Confusion in Hartshorne exercise IV.2.7

algebraic-curvesalgebraic-geometry

In page 306 of Hartshorne's book Algebraic Geometry, the exercise 2.7 asks me to show that,
for a given curve (in Hartshrone's sense, i.e., an irreducible complete nonsingular curve over an algebraically closed field) $Y$, there is a 1-1 correspondence between finite étale morphism $f\colon X\to Y$ of degree $2$, and $2$-torsion elements of $\mathrm{Pic}(Y)$.

However, in the text Hartshorne does not assume $X$ is connected (hence also a curve). If it is indeed not assumed to be connected, then

  1. How to define the degree of a finite étale morphism $f\colon X\to Y$? Maybe a reasonable way is to define $\deg f$ by the rank of $f_*\mathscr{O}_X$ as a locally free $\mathscr{O}_Y$-module via $f_\flat\colon\mathscr{O}_Y\to f_*\mathscr{O}_X$.
  2. In (a) of this exercise, Hartshorne uses the previous exercise IV.2.6(d) to show that $(\det f_*\mathscr{O}_X)^2\simeq\mathscr{O}_Y$. But exercise IV.2.6 only states for finite morphism $f$ of curves, how to solve this problem?

But if we assume $X$ is indeed a curve, then we just take $\mathscr{L}=\mathscr{O}_Y$ in (b) of this exercise, then it seems that $\mathrm{Spec}(\mathscr{O}_Y\oplus\mathscr{L})$ is not integral, because the multiplication
\begin{equation}
(a,b)\cdot(a',b')=(aa'+bb',ab'+a'b),
\end{equation}
on $\mathscr{O}_Y\oplus\mathscr{L}$ always satisfies that $(1,1)\cdot(1,-1)=(0,0)$.
Do I miss anything?

Edit: It seems that, if $X$ is not connected, then the étale morphism $f$ can be identified with the codiagonal $\nabla\colon Y\sqcup Y\to Y$, which is a trivial "étale covering". So we WLOG just consider the case $X$ is connected.

Best Answer

(1) You can define the degree to be the number of points (with multiplicity) in the preimage of a closed point of $Y$. You could also probably reduce to the connected case.

(2) If $f:X\to Y$ is an étale morphism of varieties, then $f$ is smooth of relative dimension $0$ (by definition). Consequently, if $Y$ is smooth then so is $X$. Hence, $X$ is smooth and $\Omega_{X|Y} = \Omega_X/f^*\Omega_Y$ is a locally free sheaf of rank $0$, i.e. the $0$ sheaf. However, for a smooth morphism $\dim X - \dim Y = \mathrm{rk}\:\Omega_{X|Y}$. In particular, $\dim X = \dim Y = 1$ here. Hence, $X$ and $Y$ are both curves.

Intuition: an étale morphism corresponds to a covering map in topology. A covering map (of manifolds) preserves dimension.

Related Question