Lemma 27.13.6 (Segre embedding). Let $S$ be a scheme. There exists a closed immersion

$\mathbf{P}^ n_ S \times _ S \mathbf{P}^ m_ S \longrightarrow \mathbf{P}^{nm + n + m}_ S$

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

$\mathcal{L} = \text{pr}_1^*\mathcal{O}_{\mathbf{P}^ n}(1) \otimes \text{pr}_2^*\mathcal{O}_{\mathbf{P}^ m}(1)$

The sections we take are

$s_0 = X_0Y_0, \ s_1 = X_1Y_0, \ldots , \ s_ n = X_ nY_0, \ s_{n + 1} = X_0Y_1, \ldots , \ s_{nm + n + m} = X_ nY_ m.$

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

$\varphi ^{-1}(D_{+}(Z_{i + (n + 1)j})) = D_{+}(X_ i) \times D_{+}(Y_ j).$

Hence it is an affine morphism. The corresponding ring map in case $(i, j) = (0, 0)$ is the map

$\mathbf{Z}[Z_1/Z_0, \ldots , Z_{nm + n + m}/Z_0] \longrightarrow \mathbf{Z}[X_1/X_0, \ldots , X_ n/X_0, Y_1/Y_0, \ldots , Y_ n/Y_0]$

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.) $\square$

Comment #5998 by Nick on

Something is wrong here, the coordinate $\frac{Z_{ab}}{Z_{0}}$ should go to $\frac{X_a}{X_0} \otimes \frac{Y_b}{Y_0}$.

Comment #6163 by on

OK, I think the text is fine as it is and what you say is true also (but this is besides the point).

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