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This is a result being widely used in the literature:

$f:X\rightarrow Y$ proper morphism between Noetherian schemes. $F\in Coh(X)$ flat over $Y$, if $H^i(X_y,F_y)=const$, $y\in Y$, then $R^if_\ast F$ is locally free.

The problem is that I can only find references (EGA or GTM 52, etc) of this result with a condition 'Y is reduced', which seems to be unavoidable if one try to prove it by using the canonical technique of 'Grothendieck complex'.

However, the general case is crucial in many arguments.

So my question is that: Is the general case true? When can I find the proof of the general case?

Thanks!

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up vote 11 down vote accepted

I believe that this statement is not true. Take $Y=Spec(k[t]/t^2)$ and $X=\mathbb{P}^1_Y$. On $X$, extensions of $\mathcal{O}$ by $\mathcal{O}(-2)$ are parametrized by : $$Ext^1_X(\mathcal{O},\mathcal{O}(-2))=H^1(X,\mathcal{O}(-2))=H^0(X,\mathcal{O})^{\vee}\simeq k[t]/t^2,$$ as a $k[t]/t^2$-module.

Let $0\to\mathcal{O}(-2)\to E\to\mathcal{O}\to 0$ be the extension corresponding to $t\in k[t]/t^2$. By construction, the map $H^0(X,\mathcal{O})\to H^1(X,\mathcal{O}(-2))$ in the long exact sequence associated to the above short exact sequence is multiplication by $t$. It follows that $H^0(X,E)$, that is the kernel of this map, is necessarily $(t)\subset k[t]/t^2$. Hence it is isomorphic to $k[t]/t$ as a $k[t]/t^2$-module and it is not free.

As $f_*E=H^0(X,E)$ because $Y$ is affine, and $E$ is flat over $Y$ because it is a vector bundle on $X$ that is flat over $Y$, this is a counter-example to the statement.

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This really helped, thanks –  stjc Nov 25 '12 at 12:51
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It may be useful to "explain" the counterexample as follows. Take $Z=\mathbf A^1_k$ and $X=\mathbf P^1_Z$. As in Olivier's answer, extensions of $\mathcal O$ by $\mathcal O(-2)$ are parameterized by $H^0(X,\mathcal O)^\vee=k[t]$ as a $k[t]$-module; take the extension $E$ corresponding to $t$. By the correct version of the statement quoted by stjc ($Z$ is reduced), he direct image $f_*E$ is free outside of the origin in $Z$, but not everywhere. However, if you base change at $Y$, you have only one point and can't detect the non-freeness of $f_*E$. –  ACL Dec 13 '12 at 8:10
    
In fact, the sheaf $f_*E$ is actually free on all of $\mathbb{A}^1_k$ : it is zero as the kernel of $t:k[t]\to k[t]$. But computing $f_*E$ over $Y_n=Spec(k[t]/t^n)$ (for $n\geq 2$), one gets $k[t]/t^{n-1}$, that is neither zero nor free. In particular, the formation of $f_*E$ does not commute with the base-change by any of the $k[t]\to k[t]/t^n$. –  Olivier Benoist Dec 14 '12 at 20:06
    
There is a mistake in my comment above : if $n\geq 1$, computing $f_*E$ over the base $Spec(k[t]/t^n)$, one gets $k[t]/t$ as a $k[t]/t^n$-module. –  Olivier Benoist Dec 15 '12 at 9:24
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