Suppose that $\iota: R \to S$ is an extension of rings such that $S$ is finitely generated as an $R$-algebra, and assume that $\iota$ is an injection so we can identify $R$ with its image inside $S$. Assume also that there exists $r_1,\dots,r_n \in R$ with $\sum_i r_i = 1_R$ and the induced extensions $\iota_i : R_{r_i} \to S_{r_i}$ are all integral ring extensions, then can we conclude that $\iota$ is itself integral?
For context, I am studying finite maps of algebraic varieties, and I am interested in constructing a finite map by gluing together finite maps on a finite cover of principal open subsets.
Best Answer
Unintentionally answering my own question here, although I will leave the question open to see if there is a more direct solution. But the answer to this question: Show that a morphism of algebraic sets which restricts to finite morphisms of principal sets is finite also answers my question using the following fact:
Let $\phi : R \to S$ be a ring extension such that $S$ is finitely generated as an $R$-algebra. Then $\phi$ is an integral ring extension if and only if $S$ is finitely generated as an $R$-module.