A nonalgebraizable invertible sheaf on a formal completion of the plane

Since \(Y\) is an integral projective curve over the algebraically closed field \(k\), \[ H^0(Y,\mco_Y)=k. \] For \(r\ge2\) there is an exact sequence \[ 0 \longrightarrow \mci^{r-1}/\mci^r \longrightarrow \mco_{X_r} \longrightarrow \mco_{X_{r-1}} \longrightarrow0. \] Because \(Y\) is a hypersurface of degree \(d\), \[ \mci^{r-1}/\mci^r \cong \mco_Y(-(r-1)d). \] This line bundle has negative degree \[ -(r-1)d^2<0, \] so it has no nonzero global section. Hence \[ H^0(X_r,\mco_{X_r}) \hookrightarrow H^0(X_{r-1},\mco_{X_{r-1}}). \] The constants give a copy of \(k\) in every term, so induction from \(X_1=Y\) forces equality with \(k\) at every stage.

The immersion \[ X_r\hookrightarrow X_{r+1} \] is a square-zero thickening whose ideal is \[ \mci^r/\mci^{r+1} \cong \mco_Y(-rd). \] The Picard sequence for a square-zero thickening (Stacks Project, Tag 0C6R) contains \[ H^0(X_{r+1},\mco_{X_{r+1}}^\times) \longrightarrow H^0(X_r,\mco_{X_r}^\times) \longrightarrow H^1\bigl(Y,\mco_Y(-rd)\bigr) \] \[ \longrightarrow \Pic X_{r+1} \longrightarrow \Pic X_r \longrightarrow H^2\bigl(Y,\mco_Y(-rd)\bigr). \] By step 1 the first two groups of units are both \(k^\times\), and their map is the identity. Since \(Y\) has dimension \(1\), \[ H^2\bigl(Y,\mco_Y(-rd)\bigr)=0. \] Exactness therefore gives the displayed short exact sequence.

Let \(F\) be the homogeneous equation of \(Y\). Twisting the hypersurface sequence by \(-rd\) gives \[ 0 \longrightarrow \mco_{\PP^2}(-(r+1)d) \xrightarrow{\cdot F} \mco_{\PP^2}(-rd) \longrightarrow \mco_Y(-rd) \longrightarrow0. \] By the cohomology calculation on projective space, \[ H^1(\PP^2,\mco(q))=0 \] for every \(q\), and \[ \dim_kH^2(\PP^2,\mco(-m)) = \binom{m-1}{2} \] for \(m\ge3\). Since a curve has no \(H^2\) for a quasicoherent sheaf, the long exact sequence gives, for \(r\gg0\), \[ 0 \longrightarrow V_r \longrightarrow H^2\bigl(\PP^2,\mco(-(r+1)d)\bigr) \longrightarrow H^2\bigl(\PP^2,\mco(-rd)\bigr) \longrightarrow0. \] Therefore \[ \begin{aligned} \dim_kV_r &= \binom{(r+1)d-1}{2} - \binom{rd-1}{2}\\ &= rd^2+\frac{d(d-3)}2. \end{aligned} \] This tends to infinity with \(r\). In particular \(V_r\ne0\) for all sufficiently large \(r\), so the filtration in step 3 has infinitely many strict inclusions.

Let \[ \hat\mci=\ker(\mco_{\hat X}\to\mco_Y). \] The kernel of \[ \mco_{\hat X}^\times\longrightarrow\mco_Y^\times \] is the sheaf of principal formal units \[ 1+\hat\mci. \] In characteristic zero the convergent \(I\)-adic series \[ \log(1+a) = \sum_{m\ge1}(-1)^{m+1}\frac{a^m}{m} \] and \[ \exp(a) = \sum_{m\ge0}\frac{a^m}{m!} \] are inverse isomorphisms of sheaves of abelian groups \[ 1+\hat\mci \xrightarrow{\sim} \hat\mci. \] Because the right-hand side is a sheaf of \(k\)-vector spaces, this transports a \(k\)-vector-space structure to \(1+\hat\mci\) and hence to its cohomology.

Step 1 gives \[ H^0(\hat X,\mco_{\hat X})=k, \] so \[ H^0(\hat X,\mco_{\hat X}^\times) \longrightarrow H^0(Y,\mco_Y^\times) \] is the isomorphism \(k^\times\to k^\times\). The long exact sequence for \[ 1 \longrightarrow 1+\hat\mci \longrightarrow \mco_{\hat X}^\times \longrightarrow \mco_Y^\times \longrightarrow1 \] therefore identifies \[ K \cong H^1(\hat X,1+\hat\mci) \cong H^1(\hat X,\hat\mci) \] as a \(k\)-vector space.

If this vector space were finite-dimensional, it could not contain the infinite strictly descending chain \[ K=K_1\supsetneq K_2\supsetneq\cdots \] obtained in steps <1>3–<1>4. Hence \[ \boxed{\dim_kK=\infty}. \] This proves the vector-space assertion in part (a) when \(\operatorname{char}k=0\).

Take \[ Y=\{z=0\}\subseteq\PP_k^2. \] Thus \(Y\cong\PP_k^1\). Cover \(Y\) by the standard opens where \(x\ne0\) and \(y\ne0\), and take the corresponding formal opens in \(\hat X\). On their intersection, \[ u = 1+\frac{z^2}{xy} \] is a formal unit congruent to \(1\) modulo the ideal of \(Y\). With two opens there is no triple-overlap condition, so \(u\) is a Cech transition function for an invertible sheaf \(\mathfrak L\) on \(\hat X\) whose restriction to \(Y\) is trivial.

Its leading term \[ \frac{z^2}{xy} \] represents the standard nonzero Cech class \[ x^{-1}y^{-1} \in H^1(\PP^1,\mco_{\PP^1}(-2)), \] so \(\mathfrak L\) is nontrivial. In characteristic \(p\), \[ u^p = 1+\frac{z^{2p}}{x^py^p}. \] If \(\mathfrak L^{\otimes p}\) were trivial, its first nonzero term would be a Cech coboundary in \[ H^1(\PP^1,\mco_{\PP^1}(-2p)). \] But that term is \[ x^{-p}y^{-p}, \] which is one of the standard nonzero basis monomials in \(H^1(\PP^1,\mco(-2p))\) by T-IJW1K. Therefore \[ \mathfrak L^{\otimes p}\not\cong\mco_{\hat X}. \]

The additive group of a \(k\)-vector space in characteristic \(p\) is killed by \(p\). Since \(K\) contains the class of \(\mathfrak L\) and that class is not killed by \(p\), the Picard-group law on \(K\) cannot make it a \(k\)-vector space. The correct characteristic-free statement is the filtered statement of steps <1>2–<1>4.

By steps <1>3–<1>4, \(K\) is nonzero in every characteristic. Choose a nontrivial class \[ [\mathfrak L]\in K. \] Thus \[ \mathfrak L|_Y\cong\mco_Y. \] Suppose that \(\mathfrak L\) were algebraizable. Then there would be a coherent sheaf \(\mcg\) on \(X\) with \[ \hat\mcg\cong\mathfrak L. \]

The local argument in Exercise III.11.6, steps <1>7--<1>8 applies to this single algebraization: faithfully flat completion shows that \(\mcg\) is locally free of rank one on a neighbourhood of \(Y\), and its reflexive hull \[ \mcl=\mcg^{\vee\vee} \] is a rank-one reflexive sheaf on \(X=\PP_k^2\). Since projective space is regular and locally factorial, \(\mcl\) is invertible; moreover \(\mcl\) agrees with \(\mcg\) near \(Y\), so \[ \hat\mcl\cong\mathfrak L. \]

But \[ \Pic\PP_k^2\cong\ZZ \] with generator \(\mco_X(1)\), so \[ \mcl\cong\mco_X(m) \] for some \(m\in\ZZ\). Restricting to \(Y\) gives \[ \mco_Y(m) \cong \mathfrak L|_Y \cong \mco_Y. \] Now \[ \deg\mco_Y(m)=md. \] Since \(d>0\), triviality forces \(m=0\). Hence \[ \mcl\cong\mco_X \] and therefore \[ \mathfrak L\cong\mco_{\hat X}, \] contrary to the choice of \([\mathfrak L]\ne0\).

Thus \(\mathfrak L\) is not algebraizable, proving (b).

Condition (i) of Exercise III.11.6(b) says that every locally free sheaf on \(\hat X\) is algebraizable. Step 7 gives an invertible, hence locally free, counterexample. Therefore condition (i) is false.

Exercise III.11.6(b) proves that condition (i) is equivalent to condition (ii). Hence condition (ii) is false. Consequently there is a locally free sheaf \(\mathfrak F\) on \(\hat X\) for which no integer \(n_0\) has the property that \[ \mathfrak F(n) \] is generated by global sections for every \(n\ge n_0\).

Suppose some twist \(\mathfrak F(n_1)\) were generated by global sections. For every \(r\ge0\), the sheaf \[ \mco_{\hat X}(r) \] is generated by global sections because it is the restriction of the globally generated sheaf \(\mco_{\PP^2}(r)\). Tensor products of globally generated sheaves are globally generated, so \[ \mathfrak F(n_1+r) = \mathfrak F(n_1)\otimes\mco_{\hat X}(r) \] would be globally generated for every \(r\ge0\). Taking \(n_0=n_1\) would then satisfy condition (ii) for \(\mathfrak F\), a contradiction.

Therefore \[ \boxed{\mathfrak F(n)\text{ is not generated by global sections for every } n\in\ZZ}. \] This is the phenomenon announced in Hartshorne II.9.9.1.

Steps <1>2–<1>4 prove the characteristic-free surjectivity and infinite filtered kernel in (a), step 5 gives the stated infinite-dimensional vector-space conclusion in characteristic zero, and step 6 records the necessary correction in positive characteristic. Step 7 proves (b), and step 8 proves (c).