Let \(Y\) be an integral scheme of finite type over an algebraically closed field \(k\). Let \(f: X \to Y\) be a flat projective morphism whose fibres are all integral schemes. Let \(\mcl, \mcm\) be invertible sheaves on \(X\), and assume for each \(y \in Y\) that \(\mcl_y \cong \mcm_y\) on the fibre \(X_y\).
Show that there is an invertible sheaf \(\mcn\) on \(Y\) such that \(\mcl \cong \mcm \tensor f^* \mcn\).
Hint: use the results of this section to show that \(f_*(\mcl \tensor \mcm^{-1})\) is locally free of rank \(1\) on \(Y\).
Solution
Put \[ \mcf=\mcl\tensor\mcm^{-1}. \] Then \(\mcf\) is invertible on \(X\), and the hypothesis says \[ \mcf_y\cong\mco_{X_y} \] for every \(y\in Y\).
1. The sheaf \(\mcf\) is flat over \(Y\), and\[ h^0(X_y,\mcf_y)=1 \]for every \(y\in Y\).
Since \(f\) is flat, \(\mco_X\) is flat over \(Y\). Locally on \(X\), the invertible sheaf \(\mcf\) is isomorphic to \(\mco_X\), so \(\mcf\) is also flat over \(Y\).
For every closed point \(y\in Y\), the residue field is \(k\), because \(Y\) is of finite type over the algebraically closed field \(k\). The fibre \(X_y\) is integral and projective over \(k\), hence \[ H^0(X_y,\mco_{X_y})=k. \] Thus \[ h^0(X_y,\mcf_y)=1 \] at every closed point.
By semicontinuity, the locus where \[ h^0(X_y,\mcf_y)\ge2 \] is closed. If it were nonempty, then, since a finite-type \(k\)-scheme is Jacobson, it would contain a closed point, contrary to the preceding paragraph. On the other hand constants give \[ h^0(X_y,\mco_{X_y})\ge1 \] for every fibre, and \(\mcf_y\cong\mco_{X_y}\). Hence \[ h^0(X_y,\mcf_y)=1 \] for all \(y\in Y\).
2. The sheaf\[ \mcn=f_*\mcf \]is invertible on \(Y\), and formation of \(f_*\mcf\) commutes with everyresidue-field base change:\[ \mcn\tensor\kappa(y) \xrightarrow{\sim} H^0(X_y,\mcf_y). \]
The base \(Y\) is integral, hence reduced. By step 1, \(\mcf\) is coherent and flat over \(Y\), and the function \[ y\longmapsto h^0(X_y,\mcf_y) \] is constantly \(1\). The Grauert/base-change statement in cohomology and base change therefore gives that \[ f_*\mcf \] is locally free of rank \(1\), with the displayed base-change isomorphism. A locally free sheaf of rank \(1\) is invertible, so \(\mcn=f_*\mcf\) is the required line bundle candidate on \(Y\).
3. The canonical evaluation morphism\[ \epsilon:f^*\mcn=f^*f_*\mcf\longrightarrow\mcf \]restricts to an isomorphism on every fibre \(X_y\).
By the base-change isomorphism of step 2, the restriction of \(\epsilon\) to \(X_y\) is the ordinary evaluation map \[ H^0(X_y,\mcf_y)\tensor_{\kappa(y)}\mco_{X_y} \longrightarrow \mcf_y. \] Choose an isomorphism \[ \mcf_y\cong\mco_{X_y}. \] Step 1 gives \[ H^0(X_y,\mcf_y)\cong\kappa(y), \] and under this identification the evaluation map is \[ \kappa(y)\tensor_{\kappa(y)}\mco_{X_y} \longrightarrow \mco_{X_y}, \qquad a\tensor s\longmapsto as, \] which is an isomorphism.
4. The evaluation morphism\[ \epsilon:f^*\mcn\longrightarrow\mcf \]is an isomorphism on \(X\).
Let \[ \mcc=\operatorname{coker}(\epsilon). \] This is coherent. Fix \(x\in X\), put \(y=f(x)\), and take the stalk at \(x\). By step 3, tensoring the stalk sequence with \(\kappa(y)\) kills the cokernel: \[ \mcc_x/\mfm_y\mcc_x=0, \] where \(\mfm_y\mco_{X,x}\subseteq\mfm_x\). Nakayama’s lemma therefore gives \[ \mcc_x=0. \] Since this holds for every \(x\), the map \(\epsilon\) is surjective.
Both its source and target are invertible sheaves. Locally at \(x\), \(\epsilon_x\) is therefore a surjective endomorphism between free rank-one \(\mco_{X,x}\)-modules. It is multiplication by a generator of the unit ideal, hence by a unit. Thus \(\epsilon_x\) is an isomorphism for every \(x\), so \(\epsilon\) is an isomorphism globally.
5. There is an invertible sheaf \(\mcn\) on \(Y\) such that\[ \mcl\cong\mcm\tensor f^*\mcn. \]
By step 4, \[ \mcl\tensor\mcm^{-1} =\mcf \cong f^*\mcn. \] Tensoring by \(\mcm\) gives \[ \boxed{\mcl\cong\mcm\tensor f^*\mcn}. \]
6. Q.E.D.
Step 5 is the required conclusion.