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
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Comment #5998 by Nick on
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