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My homework problem is to prove that $W^{-1} R \subseteq W^{-1} S$ is an integral extension whenever $R \subseteq S$ is an integral extension. I have solved the problem, but I did something that I'm not so sure it is possible. Here is my method.

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I'm not sure if I can simplify $[s_1/1][r_1/1]$ to $s_1 r_1$. Is this accurate and possible?

Dylan C. Beck
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Chelsea
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  • Could you write it in your post? Images are generaly less readable than LaTeX text. – Fractal Admirer Dec 16 '21 at 04:53
  • The short answer is no; you have to be a little careful. But you are correct that $W^{-1} S$ is a finitely generated $W^{-1} R$-module, so it is an integral extension. – Dylan C. Beck Dec 27 '21 at 05:49
  • In order to show that $W^{-1} S$ is a finitely generated $W^{-1} R$-module, use the fact that $s/v = t/w$ in $W^{-1} S$ if and only if there exists an element $u \in R \setminus W$ such that $u(ws - vt) = 0.$ – Dylan C. Beck Dec 27 '21 at 05:58
  • For instance, this link might help. – Dylan C. Beck Dec 27 '21 at 06:00
  • @DylanC.Beck why should it be that $W^{-1}S$ is a finitely generated $W^{-1}R$ module? It seems that the only information given is that $S$ is an integral extension of $R$, which does not imply that $S$ is finite over $R$. (As I'm sure you know, finite = integral + finite type. For example, the integral closure of $\mathbb{Z}$ in $\mathbb{C}$ is integral over $\mathbb{Z}$, but not finite over $\mathbb{Z}$.) – Alex Wertheim Dec 27 '21 at 06:34
  • @AlexWertheim, if you read the attached document, you will find that $S$ is assumed to be a finitely generated $R$-module. I think the OP is a bit misleading: the task is to prove that $W^{-1} R \subseteq W^{-1} S$ is an integral extension whenever $R \subseteq S$ is a module-finite extension. – Dylan C. Beck Dec 27 '21 at 19:24
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    @DylanC.Beck I did read the attached document. To my eye, the introduction of finiteness conditions into the problem seemed more likely to be an error made by the OP, and not part of the problem statement, considering that the statement is true absent any unnecessary finiteness conditions and also has an easy proof. I suppose only the OP can clarify what they were asked. – Alex Wertheim Dec 27 '21 at 19:57

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