Show that the image of the Segre embedding \[ \sigma:\PP^1_\CC\cross\PP^1_\CC\longrightarrow\PP^3_\CC \] is the smooth quadric \[ Q=V(x_0x_3-x_1x_2). \]
Solution
Write \[ \sigma\bigl([a_0:a_1],[b_0:b_1]\bigr) = [a_0b_0:a_0b_1:a_1b_0:a_1b_1]. \]
1. The image of \(\sigma\) is contained in \(Q\).
At a point of the image, \[ \begin{aligned} x_0x_3-x_1x_2 &=(a_0b_0)(a_1b_1)-(a_0b_1)(a_1b_0)\\ &=0. \end{aligned} \] Hence \[ \operatorname{im}\sigma\subseteq Q. \]
2. Every point of \(Q\) lies in the image of \(\sigma\).
Let \[ p=[x_0:x_1:x_2:x_3]\in Q. \] Form the matrix \[ M_p = \begin{pmatrix} x_0&x_1\\ x_2&x_3 \end{pmatrix}. \] Since \[ \det M_p=x_0x_3-x_1x_2=0, \] the matrix has rank at most \(1\). It is not the zero matrix because \(p\) is a projective point, so it has rank exactly \(1\).
Therefore there are nonzero vectors \[ \begin{pmatrix}a_0\\a_1\end{pmatrix}, \qquad \begin{pmatrix}b_0&b_1\end{pmatrix} \] such that \[ M_p = \begin{pmatrix}a_0\\a_1\end{pmatrix} \begin{pmatrix}b_0&b_1\end{pmatrix}. \] Thus \[ [x_0:x_1:x_2:x_3] = [a_0b_0:a_0b_1:a_1b_0:a_1b_1] \] and hence \[ p = \sigma\bigl([a_0:a_1],[b_0:b_1]\bigr). \] So \[ Q\subseteq\operatorname{im}\sigma. \]
3. The Segre image is exactly the quadric:\[ \boxed{ \operatorname{im}\sigma = V(x_0x_3-x_1x_2). } \]
Steps 1 and 2 give the two inclusions.
4. The quadric \(Q\) is smooth.
Put \[ F=x_0x_3-x_1x_2. \] Its partial derivatives are \[ \frac{\partial F}{\partial x_0}=x_3, \qquad \frac{\partial F}{\partial x_1}=-x_2, \qquad \frac{\partial F}{\partial x_2}=-x_1, \qquad \frac{\partial F}{\partial x_3}=x_0. \] If all four partial derivatives vanished at a point of \(Q\), then \[ x_0=x_1=x_2=x_3=0, \] which is impossible in projective space. Hence \(Q\) has no singular points. Therefore \(Q\) is smooth.
5. Q.E.D.
Step 3 identifies the image of the Segre embedding, and step 4 proves that this image is a smooth quadric surface.