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