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 $f$.

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

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