Past exam of the algebraic Geometry course of the University of Oxford / 2011 / 2 / b / Solution

The Segre embedding isRescaling the two representatives multiplies every target coordinate by the same scalar, and at least one product is nonzero, so the map is well defined. In affine source charts its coordinate ratios are regular products, so it is a morphism.

Arrange target coordinates into a matrix $Z=(z_{ij})$. Every image matrix has rank one and satisfies all homogeneous equationsConversely, a nonzero matrix satisfying these equations has rank one. Choose $z_{i_0{j}_0}\ne 0$ and set $x_i=z_{i{j}_0}$, $y_j=z_{i_{0}j}/z_{i_0{j}_0}$. The minors imply $z_{ij}=x_i{y}_j$. Thereforea projective variety. The two projective factors are uniquely recoverable from a nonzero column and row. This also shows injectivity and supplies the local inverse formulas used in the next part.