Let $C$ be a curve of genus $g$. Let $\mathcal{L}$ be a line bundle on $C$ of degree 2n.

I would like to know a proof of the following statements:

To prove that there is a line bundle $\mathcal{M}$ on $C$ such that $\mathcal{M}^{2}= \mathcal{L}$.

To prove that there are $2^{2g}$ many such line bundle which are square roots of $\mathcal{L}$.

  • $\begingroup$ Where did you find these statements? Have you tried to prove them yourself? $\endgroup$ – Michael Albanese Jun 9 '18 at 11:24

We assume that $C$ is a projective algebraic curve over an algebraically closed field $k$ of char $0$ and genus $g$. We will state some facts which will be used in the proof. We will abuse notation and write $H^i$ for etale cohomology.

1.One knows by the theory of Jacobian varities that there exists an abelian variety $J(C)$ of dimension $g$ such that $J(C)(k)$ is the group of degree zeroline bundles on $C$.

2.On any abelian variety $X$, the map $n_X : X \rightarrow X$, that is the multiplication by $X$ is a finite surjective map on $k$, points with kernel isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{2g}$.

Consider the following commutative diagram $$ \begin{array} &0 &\rightarrow & Pic^0(C) & \rightarrow & Pic(C) & \xrightarrow{deg} & \mathbb{Z} & \rightarrow & 0\\ & & \downarrow && \downarrow && \downarrow \\ 0 &\rightarrow & Pic^0(C) & \rightarrow & Pic(C) & \xrightarrow{deg} & \mathbb{Z} & \rightarrow & 0 \end{array} $$

where the downwards arrow are multiplication by $n$.

Using fact 1&2, we get that multiplication by $n$ is surjective on $Pic^0(C)$. Thus using snake lemma we get that $Pic(C)/nPic(C) \cong \mathbb{Z}/n\mathbb{Z}$, where the isomorphism is given by the degree map. Thus if $n| deg(L)$, this implies that $L \in nPic(C)$. Let $M$ be such that $M^n = L$.

Let $\lbrace L_1, \dots, L_{n^{2g}} \rbrace = Pic^0(C)[n] = Ker(J(C)(k) \xrightarrow{\times n} J(C)(k))$. Clearly $M \otimes L_i$ are the $n^{2g}$ roots of order $n$ of the line bundle $L$.

One can use the above with $n = 2$, to get the specific case asked in the question above.

You may find these illuminating :



  • $\begingroup$ A splendid answer, both mathematically and pedagogically: bravo! And it seems that random users on this site are excellent :-) $\endgroup$ – Georges Elencwajg Jun 9 '18 at 16:03
  • $\begingroup$ I have taken the liberty of correcting a typo in your great answer: I hope you don't mind. $\endgroup$ – Georges Elencwajg Jun 9 '18 at 16:08
  • $\begingroup$ @GeorgesElencwajg Thank you for the kind words. $\endgroup$ – random123 Jun 10 '18 at 6:40

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