Let \(C\) be a nonsingular affine curve. Show that two locally free sheaves \(\mathcal{E}, \mathcal{E}^{\prime}\) of the same rank are isomorphic if and only if their classes in the Grothendieck group \(K(X)\) (II, Ex. 6.10) and (II, Ex. 6.11) are the same. This is false for a projective curve.
Solution
Write \[ C=\Spec A. \] Because \(C\) is a nonsingular affine curve, \(A\) is a noetherian integrally closed domain of dimension one; hence \(A\) is a Dedekind domain.
1. Locally free sheaves of finite rank on \(C\) are equivalent to finitelygenerated projective \(A\)-modules.
For an affine scheme, the functor \[ M\longmapsto\widetilde M \] identifies finitely generated projective \(A\)-modules with locally free coherent sheaves on \(\Spec A\). Under this correspondence, direct sums, exterior powers, ranks, and isomorphisms are preserved.
2. If\[ \mathcal E\cong\mathcal E', \]then\[ [\mathcal E]=[\mathcal E'] \]in \(K(C)\).
The Grothendieck group \(K(C)\) is generated by isomorphism classes of coherent sheaves subject to the short-exact-sequence relations P-AGH2610KGROUP. Isomorphic coherent sheaves define the same generator, so the implication is immediate.
3. If\[ [\mathcal E]=[\mathcal E'] \]in \(K(C)\), then\[ \det\mathcal E\cong\det\mathcal E'. \]
Hartshorne II.6.11, proved on P-AGH2611KCURVE, constructs a homomorphism \[ \det:K(C)\longrightarrow\Pic(C) \] whose value on a locally free sheaf is its usual top exterior power. Applying this homomorphism to the assumed equality of Grothendieck classes gives \[ [\det\mathcal E]=[\det\mathcal E'] \] in \(\Pic(C)\), hence the two determinant line bundles are isomorphic.
4. Let \(P\) be a finitely generated projective module of rank \(r\ge1\)over a Dedekind domain \(A\). Then\[ \boxed{P\cong A^{r-1}\oplus\det P.} \]
This is the Steinitz classification of finitely generated projective modules over a Dedekind domain. Equivalently, every such module is a direct sum of rank-one projective modules \[ P\cong I_1\oplus\cdots\oplus I_r, \] and the standard ideal identity \[ I\oplus J\cong A\oplus IJ \] allows one to combine the summands successively: \[ P \cong A^{r-1}\oplus(I_1\cdots I_r). \] The final rank-one projective module is the top exterior power \[ I_1\cdots I_r\cong\bigwedge^rP=\det P. \]
5. If \(\mathcal E\) and \(\mathcal E'\) have the same rank and the sameclass in \(K(C)\), then\[ \boxed{\mathcal E\cong\mathcal E'.} \]
Let \(P\) and \(P'\) be the finite projective \(A\)-modules corresponding to \(\mathcal E\) and \(\mathcal E'\) by step 1, and let their common rank be \(r\).
Step 3 gives \[ \det P\cong\det P'. \] Applying step 4 twice gives \[ P \cong A^{r-1}\oplus\det P \cong A^{r-1}\oplus\det P' \cong P'. \] Sheafifying this module isomorphism gives \[ \mathcal E\cong\mathcal E'. \] Together with step 2, this proves the asserted equivalence on a nonsingular affine curve.
6. The assertion fails on the projective curve \(\PP^1\).
Fix a closed point \(P\in\PP^1\). The standard exact sequences \[ 0 \longrightarrow \OO_{\PP^1}(-1) \longrightarrow \OO_{\PP^1} \longrightarrow k(P) \longrightarrow0 \] and \[ 0 \longrightarrow \OO_{\PP^1} \longrightarrow \OO_{\PP^1}(1) \longrightarrow k(P) \longrightarrow0 \] give in \(K(\PP^1)\) \[ [\OO(-1)] =[\OO]-[k(P)] \] and \[ [\OO(1)] =[\OO]+[k(P)]. \] Hence \[ [\OO(1)\oplus\OO(-1)] =2[\OO] =[\OO\oplus\OO]. \] The two bundles have the same rank, namely \(2\), and the same Grothendieck class.
They are not isomorphic. After twisting by \(\OO(-1)\), \[ H^0\bigl(\PP^1,(\OO\oplus\OO)(-1)\bigr) =0, \] whereas \[ H^0\bigl(\PP^1,(\OO(1)\oplus\OO(-1))(-1)\bigr) = H^0(\PP^1,\OO\oplus\OO(-2)) \cong k. \] Thus the original bundles cannot be isomorphic. This proves the stated projective counterexample.
7. Q.E.D.
Steps <1>2–<1>5 prove the equivalence for a nonsingular affine curve, and step 6 proves that it fails for projective curves.