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

Choose projective embeddings $X\subseteq {\mathbb{P}}^n$ and $Y\subseteq {\mathbb{P}}^m$. Define their product as the Segre image of the set of pairs $(x,y)$, with its induced variety structure.

It is closed in the Segre variety. To see this directly, for every homogeneous $f\in I(X)$ impose the equations $f(z_{0j},\dots ,z_{nj})=0$ on every column, and for every homogeneous $g\in I(Y)$ impose $g(z_{i0},\dots ,z_{im})=0$ on every row. For a rank-one matrix $z_{ij}=x_i{y}_j$, the column equations are $y_j^{\mathrm{deg}f}f(x)=0$. A nonzero $y_j$ therefore forces $x\in X$. The row equations similarly force $y\in Y$. Thus these homogeneous equations together with the minors cut out exactly $X\times Y$, proving it is projective.

On a chart where $z_{i_0{j}_0}\ne 0$, the projections have formulasThese homogeneous coordinate tuples are nonzero on the chart and agree on overlaps, so both projections are morphisms.

Given morphisms $\alpha :Z\to X$ and $\beta :Z\to Y$, define $\gamma (z)={\sigma }_{n,m}(\alpha (z),\beta (z))$. Locally write their coordinates as $[a_0:\cdots :a_n]$ and $[b_0:\cdots :b_m]$. Then $\gamma$ has coordinates $[a_i{b}_j]$, which are regular with a nonzero coordinate near each point. Hence $\gamma$ is a morphism andThe two projections determine the underlying pair uniquely, so this morphism is unique. This is the product universal property, and it makes the construction independent up to canonical isomorphism of the chosen embeddings.