Let \(Y\) be a hypersurface in \(X = \PP_k^N\) with \(N \geq 4\). Let \(\hat{X}\) be the formal completion of \(X\) along \(Y\) (II, §9). Prove that the natural map \[ \Pic \hat{X} \to \Pic Y \] is an isomorphism.
Hint: use (II, Ex. 9.6), and then study the maps \(\Pic X_{n+1} \to \Pic X_n\) for each \(n\) using (Ex. 4.6) and (Ex. 5.5).
Solution
Let \[ d=\deg Y \] and let \[ \mci=\mci_Y\subseteq\mco_X \] be the ideal sheaf of \(Y\). Since \(Y\) is a hypersurface of degree \(d\), \[ \mci\cong\mco_X(-d). \] For \(n\ge1\), put \[ X_n=\bigl(Y,\mco_X/\mci^n\bigr). \] Thus \(X_1=Y\), and the formal completion \(\hat X\) is the inverse system of the infinitesimal neighbourhoods \(X_n\).
1. The formal Picard group is the inverse limit\[ \Pic\hat X\cong\varprojlim_n\Pic X_n. \]
Each \(X_n\) is the closed subscheme of the projective scheme \(X=\PP_k^N\) defined by \(\mci^n\), hence is projective over the common field \(k\). Therefore the Mittag–Leffler hypothesis in Exercise II.9.6(d) holds for the inverse system \[ \Gamma(X_n,\mco_{X_n}). \] Exercise II.9.6 then gives the canonical isomorphism \[ \boxed{\Pic\hat X\cong\varprojlim_n\Pic X_n}. \] Under this isomorphism, restriction to \(Y=X_1\) is the projection to the first factor.
2. For every \(n\ge1\), the closed immersion\[ X_n\hookrightarrow X_{n+1} \]is a square-zero thickening with ideal\[ \mci^n/\mci^{n+1}\cong\mco_Y(-nd). \]
Its ideal in \[ \mco_{X_{n+1}}=\mco_X/\mci^{n+1} \] is \[ \mci^n/\mci^{n+1}. \] Its square is zero because \[ \mci^{2n}\subseteq\mci^{n+1} \] for \(n\ge1\).
Moreover multiplication by \(\mci\) annihilates \(\mci^n/\mci^{n+1}\), so this ideal is naturally an \(\mco_Y\)-module. Since \(\mci\) is invertible, \[ \mci^n/\mci^{n+1} \cong \mci^n\otimes_{\mco_X}\mco_Y. \] Using \[ \mci^n\cong\mco_X(-nd) \] and restricting to \(Y\) gives \[ \boxed{\mci^n/\mci^{n+1}\cong\mco_Y(-nd)}. \]
3. For every \(n\ge1\),\[ H^1\bigl(Y,\mco_Y(-nd)\bigr) = H^2\bigl(Y,\mco_Y(-nd)\bigr) =0. \]
The hypersurface \(Y\subseteq\PP_k^N\) is a complete intersection of dimension \[ \dim Y=N-1\ge3. \] By Exercise III.5.5(c), \[ H^i(Y,\mco_Y(m))=0 \] for every \(m\in\ZZ\) and every \[ 0<i<\dim Y. \] Taking \(m=-nd\), the values \(i=1,2\) both lie in this range because \(\dim Y\ge3\). This proves the two vanishings.
4. For every \(n\ge1\), restriction induces an isomorphism\[ \Pic X_{n+1}\xrightarrow{\sim}\Pic X_n. \]
2. The relevant part of its exactsequence is\[ H^1\bigl(Y,\mci^n/\mci^{n+1}\bigr) \longrightarrow \Pic X_{n+1} \longrightarrow \Pic X_n \longrightarrow H^2\bigl(Y,\mci^n/\mci^{n+1}\bigr). \]By steps <1>2–<1>3, both outer groups are zero. Hence the middle restrictionmap is both injective and surjective:\[ \boxed{\Pic X_{n+1}\xrightarrow{\sim}\Pic X_n}. \]Equivalently, every line bundle on \(X_n\) has a unique isomorphism class oflift to \(X_{n+1}\).
5. Projection to the first term induces an isomorphism\[ \varprojlim_n\Pic X_n\xrightarrow{\sim}\Pic Y. \]
By step 4 every transition map in the inverse system \[ \cdots\longrightarrow\Pic X_3 \longrightarrow\Pic X_2 \longrightarrow\Pic X_1 \] is an isomorphism. Therefore a compatible sequence of classes is uniquely determined by its first term, and every class in \[ \Pic X_1=\Pic Y \] extends uniquely through all the inverse transitions. Hence the first projection is an isomorphism.
6. Q.E.D.
Combining steps 1 and 5 gives \[ \Pic\hat X \xrightarrow{\sim} \varprojlim_n\Pic X_n \xrightarrow{\sim} \Pic X_1 = \Pic Y. \] The composite is precisely the natural restriction map in the statement. Thus \[ \boxed{\Pic\hat X\xrightarrow{\sim}\Pic Y}. \]