Let \(f: X \to Y\) be a projective morphism, let \(\mcf\) be a coherent sheaf on \(X\) which is flat over \(Y\), and assume that \(H^i(X_y, \mcf_y) = 0\) for some \(i\) and some \(y \in Y\). Show that \(R^i f_*(\mcf)\) is \(0\) in a neighborhood of \(y\).
Solution
Under the standing Noetherian hypotheses of Hartshorne III.11, put \[ A=\mco_{Y,y}, \qquad \mfm=\mfm_y, \qquad k=\kappa(y). \] For \(n\ge1\), let \[ X_n=X\times_Y\Spec(A/\mfm^n) \] and let \(\mcf_n\) be the pullback of \(\mcf\) to \(X_n\). Thus \[ X_1=X_y, \qquad \mcf_1=\mcf_y. \]
1. For every \(n\ge1\) there is a short exact sequence\[ 0 \longrightarrow \mcf_y\otimes_k \mfm^n/\mfm^{n+1} \longrightarrow \mcf_{n+1} \longrightarrow \mcf_n \longrightarrow0, \]where the first term is regarded as a sheaf on \(X_{n+1}\) supported on\(X_y\).
Base-change first to \(X_A=X\times_Y\Spec A\) and write \(\mcf_A\) for the pullback of \(\mcf\). Flatness of \(\mcf\) over \(Y\) says that \(\mcf_A\) is flat over \(A\). Hence tensoring the exact sequence \[ 0 \longrightarrow \mfm^n/\mfm^{n+1} \longrightarrow A/\mfm^{n+1} \longrightarrow A/\mfm^n \longrightarrow0 \] with \(\mcf_A\) remains exact. The middle and right terms are \(\mcf_{n+1}\) and \(\mcf_n\). Since \(\mfm\) annihilates \(\mfm^n/\mfm^{n+1}\), the left term is \[ \mcf_y\otimes_k \mfm^n/\mfm^{n+1}. \]
2. For every \(n\ge1\),\[ H^i(X_n,\mcf_n)=0. \]
The case \(n=1\) is the hypothesis \[ H^i(X_y,\mcf_y)=0. \] Assume inductively that \(H^i(X_n,\mcf_n)=0\). Because \(A\) is Noetherian, \(\mfm^n/\mfm^{n+1}\) is a finite-dimensional \(k\)-vector space. Hence \[ \mcf_y\otimes_k\mfm^n/\mfm^{n+1} \] is the pushforward from the closed fibre of a finite direct sum of copies of \(\mcf_y\). A closed immersion has exact pushforward, so its \(i\)-th cohomology is the corresponding finite direct sum of \(H^i(X_y,\mcf_y)\) and is therefore zero. The long exact cohomology sequence of step 1 consequently contains an injection \[ H^i(X_{n+1},\mcf_{n+1}) \hookrightarrow H^i(X_n,\mcf_n)=0. \] Thus \(H^i(X_{n+1},\mcf_{n+1})=0\), and induction proves the claim.
3. The completed stalk of \(R^if_*\mcf\) at \(y\) is zero:\[ \widehat{(R^if_*\mcf)_y}=0. \]
Since \(f\) is projective, it is proper. By the theorem on formal functions, \[ \widehat{(R^if_*\mcf)_y} \cong \varprojlim_n H^i(X_n,\mcf_n). \] Every term of this inverse system is zero by step 2, so the inverse limit is zero.
4. The stalk \((R^if_*\mcf)_y\) is zero.
3.Since\[ \widehat M=\varprojlim_n M/\mfm^nM \]projects surjectively onto \(M/\mfm M\), it follows that\[ M/\mfm M=0. \]By Nakayama's lemma, \(M=0\).
5. There is an open neighborhood \(U\) of \(y\) such that\[ R^if_*\mcf|_U=0. \]
The sheaf \(R^if_*\mcf\) is coherent. Its support is therefore closed. By step 4, the point \(y\) does not belong to this support, so the complement of the support is an open neighborhood \(U\) of \(y\) on which the sheaf vanishes.
6. Q.E.D.
Step 5 is exactly the required neighborhood vanishing of \(R^if_*\mcf\).