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This is in continuation to the question, Finite morphism of varieties - morphism of sheaves .

Suppose $f:X\rightarrow Y$ is a double cover of non-singular surfaces where $E\subset Y$ is the branch locus. Then by the answer https://math.stackexchange.com/a/2107414/52991, $f_*O_X=O_Y\oplus L$ where $L$ is a line bundle such that $L^{\otimes 2}=O_Y(-E)$.

Does the above mean that the line bundle $O_Y(−E)$ admits a square root? If we start with a line bundle $L′$ the Picard group which does not admit a root , can we not construct a double cover branched along some $E'$ where $O_Y(E')=L′$? For example consider $O(1)$ on $\mathbb{P}^2$, there is no line bundle $L$ on $\mathbb{P}^2$ such that $L^{\otimes 2}=O(1)$. Then can we not have a double cover of the projective plane branched along a hyperplane? Lazarsfeld's book on Positivity mentions "Bloch Gieseker's theorem" and "Kawamata theorem" which seem to say that this is possible?

Also since $E\subset Y$ is the branch locus, $f^*E$ is a non-reduced divisor of the form $2D\subset X$. Then $D=(2D)_{red}$ is isomorphic to E. Then it looks like $π^∗L^{−1}$ is $O_Y(D)$? Is this true?

I am trying to understand what this really means. I looked up the section on cyclic covers from Lazarsfeld's book and Robert Freidman's book on Algebraic surfaces. But I am still not very clear.

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If $f:X \to Y$ is a double cover with $Y$ smooth and $B \subset Y$ is its branch locus, then $O(-B)$ is indeed a square in $Pic(Y)$. In particular, there is no double cover of $\mathbb{P}^2$ branched over line.

If $f_*O_X \cong O_Y \oplus L$, then $f^*L \cong O_X(-R)$, where $R$ is the ramification divisor. Moreover, $f^{-1}(B) = 2R$.

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    doesn't the Kawamata covering say that, given a non-singular surface $Y$ and a divisor $D\subset Y$, then we can construct a covering $f:X\rightarrow Y$ and a smooth divisor $D'\subset X$ such that $f^*D=2D'$ for example. Doesn't this mean that $f$ is a double cover? That any surface admits a double cover. – 2017-01-23
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    You probably misunderstood something. Give me a precise reference. – 2017-01-23
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    Proposition 4.1.12, Lazarsfeld, Positivity in Algebraic Geometry - I, Page 245. – 2017-01-23
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    Lazarsfeld never claims that $D_i$ are the only components of the branch divisor. In your example, given $E \subset Y$ you can always add to it some $E'$ such that $E + E'$ is divisible by 2, and consider the corresponding double covering. – 2017-01-23
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    it says given any normal crossing divisor $\Sigma D_i\subset Y$, and integers $m_i$, we can find a cover $X$ such that $D_i'=m_i D_i$ where $D_i'$ is a smooth divisor on $X$. In particular, given an irreducible divisor $E\subset Y$, there is a cover $X$ and a smooth divisor $D'\subset X$ such that $f^*E=2D$ right? – 2017-01-23
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    It is very easy to construct a double cover brached over $E + E'$. But typically it will be singular in the intersection points of $E$ and $E'$. If you want to construct a smooth cover, you can first apply the trick (called Bloch--Gieseker covering by Lazarsfeld) to make $E$ divisible by 2. But then the degree of the resulting map is something like $2^{\dim Y + 1}$. – 2017-01-23
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    ah I will have a look at that theorem. Thanks! – 2017-01-23
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    I have been thinking about the Kawamata theorem that you helped me with. If $X$ is a non-singular projective variety, and $D$ is a irreducible smooth divisor. Then Kawamata says - for any $m>0$ we can construct a non-singular variety $p:Y\rightarrow X$ with a smooth divisor $D'$ of $Y$ such that $p^*D=mD'$. Will this divisor $D'$ be irreducible, I think it will be looking at the proof. Will that mean $D'\rightarrow D$ is an isomorphism? – 2017-06-10