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Let $\mathbb C[[x]]$ be the ring of formal power series in a single indeterminate $x$. Let $\mathbb C((x))$ be the field of formal Laurent series which can be seen as the fraction field of $\mathbb C[[x]]$. It is well known that $\mathbb C((x))$ is a flat $\mathbb C[[x]]$-module. Is $\mathbb C((x))$ flat over $\mathbb C$?

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The complex numbers are a field and all modules over fields are free, therefore flat.

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  • $\begingroup$ Ok, I agree. Thanks. Can you factor $\mathbb C((x))$ by the maximal ideal $(x)$ in $\mathbb C[[x]]$? $\mathbb C((x))$is a flat $C[[x]]$-module, so the result should be $\mathbb C\otimes_{\mathbb C[[x]]}\mathbb C((x))$. $\endgroup$ Jul 25, 2022 at 0:37
  • $\begingroup$ Not sure I understand what you are asking. What are you claiming that the final tensor product should be? $\endgroup$
    – Arkady
    Jul 25, 2022 at 1:17
  • $\begingroup$ The field $\mathbb C((x))$ has (at least) two non-isomorphic $\mathbb C[[x]]$-module structures: One where $x$ acts by multiplication by $x$ and one where it acts as $0$. In the latter case, yes, we may essentially treat it as a $\mathbb C$-module since the action factors through the maximal ideal $(x)$. $\endgroup$
    – Arkady
    Jul 25, 2022 at 5:58
  • $\begingroup$ Wait. $\mathbb C\otimes_{\mathbb C[[x]]}\mathbb C((x))$ is always a $\mathbb C$-module regardless of the $\mathbb C[[x]]$-module structure of $\mathbb C((x))$, right? $\endgroup$ Jul 25, 2022 at 8:27
  • $\begingroup$ Also, $\mathbb C((x))$ has a $\mathbb C$-module structure by multiplication regardless of the $\mathbb C[[x]]$-module structure of $\mathbb C((x))$? $\endgroup$ Jul 25, 2022 at 8:29

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