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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.

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.

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

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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  • $\begingroup$ I am unable to find this statement in my copy of Lange and Birkenhake, Complex abelian varieties, 1992 edition, and it does not seem to be in the 2nd edition either at least as searchable on amazon. Is it perhaps the other book, Complex Tori? or is there a difference in our books? $\endgroup$
    – roy smith
    Commented Jun 14, 2017 at 19:16
  • $\begingroup$ or maybe I am just not understanding what is there? $\endgroup$
    – roy smith
    Commented Jun 14, 2017 at 19:26
  • $\begingroup$ It is at p. 77 of Complex Abelian varieties, second edition (2003) $\endgroup$
    – Francesco Polizzi
    Commented Jun 14, 2017 at 19:32
  • $\begingroup$ Thank you Francesco. Unfortunately in the 1st edition this prop does not appear, so I am trying to figure out what purpose it serves. In my version Th. 3.1 the decomposition theorem is stated on p.78, and then a strengthening of Bertini's theorem, Th. 3.5 on p. 81. It sounds as if the prop you mention might be a lemma for this Bertini theorem. I am wondering whether the original proof was correct, and just which result lacks correct details. I.e. I wonder whether the flaw you observe was an attempt to improve or correct the original version. In my book section 4.3 is from pp 77 thru 82. $\endgroup$
    – roy smith
    Commented Jun 14, 2017 at 21:02

1 Answer 1

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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 $C\cap C'$.

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    $\begingroup$ For abelian varieties of any dimension, I think $f^*D$ will be irreducible if $D$ is normal. In fact I think all counterexamples to irreducibility are similar to the one you give, and so are singular in codimension 1. $\endgroup$
    – rita
    Commented Jun 14, 2017 at 19:40
  • $\begingroup$ If $D$ is normal, the same is true for every finite, étale cover of $D$. In particular, the divisor $f^*D$ cannot have singularities in codimension $1$, hence $D_i \cdot D_j=0$ for $i \neq j$ and the proof given in the book should work. $\endgroup$
    – Francesco Polizzi
    Commented Jun 15, 2017 at 7:46
  • $\begingroup$ Yes, I agree... $\endgroup$
    – abx
    Commented Jun 15, 2017 at 9:42
  • $\begingroup$ @abx: thank you for the counterexample. I was thinking about something similar involving a $(1, \, 2)$-polarization, but your construction is simpler. $\endgroup$
    – Francesco Polizzi
    Commented Jun 15, 2017 at 10:06

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