Multiplication of (pluri-)canonical ring

Question

Let $X$ be a smooth projective variety. One uses the so-called (pluri-)canonical ring $R(X) = \bigoplus_{m \geq 0} H^0(X,\omega_X^{\otimes m})$ to define the canonical model $X^{\text{can}}$ of $X$. Just because of the resemblence with the tensor algebra, I had always thought that the multiplication on $R(X)$ was given by $$ \left( \sum_i s_i \right) \cdot \left( \sum_j t_j\right ) = \sum_n \sum_{i+j= n} (s_i \cdot t_j),$$ where the multiplication $s_i \cdot t_j$ is induced by the tensor product. Since that would not result in a commutative ring, I was certainly mistaken and am now looking for the actual multiplication. I checked several sources (including Mumford's original paper where he first defines this ring), but all these sources never spell out the multiplication or just claim that it has an obvious one. While I will probably agree a posteriori that there was an obvious multiplication that I seem to overlook, I just do not see it at the moment and would therefore be grateful for someone to clarify this little problem.

Answer 1

I think the point is that a section of a line bundle is uniquely determined by the divisor given by $\{s=0\}$. This is the divisor line bundle correspondence.

Given the sections, we get a divisor $D_{1}$ linearly equivelant to $nK_X$ and a divisor $D_2$ linearly equivelant to $mK_{X}$. Then, the divisor $D_1+ D_2$ is linearly equivelant to $(n+m)K_X$. This corresponding section is determined by the divisor $$D_1 + D_2$$ (not the class of the divisor in the picard group but the class in the formal abelian group of divisors). Then, the machinery of the divisor-line bundle correspondence gives a section of $\otimes^{m+n}K_{X}$ whose vanishing is exactly $D_1+D_{2}$.

Matt E's answer gives a very nice explanation of this correspondance here Divisor -- line bundle correspondence in algebraic geometry. See the last few paragraphs of the answer.