Let $X$ be a smooth algebraic variety over $\mathbb{C}$. Is the sheaf of smooth functions on $X$ flat as an $\mathcal{O}_X$ module?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Do you mean smooth function to $\mathbb{R}$ or holomorphic functions to $\mathbb{C}$... or what? $\endgroup$– Mattia TalpoCommented Dec 15, 2015 at 17:13
-
$\begingroup$ Sorry, I mean complex valued smooth functions. $\endgroup$– Ying XieCommented Dec 16, 2015 at 1:59
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Yes, it is. First of all, the ring of germs of holomorphic functions $\mathcal{O}^h_x$ is flat over the ring of germs regular functions $\mathcal{O}_x$ at some point $x \in X$, see for example Taylor, "Several complex variables with Connections to Algebraic Geometry and Lie Groups", Theorem 13.3.5.
Secondly, $\mathcal{O}^r_x$ is flat over the ring $\mathcal{O}^h_x$ of real analytic functions. This is so, since germs of real analytic functions in $\mathbb{C}^n$ is isomorphic to germs of holomorphic functions in $\mathbb{C}^{2n}$, and $\mathcal{O}^h_{\mathbb{C}^{2n},0}$ is flat over $\mathcal{O}^h_{\mathbb{C}^n,0}$, see for example Fischer, "Complex Analytic Geometry", Proposition 3.17.
Finally, the ring of germs of smooth functions is flat over the ring of real analytic functions, which can be found in Malgrange, "Ideals of Differentiable Functions", Corollary VI.1.12.
answered Dec 16, 2015 at 9:20
-
-
$\begingroup$ I guess that you mean to write "$\mathcal{O}^r_x$ is flat over the ring $\mathcal{O}^h_x$" at the start of the second paragraph, since the inclusion goes the other way. Also, it looks like you meant to refer to Corollary VI.1.12 of Malgrange at the end of the last paragraph. $\endgroup$– MichaelCommented Oct 13, 2018 at 23:46
-
$\begingroup$ You were right @Michael, corrected now. $\endgroup$ Commented Oct 14, 2018 at 7:14