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Let $X=P^{k_1}\times \ldots \times P^{k_r}$ be the product of several complex projective spaces ($P^k$ is the projectivization of $\mathbb{C}^{k+1}$) in the Segre embedding into $P=P^{(k_1+1)\cdots(k_r+1)-1}$. The $X$-discriminant is the equation of the projective dual variety of $X$ in $P^*$. In the book of Gelfand, Kapranov and Zelevinsky the hyperdeterminant Det (of format $(k_1+1)\times \ldots \times (k_r+1)$) is defined to be the $X$-discriminant, which is a homogeneous polynomial. Now they say that this is determined uniquely (up to sign) by the requirement that Det has integral coefficients and is irreducible over $\mathbb{Z}$.

Why is this so? Or more precisely, why can we require that it has integral coeffients?

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  • $\begingroup$ "uniquely" only up to sign, right? (Good question.) $\endgroup$ – darij grinberg Oct 9 at 16:36
  • $\begingroup$ $X$ and its dual hypersurface are defined over $\Bbb{Z}$. $\endgroup$ – abx Oct 9 at 16:49
  • $\begingroup$ Yes, uniquely up to sign. $X$ and its dual surface are defined over $\mathbb{C}$. Thanks. $\endgroup$ – Sophie Oct 9 at 17:15
  • $\begingroup$ I don't understand your comment: the projective space is defined over $\Bbb{Z}$, and so is the Segre embedding. $\endgroup$ – abx Oct 9 at 18:27
  • $\begingroup$ The question is non-trivial and boils down to the elimination of variables - how do you know that the projection of something defined over $\mathbb{Z}$ is also defined over $\mathbb{Z}$. It is true, yet not automatically clear. $\endgroup$ – Lev Soukhanov Oct 9 at 19:45

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