5
$\begingroup$

Given a projective variety $X$ over a field of any characteristic, consider a line bundle $\mathcal{L}$ over $X$.

  • The existence of a line bundle $\mathcal{L}^\prime$ with an isomorphism ${\mathcal{L}^\prime}^{\otimes 2} \simeq \mathcal{L}$ is equivalent to the existence of a simple cyclic cover $Y \rightarrow X$ of degree $2$ which is branched on a divisor $D$ associated to $\mathcal{L}$. Such a cyclic cover depends on the choice of the rational section $s$ of $\mathcal{L}$ such that $D = \text{div}(s)$. See section 3.3 of https://arxiv.org/abs/2009.01831v2 for a proof. In the related question divisors and powers of line bundles, Francesco Polizzi gives an example of a line bundle on a K3 surface which does not admit a square root.

  • There is also the root stack construction that gives a Deligne-Mumford stack $f : {}^{2}\sqrt{(X,\mathcal{L})} \rightarrow X$ which is the moduli space of square roots of $\mathcal{L}$.

I am wondering if there exists a finite surjective morphism of projective varieties $g : X^\prime \rightarrow X$ such that $g^*\mathcal{L}$ admits a square root over $X^\prime$. If such a map exists, it should factor through $f$.

Improve this question
$\endgroup$
1
  • $\begingroup$ Maybe as a remark: a possible alternative approach to Sasha's great answer is to note that the root stack ${}^{2}\sqrt{(X,\mathcal{L})}$ is quasifinite and proper over $X$. The one can use Theorem B in arxiv.org/pdf/0904.0227.pdf to get a finite surjective morphism $X' \to {}^{2}\sqrt{(X,\mathcal{L})}$ where $X'$ is a scheme. The composition $X' \to X$ is proper and quasifinite, so it is finite (+surjective). This yields the desired $X'$, and it works for any quasicompact scheme $X$. Of course this is probably an overkill. $\endgroup$
    – afh
    Nov 7, 2021 at 0:02

1 Answer 1

Reset to default
7
$\begingroup$

Assume $\mathcal{L}$ is associated with an effective Cartier divisor $D$. Let $D'$ be another Cartier divisor such that $D + D'$ is divisible by 2 in $\mathrm{Pic}(X)$. Let $$ g \colon X' \to X $$ be the double covering branched at $D + D'$. Then $g^{-1}(D) = 2R$ for a Cartier divisor $R$ on $X'$, hence $g^*\mathcal{L} \cong \mathcal{O}_{X'}(2R)$ has a square root.

If $\mathcal{L}$ is not associated with an effective divisor, you can replace $\mathcal{L}$ by $\mathcal{L} \otimes \mathcal{O}_X(2N)$ (where $\mathcal{O}_X(1)$ is an ample line bundle) with $N \gg 0$, so that this bundle is associated with an effective divisor, and apply the previous construction.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Just to be sure : in the effective case, is it sufficient to take simply $D^\prime = D$ ? and the resulting double cover $g$ will be singular at $g^{-1}(D)$ ? In general, if I take $D^\prime$ linearly equivalent to $D$ such that the intersection $D \cap D^\prime$ is transverse, the map $g$ will be singular at $g^{-1}(D \cap D^\prime)$ ? $\endgroup$
    – user158892
    Nov 7, 2021 at 7:40
  • 1
    $\begingroup$ If you take $D' = D$ the double covering will be reducible --- it will be the union of two components isomorphic to $X$ and intersecting along $D$. I am not sure whether this is fine for you. If $D$ intersects $D'$ then $X'$ is indeed singular over the intersection points, but you want $X'$ to be smooth (you didn't ask $X'$ to be smooth in the question) you can just resolve these singularities. $\endgroup$
    – Sasha
    Nov 7, 2021 at 9:11
  • $\begingroup$ I am trying to understand why $D^\prime = D$ yields a reducible scheme. The double cover is locally the zero locus of $t^2 - s^2$ in $X \times \mathbb{A}^1$ where $s \in H^0(X,\mathcal{L})$ defines $D$. As $t^2 - s^2$ is reducible, so is the cover. Is it correct ? $\endgroup$
    – user158892
    Nov 7, 2021 at 10:39
  • 1
    $\begingroup$ Yes, precisely, there are two components: $t - s$ and $t + s$. $\endgroup$
    – Sasha
    Nov 7, 2021 at 11:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.