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Let $(S,\mathcal O_S)$ be a scheme. An $\mathcal O_S$-algebra $\mathcal B$ is an $\mathcal O_S$-module together with a morphism $\varphi:\mathcal B\otimes_{\mathcal O_S}\mathcal B\to \mathcal B$ which satisfies two following conditions:

$$(\mathcal B\otimes_{\mathcal O_S}\mathcal B)\otimes_{\mathcal O_S}\mathcal B\stackrel{\varphi\otimes \operatorname{Id}_{\mathcal B}}\longrightarrow\mathcal B\otimes_{\mathcal O_S}\mathcal B\stackrel{\varphi}\longrightarrow\mathcal B$$ and $$\mathcal B\otimes_{\mathcal O_S}(\mathcal B\otimes_{\mathcal O_S}\mathcal B)\stackrel{ \operatorname{Id}_{\mathcal B}\otimes\varphi}\longrightarrow \mathcal B\otimes_{\mathcal O_S}\mathcal B\stackrel{\varphi}\longrightarrow\mathcal B$$ are the same, $$\mathcal B\otimes_{\mathcal O_S}\mathcal B\stackrel{\textrm{commute two factors}}\longrightarrow\mathcal B\otimes_{\mathcal O_S}\mathcal B\stackrel{\varphi}\longrightarrow \mathcal B$$ and $$\mathcal B\otimes_{\mathcal O_S}\mathcal B\stackrel{\varphi}\longrightarrow \mathcal B$$ are the same.

Then for any open set $U\subset S$, we can define multiplication on $\mathcal B(U)$ which is associative and commutative. But in order to let $\mathcal B(U)$ form a commutative ring, how to obtain the identity of $\mathcal B(U)$ from the above definition?

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You can't. The definition you've stated is the definition of a not-necessarily-unital commutative $\mathcal{O}_S$-algebra. To get a unital algebra, you need to also include a morphism $\mathcal{O}_S\to \mathcal{B}$ which acts as an identity with respect to $\varphi$.

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  • $\begingroup$ Which page can I find the definition of $\mathcal O_S$-algebra in "Stacks Project"? $\endgroup$ – Born to be proud Oct 19 '18 at 6:27

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