Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
It seems to me that you're saying your map factors through an open affine $U \subset X$ such that the map to $U$ is flat. But open immersions are flat and a composition of flat maps is flat, so that would seem to do it. –  Mike Skirvin Jun 3 '11 at 16:35
    
@Mike: This argument is not quite correct, see the link to the counterexample in Lei's answer. –  Martin Brandenburg Jun 4 '11 at 9:13
add comment

1 Answer

up vote 3 down vote accepted

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

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

share|improve this answer
    
Lei: Such an example was given by Jason Starr under mathoverflow.net/questions/65267/… –  Philipp Hartwig Jun 4 '11 at 7:21
    
Thanks Philipp, I just find it:) –  Lei Jun 4 '11 at 8:07
    
Lei, many thanks for this. I was aware of the subtlety that flat sheaves are not preserved under flat maps (see also mathforum.org/kb/…) but was unsure of what assumptions were needed in my situation to make it work. –  Michael Jun 6 '11 at 7:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.