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 $Y$.
Best Answer
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 $\mathfrak p=\phi^{-1}(\mathfrak q)$.