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$$(\mathcal B\otimes_{\mathcal O_S}\mathcal B)\otimes_{\mathcal O_S}\mathcal B\to\mathcal B\otimes_{\mathcal O_S}\mathcal B\to\mathcal B$$ 与 $$\mathcal B\otimes_{\mathcal O_S}(\mathcal B\otimes_{\mathcal O_S}\mathcal B)\to \mathcal B\otimes_{\mathcal O_S}\mathcal B\to\mathcal B$$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）
这是比较范畴论的看法
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.