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I have a confusion about the definition of flat sheaf of module.

Let $f: X \rightarrow Y$ be a morphism of schemes and $\mathcal{F}$ be a sheaf of $\mathcal{O}_X$ module. Then $\mathcal{F}$ is flat over $Y$ at a point $x\in X$ if $\mathcal{F_x}$ is flat $\mathcal{O}_{y,Y}$ -module where $y=f(x)$ and $\mathcal{F_x}$ is considered as an $\mathcal{O}_{y,Y}$ -module by the natural map $f^{\sharp}:\mathcal{O}_{y,Y}\rightarrow \mathcal{O}_{x,X}$.

Now, my question is: is the above definition equivalent as saying $(f_{*} \mathcal{F})_{y}$ is flat as $\mathcal{O}_{y,Y}$ module?

Similarly, is $\mathcal{F}$ flat at every point of $X$ over $Y$ equivalent as saying $f_*\mathcal{F}$ is a flat $\mathcal{O}_Y$-module?'

This question came in my mind when I tried to prove the following result:

'Let $f:X\rightarrow Y$ be a finite morphism of noetherian schemes and $\mathcal{F}$ be a coherent sheaf on $X$. Then $\mathcal{F}$ is flat over $Y$ if and only if $f_*\mathcal{F}$ if locally free on $Y$.'

Let's assume $\mathcal{F}$ is flat first. Now, since $f$ is finite and $\mathcal{F}$ is coherent, therefore $f_*\mathcal{F}$ is coherent on $Y$.

At this point it appears to me that, if flatness of $\mathcal{F}$ implies that $f_*\mathcal{F}$ is flat over $Y$ then localizing at a point of $Y$ and then using the equivalence of free module and flat module over a noetherian local ring I can complete the proof.

Similarly, assuming $f_*\mathcal{F}$ is free on $Y$ implies it's flat on $Y$, then it seems to me that this should be same as saying $\mathcal{F}$ is flat over $Y$.

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up vote 3 down vote accepted

The answer to the first part of your question is no. See this thread.

However, in the case of finite (or more generally affine) morphism, $\mathcal F$ is flat over $Y$ if and only if $f_*\mathcal F$ is flat over $Y$. This is because is $\phi: A\to B$ is a ring homomorphism and if $M$ is a $B$-module, then $M$ if flat over $A$ if and only if for any prime ideal $\mathfrak q$ of $B$, $M_\mathfrak q$ is flat over $A_{\mathfrak p}$ where $\mathfrak p=\phi^{-1}(\mathfrak q)$.

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Thank you very much for your reply! I came to exactly the same point as you for finite or affine morphism case but then I could not prove the claim that M is flat over A if and only if $M_\mathfrak{q}$ is flat over $A_\mathfrak{p}$ for every $\mathfrak{p}$ and $\mathfrak{q}$. –  Omprokash Das Nov 18 '12 at 20:43
    
Have a look at math.stackexchange.com/questions/114104. –  Qing Liu Nov 18 '12 at 22:41
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this might also be relevant math.stackexchange.com/questions/133876/… –  Jacob Bell Nov 19 '12 at 8:47
    
Thanks to all of you guys for your valuable comments! –  Omprokash Das Nov 20 '12 at 17:29
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