Commutative Algebra 64

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.


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.


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. ♦

warningAs 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).


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.


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


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.


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.


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.


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. ♦


This entry was posted in Advanced Algebra and tagged , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s