algebra?

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    $\begingroup$ By the universal property of $\mathrm{Spec}$, given $f$ you have that $\Gamma(X,\mathcal{O}_X)$ is an $R$-algebra. Composing with the restriction map from $X$ to $U$ you get the answer. $\endgroup$
    – Aurelio
    Oct 8, 2020 at 14:20
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    $\begingroup$ @Aurelio Could we have also restricted $f|_U : U \to \mathrm{Spec}(R)$ and that would have given us a map $\mathrm{Spec}(A) \to \mathrm{Spec}(R)$ which in turn gives a ring hom $R \to A$? $\endgroup$ Oct 8, 2020 at 14:50
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    $\begingroup$ Yes, of course! Thar argument works too. If you write down the final $R$-action on $A$, you will see that the two approaches are basically the same. $\endgroup$
    – Aurelio
    Oct 8, 2020 at 14:50
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    $\begingroup$ @Aurelio Please consider compiling your comments in to an answer below :) $\endgroup$
    – KReiser
    Oct 8, 2020 at 18:21
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    $\begingroup$ I'm surprised that a 10k+ rep user has not yet learned that using images for significant portions of your post is a bad idea!! $\endgroup$
    – K.defaoite
    Jun 1, 2021 at 1:07

1 Answer 1

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Following @KReiser's suggestion, I bundle my comments above into a proper answer. Given a morphism $\DeclareMathOperator{Spec}{Spec}f:X\to \Spec R$, we obtain a map on global sections $f^\sharp\colon R\to \Gamma(X,\mathcal{O}_X)$, hence the latter ring is an $R$-algebra on the nose. Moreover, by the universal property of $\Spec$, there is a natural bijection between $R$-algebra structures on $\Gamma(X,\mathcal O_X)$ and morphisms $X\to \Spec R$.

This construction localises easily. For any open subset $U\subseteq X$, the restriction map $\Gamma(X,\mathcal O_X)\to\Gamma(U,\mathcal O_X)$ presents the ring of sections over $U$ as an $R$-algebra. In particular, when $U=\Spec A$, this holds for $A\simeq \Gamma(U,\mathcal O_X)$. Equivalently, we could simply consider the restriction $f_{\vert U}\colon U \to \Spec R$ and apply the universal property.

This discussion shows why we often work in the category $\mathrm{Sch}/B$ of schemes over a fixed scheme $B$, also called "$B$-schemes": they are schemes with a given morphism $X\to B$, and morphisms of $B$-schemes preserve the map to the base.

Consider for starters $B=\Spec R$ affine, for example $R=k$ a field: we are saying that a $B$-scheme is covered by spectra of $R$-algebras, not just rings. Similarly, a map between affine $B$-schemes is equivalent to a map of $R$-algebras. Now look at $X\to B$, when $B$ is not necessarily affine. We pull back an affine cover $\{\Spec A_i\}$ of $B$ to $X$, and up to refining we obtain that $X$ is covered by algebras over different rings $\{A_i\}$.

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