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?

Question

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 $B$.

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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?

Answer 1

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 $\{A_i\}$.