Remark 114.5.7. Let $R$ be a ring. Suppose that we have $F \in R[X, Y]_ d$ and $G \in R[X, Y]_ e$ such that, setting $S = R[X, Y]/(F)$ we have (1) $S_ n$ is finite locally free of rank $d$ for all $n \geq d$, and (2) multiplication by $G$ defines isomorphisms $S_ n \to S_{n + e}$ for all $n \geq d$. In this case we may define a finite, locally free $R$-algebra $A$ as follows:

as an $R$-module $A = S_{ed}$, and

multiplication $A \times A \to A$ is given by the rule that $H_1 H_2 = H_3$ if and only if $G^ d H_3 = H_1 H_2$ in $S_{2ed}$.

This makes sense because multiplication by $G^ d$ induces a bijective map $S_{de} \to S_{2de}$. It is easy to see that this defines a ring structure. Note the confusing fact that the element $G^ d$ defines the unit element of the ring $A$.

## Comments (0)