Let \(Y\) be a curve over a field \(k\) of characteristic \(\neq 2\). We show there is a one-to-one correspondence between finite étale morphisms \(f: X \to Y\) of degree 2, and 2-torsion elements of \(\Pic Y\), i.e., invertible sheaves \(\mcl\) on \(Y\) with \(\mcl^2 \cong \OO_Y\).
Given an étale morphism \(f: X \to Y\) of degree 2, there is a natural map \(\OO_Y \to f_* \OO_X\). Let \(\mcl\) be the cokernel. Then \(\mcl\) is an invertible sheaf on \(Y\), \(\mcl \cong \det f_* \OO_X\), and so \(\mcl^2 \cong \OO_Y\) by (Ex. 2.6). Thus an étale cover of degree 2 determines a 2-torsion element in \(\Pic Y\).
Conversely, given a 2-torsion element \(\mcl\) in \(\Pic Y\), define an \(\OO_Y\dash\)algebra structure on \(\OO_Y \oplus \mcl\) by \(\langle a, b\rangle \cdot\left\langle a', b'\right\rangle=\left\langle a a'+\varphi(b \tensor b'), a b'+a' b\right\rangle\), where \(\varphi\) is an isomorphism of \(\mcl \tensor \mcl \to \OO_Y\). Then take \(X=\Spec(\OO_Y \oplus \mcl)\) (II, Ex. 5.17). Show that \(X\) is an étale cover of \(Y\).
Show that these two processes are inverse to each other. Hint: Let \(\tau: X \to X\) be the involution which interchanges the points of each fibre of \(f\). Use the trace map \(a \mapsto a+\tau(a)\) from \(f_* \OO_X \to \OO_Y\) to show that the sequence of \(\OO_Y\dash\)modules in a. is split exact: \[ 0 \to \OO_Y \to f_* \OO_X \to \mcl \to 0 \]
Note. This is a special case of the more general fact that for \((n, \characteristic k)=1\), the étale Galois covers of \(Y\) with group \(\ZZ/n\ZZ\) are classified by the étale cohomology group \(H_{\text{et}}^1(Y, \ZZ/n\ZZ)\), which is equal to the group of \(n\)-torsion points of \(\Pic Y\). See Serre.
Solution
The correspondence below is between isomorphism classes of degree-\(2\) finite étale covers over \(Y\) and \(2\)-torsion classes in \(\Pic Y\).
1. Let\[ f:X\longrightarrow Y \]be finite étale of degree \(2\), and put\[ A=f_*\OO_X. \]Then the unit inclusion\[ \OO_Y\longrightarrow A \]is split as a map of \(\OO_Y\)-modules, and its cokernel \(\mcl\) is invertible.
The finite étale algebra \(A\) is locally free of rank \(2\). Its trace map \[ \Tr_{A/\OO_Y}:A\longrightarrow\OO_Y \] satisfies \[ \Tr(1)=2. \] Because \(\operatorname{char}k\ne2\), multiplication by \(1/2\) gives a retraction \[ \frac12\Tr:A\longrightarrow\OO_Y \] of the unit inclusion. Hence \[ A\cong\OO_Y\oplus M, \qquad M=\ker\Tr. \] Since \(A\) has rank \(2\), \(M\) is locally free of rank \(1\). The quotient \[ \mcl=A/\OO_Y \] is naturally isomorphic to \(M\), and is therefore invertible.
2. The invertible sheaf from step 1 satisfies\[ \mcl\cong\det f_*\OO_X \qquad\text{and}\qquad \mcl^{\tensor2}\cong\OO_Y. \]
The split exact sequence \[ 0 \longrightarrow \OO_Y \longrightarrow A \longrightarrow \mcl \longrightarrow0 \] gives \[ \det A \cong \det\OO_Y\tensor\det\mcl \cong \mcl. \]
Because \(f\) is étale, its ramification divisor and hence its branch divisor are zero. Applying Exercise IV.2.6, part (d), gives \[ (\det A)^2\cong\OO_Y. \] Thus \[ \mcl^{\tensor2}\cong\OO_Y. \] This proves part (a).
3. Conversely, let \(\mcl\) be invertible and choose an isomorphism\[ \varphi:\mcl\tensor\mcl\xrightarrow{\sim}\OO_Y. \]The multiplication in the statement makes\[ A_\varphi=\OO_Y\oplus\mcl \]into a commutative finite locally free \(\OO_Y\)-algebra of rank \(2\).
The unit is \((1,0)\). Commutativity is immediate from symmetry of the tensor product of line bundles. Associativity can be checked after locally trivializing \(\mcl\). On an open set where \(\mcl=\OO_Ye\), write \[ \varphi(e\tensor e)=u\in\OO_Y^\times. \] Then the map sending a formal variable \(t\) to \((0,e)\) identifies the algebra with \[ A_\varphi \cong \OO_Y[t]/(t^2-u). \] This is manifestly an associative commutative algebra, free with basis \(1,t\). The local descriptions glue because they came from the globally defined multiplication in the statement.
4. The morphism\[ f_\varphi:X_\varphi=\Spec_Y(A_\varphi)\longrightarrow Y \]is finite étale of degree \(2\).
Finiteness and degree \(2\) follow from the fact that \(A_\varphi\) is locally free of rank \(2\). In the local description of step 3, \[ A_\varphi=R[t]/(t^2-u), \qquad u\in R^\times. \] The element \(t\) is a unit because \(t^2=u\), and \(2\) is a unit because the characteristic is not \(2\). Therefore \[ 2t\in A_\varphi^\times. \] The relative differentials are \[ \Omega_{A_\varphi/R} \cong A_\varphi\,dt/(2t\,dt) =0. \] Thus the finite flat morphism \(f_\varphi\) is unramified, hence étale. This proves part (b).
5. For a quadratic étale cover \(f:X\to Y\), the trace-zero summand\[ M=\ker\Tr\subseteq A=f_*\OO_X \]is the \((-1)\)-eigensheaf for the nontrivial deck involution \(\tau\), andmultiplication induces an isomorphism\[ M\tensor M\xrightarrow{\sim}\OO_Y. \]
The deck involution interchanges the two points of each geometric fibre. On the rank-\(2\) algebra it satisfies \[ \Tr(a)=a+\tau(a). \] Hence an element of \(M\) obeys \[ \tau(a)=-a, \] while \(\OO_Y\cdot1\) is the \((+1)\)-eigensheaf. If \(a,b\in M\), then \[ \tau(ab)=\tau(a)\tau(b)=ab, \] so multiplication lands in the invariant summand \(\OO_Y\) and gives a map \[ M^{\tensor2}\longrightarrow\OO_Y. \]
This map is an isomorphism fibrewise. Indeed, over a geometric point the étale quadratic algebra is \[ k\times k, \] the involution swaps the two factors, and the trace-zero line consists of \((a,-a)\). Multiplication sends \[ (a,-a)(b,-b)=(ab,ab), \] which is a nonzero perfect pairing on that one-dimensional line. A map between line bundles that is an isomorphism on every fibre is an isomorphism.
6. Starting with a degree-\(2\) finite étale cover, applying part (a) andthen the construction of part (b) recovers the original cover up toisomorphism over \(Y\) once the multiplication pairing is used as the chosentrivialization.
By step 1 and step 5, \[ A=f_*\OO_X \cong \OO_Y\oplus M \] as an \(\OO_Y\)-module, and the algebra multiplication is completely described by \[ M\tensor M\xrightarrow{\sim}\OO_Y: \] products involving the first summand are scalar multiplication, while the product of two trace-zero elements is exactly this pairing. Therefore the algebra reconstructed in part (b) is isomorphic to \(A\) as an \(\OO_Y\)-algebra. Relative Spec then recovers \(X\) and its morphism to \(Y\).
7. Starting with a \(2\)-torsion line bundle \(\mcl\), the cover constructedin part (b) returns the same class \([\mcl]\in\Pic Y\) under part (a).
For \[ A_\varphi=\OO_Y\oplus\mcl, \] the unit map is inclusion of the first summand. Its cokernel is therefore exactly \(\mcl\). Thus part (a) recovers the original line-bundle class.
8. The isomorphism class of the cover constructed from \(\mcl\) isindependent of the chosen trivialization\[ \varphi:\mcl^{\tensor2}\xrightarrow{\sim}\OO_Y. \]
Because \(Y\) is a projective integral curve over the algebraically closed field \(k\), \[ H^0(Y,\OO_Y)^\times=k^\times. \] Thus any two trivializations satisfy \[ \varphi'=c\varphi \] for some \(c\in k^\times\). Choose \(\lambda\in k^\times\) with \[ \lambda^2c=1, \] which is possible because \(k\) is algebraically closed. Then \[ \OO_Y\oplus\mcl \longrightarrow \OO_Y\oplus\mcl, \qquad (a,b)\longmapsto(a,\lambda b) \] is an isomorphism from the algebra defined by \(\varphi\) to the algebra defined by \(\varphi'\). Hence the associated covers are isomorphic over \(Y\).
The same argument shows that replacing \(\mcl\) by an isomorphic line bundle does not change the cover. Therefore the construction depends only on the class of \(\mcl\) in \(\Pic Y\).
9. The two constructions give mutually inverse bijections\[ \left\{ \begin{array}{c} \text{degree-$2$ finite étale covers of $Y$}\\ \text{up to isomorphism over $Y$} \end{array} \right\} \longleftrightarrow \Pic(Y)[2]. \]
Step 6 proves that cover \(\to\) line bundle \(\to\) cover is the identity on isomorphism classes. Steps <1>7–<1>8 prove that line bundle \(\to\) cover \(\to\) line bundle is the identity on \(2\)-torsion classes and that the cover does not depend on auxiliary choices. Thus the two maps are inverse.
10. Q.E.D.
Steps <1>1–<1>2 prove part (a), steps <1>3–<1>4 prove part (b), and steps <1>5–<1>9 prove part (c).