Is the field of formal Laurent series flat over the ground ring?

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 $$\mathbb C$$?

• 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 $\mathbb C\otimes_{\mathbb C[[x]]}\mathbb C((x))$. Commented Jul 25, 2022 at 0:37
• 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 $(x)$. Commented Jul 25, 2022 at 5:58
• 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? Commented Jul 25, 2022 at 8:27
• Also, $\mathbb C((x))$ has a $\mathbb C$-module structure by multiplication regardless of the $\mathbb C[[x]]$-module structure of $\mathbb C((x))$? Commented Jul 25, 2022 at 8:29