# How to obtain the identity of $\mathcal B(U)$ from the above definition?

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?

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