Let $B$ be a graded ring and $X=\mathrm {Proj}\: B$. Let $B(m)$ denote the twist of $B$ (i.e. a graded $B$-module such that $B(m)_d=B_{m+d}$,) $\mathcal O_X(m)$ denote the quasi-coherent sheaf $\tilde{B(m)}$. Then how to prove the statement $\mathcal O_X(m)\otimes_{\mathcal O_X} \mathcal O_X(n)= \mathcal O_X(m+n)$? (The statement is from Liu’s book Algebraic Geometry and Arithmetic Curves.)
Let $B(m)_{(g)}$ denote the set of elements of degree 0 in the localized ring $B(m)_{g} $ for any homogeneous $g$ with a positive degree. Then it is equivalent to $$B(m)_{(g)}\otimes_{B_{(g)}} B(n)_{(g)}\simeq B(m+n)_{(g)} $$ for any homogeneous $g$ with a positive degree. The isomorphism is obvious if $\mathrm D_+(g)\subseteq \mathrm D_+(f)$ for some $f$ of degree 1, but such an $f$ may not exist.
Set $m=n=1$ and $\mathrm{deg}\: g =2$, what is the isomorphism above exactly? I can not find it.
There's no reason for this map to be an isomorphism in general - in fact, it's not even true that $\mathcal{O}_X(n)\otimes \mathcal{O}_X(m) = \mathcal{O}_X(n+m)$ in general! See Stacks 01MJ for counterexamples, and Stacks 01MM, specifically 01MT for more details about when this is true - it's true for instance if $B$ is generated by $B_1$ as a $B_0$ algebra.