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Let $Y$ be the abelian variety $\mathbb{C}/\mathbb{Z}[i] \times \mathbb{C}/\mathbb{Z}[i] $ where $\mathbb{Z}[i]$ denote the set of Gaussian integers in $\mathbb{C}$. Let $X$ be the quotient of $Y$ by action of the group generated by the map $\eta(x,y)=(ix,iy)$. This group generated is of order 4, and is given by $\{e, -e, \eta, -\eta\}$ where $e$ is the identity map.

How can we show that $X$ is in fact a rational surface and has 10 singularities? I have perused several resources in regarding Kummer surfaces and could not find any literature using such a construction.

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    $\begingroup$ This is just basic surface theory, nothing particular to do with Kummer surfaces. Just list the fixed points of $\eta$ and $\eta^2$ and look at the action of the group on them. $\endgroup$
    – abx
    Commented Jan 29, 2019 at 11:17
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    $\begingroup$ The fact that it's rational needs a justification too (probably Castelnuovo's criterion). One can do the same contruction in any dimension $n$ (take the quotient of $\mathbb C/\mathbb Z[i])^n$ by the diagonal action of $i$) and I believe it's unknown for what values of $n$ the quotient is a rational variety. $\endgroup$
    – user47305
    Commented Jan 30, 2019 at 15:36
  • $\begingroup$ @Mark Are there any papers giving a detailed study of this particular construction? I can't seem to find any. I can only find papers detailing constructions where they consider the quotient by an involution. $\endgroup$
    – Soby
    Commented Jan 30, 2019 at 16:50
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    $\begingroup$ Check out Catanase-Oguiso-Truong, "Unirationality of Ueno-Campana’s threefold", where they prove the three-dimensional one is unirational. You might also like Oguiso & Truong, "Explicit Examples of rational and Calabi-Yau threefolds with primitive automorphisms of positive entropy", which proves the three-dimensional one is rational if you use the curve with an order 6 auto instead. For the 2-diml one I don't know a reference offhand, but I bet you can figure it out from these other papers. $\endgroup$
    – user47305
    Commented Jan 30, 2019 at 16:57
  • $\begingroup$ @Mark Alright I will take a look at those papers. Thank you very much! $\endgroup$
    – Soby
    Commented Jan 31, 2019 at 3:18

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