The Segre embedding of $\PP^r \times \PP^s$

Let $k$ be algebraically closed and let $r,s\ge0$. Let $\psi: \PP^r \times \PP^s \to \PP^N$ be the map sending the ordered pair $\tv{a_0 : \cdots : a_r} \times \tv{b_0 : \cdots : b_s}$ to the point with coordinates $a_i b_j$ in lexicographic order, where $N = rs + r + s$. The map $\psi$ is well defined and injective; it is called the Segre embedding. Show that the image of $\psi$ is a subvariety of $\PP^N$.

Hint

Let the homogeneous coordinates of $\PP^N$ be $\ts{z_{ij} \st 0 \leq i \leq r,\ 0 \leq j \leq s}$, and let $\mfa$ be the kernel of the homomorphism $k[\ts{z_{ij}}] \to k[x_0,\ldots,x_r,y_0,\ldots,y_s]$ sending $z_{ij} \mapsto x_i y_j$. Then show $\im \psi = Z(\mfa)$.

Solution

Write $T=k[z_{ij}:0\le i\le r,\ 0\le j\le s]$ and $S=k[x_0,\ldots,x_r,y_0,\ldots,y_s]$, and let $\theta:T\to S$ be the substitution in the hint. Put $\mfa=\ker\theta$.

1. The map $\psi$ is well-defined and its image is contained in $Z(\mfa)$.

For nonzero vectors $a,b$, choose indices $i,j$ with $a_i\ne0$ and $b_j\ne0$. Then $a_ib_j\ne0$, so the matrix $(a_ib_j)$ gives a projective point. Replacing the two vectors by $\lambda a$ and $\mu b$ multiplies the entire matrix by $\lambda\mu$, leaving that point unchanged. This proves well-definedness. For every $F\in\mfa$, the polynomial $F((x_iy_j)_{ij})$ is identically zero. It therefore vanishes at every pair $a,b$, so the image satisfies all the equations of $\mfa$.

2. The ideal $\mfa$ is homogeneous and prime, and every point of $Z(\mfa)$ is in the image of $\psi$.

The map $\theta$ sends a homogeneous polynomial of degree $d$ to a homogeneous polynomial of degree $2d$, or to zero. Distinct source degrees have distinct target degrees, so the kernel is homogeneous. The quotient $T/\mfa\cong\im\theta$ is a subring of the domain $S$, hence is a domain. Thus $\mfa$ is prime.

It contains every two-by-two minor \[ z_{ij}z_{ab}-z_{ib}z_{aj}, \] since both monomials have image $x_ix_ay_jy_b$. Take a point $[c_{ij}]\in Z(\mfa)$ and choose a nonzero entry $c_{ab}$. The minor equations give \[ c_{ij}=\frac{c_{ib}c_{aj}}{c_{ab}}\qquad\text{for all }i,j. \] Set $u_i=c_{ib}$ and $v_j=c_{aj}/c_{ab}$. These vectors are nonzero because $u_a=c_{ab}\ne0$ and $v_b=1$. The displayed equations say $c_{ij}=u_iv_j$, so $[c_{ij}]=\psi([u],[v])$. This proves the reverse inclusion, and hence $\im\psi=Z(\mfa)$.

3. The image is a projective variety, and the recovered factors are unique.

The zero set $Z(\mfa)$ is nonempty by step 1, since both projective factors contain points. The homogeneous prime correspondence in P-AGH24CORRESPONDENCE therefore makes it irreducible and closed in $\PP^N$. It is consequently a projective subvariety, as required.

If $[c_{ij}]=\psi([u],[v])$ and $c_{ab}\ne0$, then its $b$th column is a nonzero scalar multiple of $u$, and its $a$th row is a nonzero scalar multiple of $v$. Thus those column and row determine $[u]$ and $[v]$ uniquely. Equivalently, on this chart the inverse coordinates are $u_i/u_a=c_{ib}/c_{ab}$ and $v_j/v_b=c_{aj}/c_{ab}$. This verifies the stated injectivity and identifies the inverse on every nonzero-entry chart. The description also includes $r=0$ or $s=0$, when every nonzero matrix automatically has rank one.

4. Q.E.D.

Steps <1>1–<1>2 identify the image with the zero set from the hint, and step 3 proves it is a projective variety.

Source Collections

Hartshorne, Algebraic Geometry, I.2.14

Backlinks

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