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!
Best Answer
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.