Lemma 27.13.6 (01WD): 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 March 22, 2021 at 14:44

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 Johan on May 01, 2021 at 21:41

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

Comment #10924 by Elías Guisado on October 20, 2025 at 08:55

For anyone interested, here I wrote a detailed proof of the equality $\varphi^{-1}(D_{+}(Z_{i + (n + 1)j})) = D_{+}(X_i) \times D_{+}(Y_j).$

Let's add more details to the assertion “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 $Y_j/Y_0$ .”

Lemma. Let $X=\operatorname{Spec}A$ and $Y=\operatorname{Spec}B$ be affine schemes over $S=\operatorname{Spec}R$ , and write $Z=X\times_SY$ . Then $\widetilde{M\otimes_RN}=\pi^*_X\widetilde{M}\otimes_{\mathcal{O}_Z}\pi^*_Y\widetilde{N}$ as $\mathcal{O}_Z$ -modules, where $X\xleftarrow{\pi_X}X\times_SY\xrightarrow{\pi_Y}Y$ are the projections.

Proof. Denote $C=A\otimes_RB$ . By 26.7.3, 26.7.2, it follows from $\begin{align} (M\otimes_AC)\otimes_C(C\otimes_BN) &=M\otimes_A(C\otimes_C(C\otimes_BN))\\ &=M\otimes_A([A\otimes_RB]\otimes_BN)\\ &=M\otimes_A(A\otimes_R[B\otimes_BN])\\ &=M\otimes(A\otimes_RN)\\ &=(M\otimes_AA)\otimes_RN\\ &=M\otimes_RN. &&&\square \end{align}$ By the lemma, and since pullbacks commute with restrictions, we deduce that $\mathcal{L}_{|D_+(X_i)\times D_+(Y_j)} =(\mathbf{Z}[X_0,\dots,X_n](1)_{(X_i)}\otimes_{\mathbf{Z}} \mathbf{Z}[Y_0,\dots,Y_m](1)_{(Y_j)})\tilde{}.$

In general, $\mathcal{L}_{|D_+(X_i)\times D_+(Y_j)}^{\otimes a} =(\mathbf{Z}[X_0,\dots,X_n](a)_{(X_i)}\otimes_{\mathbf{Z}} \mathbf{Z}[Y_0,\dots,Y_m](a)_{(Y_j)})\tilde{},$ by 27.10.3.

I will tweak the proof notation a little bit. We will write $Z_{i,j}=Z_{i+(n+1)j}$ and $s_{i,j}=s_{i+(n+1)j}=\operatorname{pr}_1^*(X_i)\otimes\operatorname{pr}_2^*(Y_j)$ .

Therefore, the map $(-)/s_{i,j}^a:\mathcal{L}^{\otimes a}_{|D_+(X_i)\times D_+(Y_j)}\to\mathcal{O}_{D_+(X_i)\times D_+(Y_j)}$ from the proof of 27.12.1 corresponds to the map $\begin{align*} (-)/s_{i,j}^a:\mathbf{Z}[X_0,\dots,X_n](a)_{(X_i)} \otimes_{\mathbf{Z}} \mathbf{Z}[Y_0,\dots,Y_m](a)_{(Y_j)} &\to\mathbf{Z}[X_0,\dots,X_n]_{(X_i)}\otimes_{\mathbf{Z}} \mathbf{Z}[Y_0,\dots,Y_m]_{(Y_j)}\\ \frac{f}{X_i^k}\otimes\frac{g}{Y_j^\ell} &\mapsto \frac{f}{X_i^{k+a}}\otimes\frac{g}{Y_j^{\ell+a}}. \end{align*}$ Thus, and with notations from the proof of 27.12.1, the map $\begin{align*} \psi:\mathbf{Z}[Z_{00},\dots,Z_{(n+1)(m+1)}] &\to\Gamma_*(Y,\mathcal{L})\\ Z_{ij}&\mapsto s_{ij}\in\Gamma_1(Y,\mathcal{L})=\Gamma(Y,\mathcal{L}) \end{align*}$ gives rise to a map $\begin{align*} \psi_{(Z_{ij})}:\mathbf{Z}[Z_{00}/Z_{ij},\dots,Z_{(n+1)(m+1)}/Z_{ij}] &\to\mathbf{Z}[X_0/X_i,\dots,X_n/X_i]\otimes_{\mathbf{Z}} \mathbf{Z}[Y_0/Y_j,\dots,Y_m/Y_j]\\ \frac{Z_{i'j'}}{Z_{ij}} &\mapsto \frac{X_{i'}\otimes Y_{j'}}{s_{ij}} =\frac{X_{i'}}{X_i}\otimes\frac{Y_{j'}}{Y_j}. \end{align*}$

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