# Segre Embedding

Throughout this article, k is a fixed algebraically closed field. We wish to construct the product in the category of quasi-projective varieties.

For our first example, let $V\subset \mathbb P^3_k$ be the projective variety defined by the homogeneous equation $T_0 T_3 - T_1 T_2 = 0$. We define maps $\pi_1, \pi_2 : V\to \mathbb P^1_k$ as follows

$\pi_1 : (t_0 : t_1 : t_2 : t_3) \mapsto \begin{cases} (t_0 : t_2), \text{ if } (t_0, t_2) \ne (0, 0), \\ (t_1 : t_3), \text{ if } (t_1, t_3) \ne (0, 0),\end{cases} \\ \pi_2 : (t_0 : t_1 : t_2 : t_3) \mapsto \begin{cases} (t_0 : t_1), \text{ if } (t_0, t_1) \ne (0, 0), \\ (t_2 : t_3), \text{ if } (t_2, t_3) \ne (0, 0).\end{cases}$

Note that the maps are well-defined: if $(t_0, t_2), (t_1, t_3) \ne (0, 0)$ then since $t_0 t_3 = t_1 t_2$ we have $(t_0 : t_2) = (t_1 : t_3)$.

Proposition 1.

The triplet $(V, \pi_1, \pi_2)$ is a product in the category of quasi-projective varieties.

Proof

Let $W\subseteq \mathbb P^n$ be a quasi-projective variety and $\psi_1, \psi_2 : W \to \mathbb P^1$ be morphisms. We will define the corresponding $f : W \to V$ as follows. For each $\mathbf w\in W$, there is an open neighbourhood U of w such that $\psi_1|_U = (F_0 : F_1)$ and $\psi_2|_U = (G_0 : G_1)$ where $F_0, F_1 \in k[T_0, \ldots, T_n]$ are homogeneous of the same degree and either $F_0$ or $F_1$ has no zero in U. Same holds for $G_0, G_1 \in k[T_0, \ldots, T_n]$.

Now define $f : U \to \mathbb P^3$ by $(F_0 G_0 : F_0 G_1 : F_1 G_0 : F_1 G_1)$. Clearly the image of f lies in V so we get a morphism $f: U \to V$. It is easy to see that $\pi_1|_U \circ f = \psi_1|_U$ and $\pi_2|_U \circ f = \psi_2|_U$. Repeating this construction over an open cover of W, we obtain our desired $f:W \to V$. ♦

Using similar techniques, we can show the following.

Proposition 2.

For any $m, n \ge 0$, the product $\mathbb P^n \times \mathbb P^m$ exists in the category of quasi-projective varieties and is a projective variety.

Specifically, the product is the image of the Segre embedding

$\mathbb P^n \times \mathbb P^m \to \mathbb P^{mn + n + m}, \quad (a_0 : \ldots : a_n), (b_0 : \ldots : b_m) \mapsto (a_i b_j)_{0\le i \le n, 0\le j \le m}$,

where the projective coordinates of $\mathbb P^{mn + n + m}$ are indxed by $(i, j)$ with $0\le i \le n$ and $0\le j \le m$.

We denote the image of this map by $\mathbb P^{n, m}$.

Exercise A

Prove that $\mathbb P^{n, m}$ is the closed subspace of $\mathbb P^{mn + n + m}$ defined by

$T_{ij}T_{kl} - T_{il} T_{kj}$ over all $(i, j), (k, l) \in \{0, \ldots, n\} \times \{0 ,\dots, m \}$

# Products of Quasi-Projective Varieties

Proposition 3.

If $W_1 \subseteq \mathbb P^n$ and $W_2\subseteq \mathbb P^m$ are open (resp. closed), so is the image of $W_1\times W_2$ in $\mathbb P^{n,m}$.

In particular, the topology on $\mathbb P^n \times \mathbb P^m$ is at least as fine as the product topology.

Proof

It suffices to prove the case where $W_1\subseteq \mathbb P^n$ and $W_2\subseteq \mathbb P^m$ are open. Pick any $\mathbf w_1 = (a_0 : \ldots : a_n) \in W_1$ and $\mathbf w_2 = (b_0 : \ldots : b_m) \in W_2$; without loss of generality say $a_0, b_0 \ne 0$.

Since $W_1$ is open in $\mathbb P^n$ there exists a homogeneous $F \in k[A_0, \ldots, A_n]$ such that $\mathbf w_1 \in D(F) \subseteq W_1$. Similarly, there exists a homogeneous $G \in k[B_0, \ldots, B_m]$ such that $\mathbf w_2 \in D(G) \subseteq W_2$. Then

$(\mathbf w_1, \mathbf w_2) \in \overbrace{D(A_0 F) \times D(B_0 G)}^{\subseteq W_1 \times W_2} \stackrel \cong \longrightarrow \overbrace{D(T_{00}F(T_{00}, \ldots, T_{n0})G(T_{00}, \ldots, T_{0m})) \cap V}^{\text{open in } W}$

so the image of $W_1 \times W_2$ in V is open. ♦

As in the product of affine varieties, the topology on $\mathbb P^n \times \mathbb P^m$ is in general strictly finer than the product topology. This is already clear in the case mn = 1, since $\mathbb P^1$ has the cofinite topology.

Corollary 1.

The product of two projective (resp. quasi-projective) varieties exists and is projective (resp. quasi-projective).

Note

In the following proof, we say a subset of a topological space is locally closed if it is an intersection of an open subset and a closed subset. Thus every quasi-projective variety (resp. quasi-affine variety) is a locally closed subspace of some $\mathbb P^n_k$ (resp. $\mathbb A^n_k$).

Prove the following properties as a simple exercise:

• an intersection of two locally closed subsets is locally closed;
• if Y is a locally closed subset of X and Z is a locally closed subset of Y then Z is a locally closed subset of X;
• a subset Y of X is locally closed if and only if Y is open in its closure in X.

Proof

If $W_1 \subseteq \mathbb P^n$ and $W_2 \subseteq \mathbb P^m$ are closed (resp. locally closed), so is the image W of $W_1 \times W_2$ in $\mathbb P^{n, m}$ by proposition 3. The projections $\mathbb P^{n,m} \to\mathbb P^n$ and $\mathbb P^{n,m}\to \mathbb P^m$ then restrict to $\pi_1: W \to W_1$ and $\pi_2 : W\to W_2$.

Let  us show that $(W, \pi_1, \pi_2)$ is the product of $W_1$ and $W_2$ in the category of quasi-projective varieties.

If X is any quasi-projective variety and $\psi_1 : X\to W_1$, $\psi_2 : X\to W_2$ are any morphisms then $\psi_1 : X \to \mathbb P^n$ and $\psi_2 : X \to \mathbb P^m$ induce $f : X\to \mathbb P^{n,m}$; the image of f lies in W so we obtain an induced $X\to W$. ♦

Exercise B

1. Let $W \subset \mathbb P^2 \times \mathbb P^1$ be the set of points $((a_0 : a_1 : a_2), (b_0 : b_1))$ satisfying $a_0^2 b_0 - a_1 a_2 b_1 = 0$. Find a set of homogeneous polynomials in $\mathbb P^{2, 1} \subset \mathbb P^5$ which define the image of W.

2. More generally prove that a subset $V \subseteq \mathbb P^{n, m}$ is closed if and only if its corresponding subset $V' \subseteq \mathbb P^n \times \mathbb P^m$ is the set of solutions of some bihomogeneous polynomials

$F(T_0, \ldots, T_n; U_0, \ldots, U_m) = 0$,

i.e. F is homogeneous as a polynomial in $T_0, \ldots, T_n$ as well as $U_0, \ldots, U_m$

# Dimensions

Lemma 1.

For any point $\mathbf v$ in a quasi-projective variety V, there is an open neighbourhood U, $\mathbf v \in U \subseteq V$, which is affine.

Proof

Suppose $V\subseteq \mathbb P^n_k$ is a locally closed subset. Without loss of generality, $\mathbf v \in U_0$ so $\mathbf v$ is contained in $W := U_0 \cap V$, a locally closed subset of $\mathbb A^n$. Now W is open in $\overline W$, its closure in $\mathbb A^n$. By an analogue of proposition 1 here, we can pick a basis of the topological space $\overline W$ in the form of $\{D(f) : f\in k[\overline W]\}$, where

$D(f) = \{ \mathbf w \in \overline W : f(\mathbf w) \ne 0\}.$

Thus for some $f\in k[\overline W]$ we have $\mathbf v \in D(f)\subseteq W$. Now we are done since $D(f)$ is isomorphic to the affine variety with coordinate ring $k[\overline W][T]/(T\cdot f - 1)$. ♦

Exercise C

Prove that if V and W are irreducible quasi-projective varieties, then $V\times W$ is also irreducible. Again, please be reminded that $V\times W$ is not the product topology.

Proposition 4.

if V and W are quasi-projective varieties, then

$\dim (V \times W) = \dim V + \dim W$.

Proof

Suppose V and W are irreducible; by lemma 1 we can pick open affine subsets $U_1 \subseteq V$ and $U_2 \subseteq W$. Then $U_1 \times U_2$ is an open affine subset of the quasi-projective variety $V\times W$, which is irreducible by exercise C. Now

$\dim (V \times W) = \dim (U_1 \times U_2) = \dim U_1 + \dim U_2 = \dim V + \dim W$.

where the first and third equalities follow from proposition 3 here and the second is from proposition 2 here

The general case is left as an exercise (write V and W as unions of irreducible components). ♦

Finally, we consider the dimension of the cone of a projective variety.

Proposition 5.

Let $V\subseteq \mathbb P^n$ be a non-empty closed subset. Then

$\dim (\mathrm{cone} V) = \dim V + 1$.

Proof

Without loss of generality, suppose $V' := U_0 \cap V \ne \emptyset$; by proposition 3 here, $\dim V = \dim V'$ and V’ is a closed subset of $\mathbb A^n$. Also since cone(V’) is open in cone(V) we have $\dim (\mathrm{cone} V) = \dim(\mathrm{cone} V')$. Now there is an isomorphism

$V' \times (\mathbb A^1 - \{0\}) \longrightarrow \mathrm{cone}(V') - \{\mathbf 0\}, \quad ((1 : t_1 : \ldots : t_n), \lambda) \mapsto (\lambda, \lambda t_1, \ldots, \lambda t_n).$

Hence $\dim(\mathrm{cone} V') = \dim V' + \dim(\mathbb A^1 - \{0\}) = \dim V' + 1$. ♦

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