4

Let $f:X\to Y$ be a morphism, $\mathcal{F}$ be an $O_X$ module which is flat over $Y$, let $g:Y'\to Y$ be any morphism. Let $X'=X\times_YY'$, let $f':X'\to Y'$ the second projection, and $\mathcal{F'}=p_1^*(\mathcal{F})$ .Then $\mathcal{F'}$ is flat over $Y'$. (Hartshorne p254 9.2)

I think we have to show: $\forall x'\in X'$, $p_1^*(\mathcal{F})_{x'}$ is flat over $O_{Y,f'(x')}$.

And $p_1^*(\mathcal{F})_{x'}\cong{(p_1^{-1}\mathcal{F}})_{x'}\otimes_{(p_1^{-1}O_{X})_{x'}}O_{X',x'}\cong\mathcal{F_{p_1(x')}}\otimes_{O_{X,p_1(x)}}O_{X',x'}$ .

We have $\mathcal{F}_{p_1(x')}$ is flat over $O_{Y,f(p_1(x'))}$.

Is it meaningful to discuss flatness of non-qcoh sheaves?

By the module version, we have $\mathcal{F_{p_1(x')}}\otimes_{O_{Y,f(p_1(x))}}O_{Y',f'(x')}$ flat over $O_{Y',f'(x')}$

Is this the stalk desired? (since $A_p\otimes_{C_p}B_p\cong(A\otimes_C B)_p $ does not hold.) How to make it right?

  • Do you want to assume $\mathscr{F}$ is quasi-coherent? It might not be necessary (i.e. this result might hold for general locally ringed spaces, but I'm not sure)? – Keenan Kidwell Jan 17 '14 at 13:34
  • Thanks, your explication below is clear, I wonder if it is true for sheaves not quasicoherent? Or is it meaningful to discuss flatness of non-qcoh sheaves? –  Jan 18 '14 at 00:13
  • Are there other commonly used non-qcoh sheaf besides skyscraper sheaf? –  Jan 18 '14 at 00:17
  • The skyscraper sheaf is quasi-coherent ... – Martin Brandenburg Jan 18 '14 at 00:17
  • It's definitely meaningful. The definition is the same: $\mathscr{F}$ is flat over $Y$ if it is flat as an $f^{-1}\mathscr{O}Y$-module (via the comorphism $f^\sharp:f^{-1}\mathscr{O}_Y\rightarrow\mathscr{O}_X$), meaning $\mathscr{F}\otimes{f^{-1}\mathscr{O}_Y}-$ is exact. I'm not sure if it's true in this generality though. – Keenan Kidwell Jan 18 '14 at 00:18
  • Dear @Martin, I was hoping you would see this thread. I figured if anyone knows whether this works for LRS it would be you ;) – Keenan Kidwell Jan 18 '14 at 00:18
  • Yes it works for LRS (and even RS if we take the fiber product there), and as always one reduces it to usual commutative algebra. We don't need any quasi-coherence assumption. Sorry but I won't write an answer, because it is really straight forward. Actually it is better and more conceptual to use the global and stalk-free definition of flatness. Most often stalks complicate things. – Martin Brandenburg Jan 18 '14 at 00:19
  • What is LRS and RS? –  Jan 18 '14 at 01:09
  • Dear @mqx, Locally ringed spaces and ringed spaces. – Keenan Kidwell Jan 18 '14 at 01:55

1 Answers1

2

Let $x^\prime\in X^\prime$, $x=p_1(x^\prime)$, $y^\prime=f^\prime(x^\prime)$, and $y=f(x)$. Let $V\subseteq Y$ be an affine open around $y$, $x\in U\subseteq f^{-1}(V)$ an affine open around $x$, and $y^\prime\in V^\prime\subseteq g^{-1}(V)$ an affine open around $y^\prime$. Then $U^\prime=U\times_VV^\prime$ is an affine open in $X^\prime$ around $x^\prime$ (it is $p_1^{-1}(U)\cap (f^\prime)^{-1}(V^\prime)$). Restricting $\mathscr{F}$ to $U$ and then pulling back to $U^\prime$ is the same as pulling $\mathscr{F}$ back to $X^\prime$ and restricting to $U^\prime$, and since stalks can be computed in any open neighborhood of a point, we may assume all the schemes involved are affine: $X=\mathrm{Spec}(B),Y=\mathrm{Spec}(A),Y^\prime=\mathrm{Spec}(A^\prime),X^\prime=\mathrm{Spec}(B^\prime)$, where $B^\prime=B\otimes_AA^\prime$. If $\mathscr{F}$ is a quasi-coherent sheaf on $X$, then $\mathscr{F}$ is associated to a $B$-module $M$, and $p_1^*\mathscr{F}$ is associated to

$M\otimes_BB^\prime=M\otimes_B(B\otimes_AA^\prime)=M\otimes_AA^\prime$,

and this is $A^\prime$-flat if $M$ is $A$-flat.

Keenan Kidwell
  • 26,020