Let $(S,\mathcal O_S)$ be a scheme. What's the definition of $\mathcal O_S$-algebra?

I got the following answer:

It's a sheaf $\mathcal{A}$ of $\mathcal{O}_S$-modules on $S$ which is also a sheaf of rings, such that the $\mathcal{O}_S$-module map $\mathcal{O}_S \to \mathcal{A}$ induced by $1 \mapsto 1$ is a ring map. (Equivalently, a sheaf of rings $\mathcal{A}$ on $S$ together with a ring homomorphism $\mathcal{O}_S \to \mathcal{A}$.)

How to show the above two definitions are equivalent?

The $\mathcal{O}_S$-module map $\mathcal{O}_S \to \mathcal{A}$ induced by $1 \mapsto 1$ is a ring map, we denote it by $f$. For any open subset $U$ of $S$, $f(U): \mathcal O_S(U)\to \mathcal A(U)$, $\forall t\in \mathcal O_S(U),x\in \mathcal A(U)$, how to show $tx=f(U)(t)x$?

可能他把$\mathcal O_S$-algebra定义成$\mathcal B\otimes_{\mathcal O_S}\mathcal B\to \mathcal B$且满足$$(\mathcal B\otimes_{\mathcal O_S}\mathcal B)\otimes_{\mathcal O_S}\mathcal B\to\mathcal B\otimes_{\mathcal O_S}\mathcal B\to\mathcal B$$$$\mathcal B\otimes_{\mathcal O_S}(\mathcal B\otimes_{\mathcal O_S}\mathcal B)\to \mathcal B\otimes_{\mathcal O_S}\mathcal B\to\mathcal B$$一样(结合律),$$\mathcal B\otimes_{\mathcal O_S}\mathcal B\stackrel{交换两factors}\longrightarrow\mathcal B\otimes_{\mathcal O_S}\mathcal B\to \mathcal B$$与原来的$$\mathcal B\otimes_{\mathcal O_S}\mathcal B\to \mathcal B$$一样(交换律)


等价的 我上面所说,应该还要加上模层morphism $\mathcal O_S\to\mathcal B$使$$\mathcal B\cong \mathcal O_S\otimes_{\mathcal O_S} \mathcal B \to \mathcal B\otimes_{\mathcal O_S} \mathcal B\to \mathcal B$$为identity.

  • 2
    $\begingroup$ Do you understand the result for algebras over a ring? A set $A$ is said to be an algebra over a ring $R$ of $A$ admits an $R$-module structure together with a compatible ring structure. Equivalently, the algebra structure on $A$ can be described by a ring structure on $A$ together with a ring map $R \to A$. $\endgroup$ – Sofie Verbeek Oct 17 '18 at 17:09
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    $\begingroup$ Now apply this description for every open $U$. The set $\mathcal{A}(U)$ is an $\mathcal{O}_S(U)$-algebra if it is a ring with a compatible $\mathcal{O}_S(U)$-module structure, or equivalently, if it is a ring together with a ring morphism $\mathcal{O}_S(U) \to \mathcal{A}(U)$. $\endgroup$ – Sofie Verbeek Oct 17 '18 at 17:10
  • $\begingroup$ Which would lead to yet another representation: an $\mathcal{O}_S$ algebra would be a function which for every open $U$, gives an $\mathcal{O}_S(U)$-algebra $\mathcal{A}(U)$, along with restriction maps $\mathcal{A}(U)\to \mathcal{A}(V)$ for each pair $V \subseteq U$, such that for any $x \in \mathcal{O}_S(U)$, $a \in \mathcal{A}(U)$, we have $(a \cdot x)|_V = (a|_V) \cdot (x|_V)$ (and also $(x \cdot a)|_V = (x|_V) \cdot (a|_V)$ if we allow sheaves of non-commutative algebras). $\endgroup$ – Daniel Schepler Oct 19 '18 at 19:31

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