The Stacks project

Remark 115.5.7. 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:

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

  2. 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)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00Q7. Beware of the difference between the letter 'O' and the digit '0'.