In the (wonderful) book by C. Birkenhake and H. Lange *Complex Abelian Varieties* we can find the following result, see Corollary 4.3.4 page 77. It is stated in any dimension $g \geq 2$, but let us consider only the case of abelian surfaces for the sake of simplicity.



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> **Proposition.** Let $f \colon X \to Y$ be an isogeny of abelian surfaces and let $D$ be a positive definite and irreducible divisor on $Y$. Then $f^\*D$ is also irreducible.
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The proof starts as follows: 


*Assume the contrary, then $f^\*D$ is a sum of effective divisors $D\_1+ \cdots + D\_n$. But necessarily $D\_i \cdot D\_j=0$ and $D\_i$ is numerically equivalent to $D\_j$ for all $i \neq j$, the map $f$ being an étale Galois covering $\ldots$* 


and then a contradiction is reached by using the Nakai-Moishezon theorem.


Now, I do not understand the part $D\_i \cdot D\_j=0$ for all $i \neq j$. 


If $D$ is smooth then this is immediate: in fact, an étale cover of a smooth curve must be smooth, in particular its irreducible components do not intersect. 


However, if $D$ is singular (and in the Proposition there is no smoothness assumption) the same argument do not work. Indeed, it is well-known that there are irreducible, $1$-nodal curves with a connected, étale double covering consisting of two copies of the normalization, see for instance example 10.6 in Chapter III of Hartshorne's *Algebraic Geometry*. The two components intersect at two points, that are the two nodes of the covering.


So my question is



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> **Q.** Is the result stated in the Proposition above also true when $D$ is singular? If not, what is a counterexample? Or maybe am I missing something? 
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Pull-back of an irreducible ample divisor via an isogeny of abelian varieties
I think the Proposition is not true if $D$ is singular. Take a smooth curve $C$ of genus 2, and $X=JC$; embed $C$ in $X$ (say, by choosing a point of $C$). Let $\alpha$ be a point of order 2 in $X$; take for $Y$ the quotient of $X$ by the translation $x\mapsto x+\alpha $, and put $D=f(C)$. Then $f^\*D=C+C'$, with $C':=C+\alpha $, and $C\cdot C'=C^2=2$. Here $D$ has one singular point, image of the two points of $C\cap C'$. 


