Lemma 27.13.6 (Segre embedding). Let $S$ be a scheme. There exists a closed immersion
called the Segre embedding.
Lemma 27.13.6 (Segre embedding). Let $S$ be a scheme. There exists a closed immersion
called the Segre embedding.
Proof. It suffices to prove this when $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Hence we will drop the index $S$ and work in the absolute setting. Write $\mathbf{P}^ n = \text{Proj}(\mathbf{Z}[X_0, \ldots , X_ n])$, $\mathbf{P}^ m = \text{Proj}(\mathbf{Z}[Y_0, \ldots , Y_ m])$, and $\mathbf{P}^{nm + n + m} = \text{Proj}(\mathbf{Z}[Z_0, \ldots , Z_{nm + n + m}])$. In order to map into $\mathbf{P}^{nm + n + m}$ we have to write down an invertible sheaf $\mathcal{L}$ on the left hand side and $(n + 1)(m + 1)$ sections $s_ i$ which generate it. See Lemma 27.13.1. The invertible sheaf we take is
The sections we take are
These generate $\mathcal{L}$ since the sections $X_ i$ generate $\mathcal{O}_{\mathbf{P}^ n}(1)$ and the sections $Y_ j$ generate $\mathcal{O}_{\mathbf{P}^ m}(1)$. The induced morphism $\varphi $ has the property that
Hence it is an affine morphism. The corresponding ring map in case $(i, j) = (0, 0)$ is the map
which maps $Z_ i/Z_0$ to the element $X_ i/X_0$ for $i \leq n$ and the element $Z_{(n + 1)j}/Z_0$ to the element $Y_ j/Y_0$. Hence it is surjective. A similar argument works for the other affine open subsets. Hence the morphism $\varphi $ is a closed immersion (see Schemes, Lemma 26.4.2 and Example 26.8.1.) $\square$
Comments (2)
Comment #5998 by Nick on
Comment #6163 by Johan on