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Given that $R\to S$ is a flat local ring homomorphism of two Noetherian local rings. Then is $S$ always a finitely generated $R$-module?

This question stems from a small detail in a proof I am currently reading, which asserts that given the above hypothesis and $Y$ is a finitely generated $R$-module, then they conclude that $S\otimes_R Y$ is a finitely generated $R$-module. This is why I have the above question.

Can you explain this for me? Thank you

T C
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  • For a Noetherian local ring $(R,\mathfrak{m})$ the map $R\to \widehat{R}$, the $\mathfrak{m}$-adic completion is always flat, but the latter is seldom a finitely generated $R$-module. – Mohan Mar 16 '18 at 19:57
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    $S \otimes_R Y$ should be finitely generated as an $S$-module, not necessarily as an $R$-module. – D_S Mar 16 '18 at 23:01

1 Answers1

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$$\mathbb Q \hookrightarrow \mathbb C$$

Georges Elencwajg
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