Why can we require that the hyperdeterminant has integral coefficients?

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?

• "uniquely" only up to sign, right? (Good question.) – darij grinberg Oct 9 at 16:36
• $X$ and its dual hypersurface are defined over $\Bbb{Z}$. – abx Oct 9 at 16:49
• Yes, uniquely up to sign. $X$ and its dual surface are defined over $\mathbb{C}$. Thanks. – Sophie Oct 9 at 17:15
• I don't understand your comment: the projective space is defined over $\Bbb{Z}$, and so is the Segre embedding. – abx Oct 9 at 18:27
• 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. – Lev Soukhanov Oct 9 at 19:45