This is a bit of a bizzare question, but I'm going to ask it anyway.


If $X={\rm Spec}R$ is an affine variety, and $\mathfrak{m}$ a closed point, then the localization $R\to R\_{\mathfrak{m}}$ gives a morphism 


$f:{\rm Spec}R\_{\mathfrak{m}}\to{\rm Spec}R$, 


yielding an adjunction between ${\rm Qcoh} R$ and ${\rm Qcoh} R\_{\mathfrak{m}}$. This is just a very long-winded way of saying localization and extension of scalars between module categories. Note that $f\_\*(R\_{\mathfrak{m}})$ is flat as an $R$-module, since localization is exact.


I'm wondering if this just generalizes as is to the scheme world. Take a scheme $X$ (with nice properties as you like, if necessary), and a closed point $x\in X$. We still have a map 


$f:{\rm Spec }\mathcal{O}\_{X,x}\to X$,


given as factorization through an open affine. Is $f\_\*(\mathcal{O}\_{X,x})$ flat as a $\mathcal{O}\_X$-module? I can see that it is flat at all points $y$ for which $x$ and $y$ both live in a common open affine (basically by above), but does this extend to all points? Morally, I'd like to think so, but I can't write down a proof. 



Local to global flatness question
This is of course true, for any semi-separated scheme (i.e. the diagonal is affine), or maybe you assume $X$ is separated if you like, and you can take any point (not necessarily closed). The reason that the sheaf is flat is that ${\rm Spec}(O\_{X,x})\to X$ is affine and flat. In general if $f: X\to Y$ is flat and affine then $f\_\*O\_X$ is $O\_Y-$flat. This is obvious.


But I don't think it is true that $f: X\to Y$ is flat implies $f\_\*O\_X$ is $O\_Y-$flat. See [link text](https://mathoverflow.net/questions/65267/global-sections-of-flat-scheme-also-flat)


In the answer of Jason Starr, the map is flat, while the 0-th direct image is not flat at all.


