# Square root of a line bundle up to a finite surjective morphism

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 $$f$$.

• 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.
– afh
Nov 7, 2021 at 0:02

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.
• 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 $g^{-1}(D \cap D^\prime)$ ? Nov 7, 2021 at 7:40
• 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. Nov 7, 2021 at 9:11
• 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 ? Nov 7, 2021 at 10:39
• Yes, precisely, there are two components: $t - s$ and $t + s$. Nov 7, 2021 at 11:18