Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 24F (Algebraic Geometry)

Because $k$ is algebraically closed, the smooth quadric surface $V$ has a $k$-point $p$. Choose a plane $\Pi \cong {\mathbb{P}}_k^2$ not containing $p$ and project from $p$:For a general point $q\in \Pi$, the line $\overline{pq}$ meets the degree-two surface $V$ in $p$ and exactly one further point. Sending $q$ to that second intersection is a rational map inverse to ${\pi }_p$ on dense open subsets. Therefore

Over an algebraically closed field of characteristic different from two, a projective change of coordinates puts every smooth quadric surface in the formThe Segre embedding identifies this surface with ${\mathbb{P}}^1\times {\mathbb{P}}^1$ byTwo distinct fibres of either coordinate projection are disjoint lines in one of the rulings of a smooth quadric surface. Explicitly,both lie on $V$ and have no common projective point. Thus $V$ contains a pair of disjoint lines.

Suppose an affine line lies in $W$. Parametrize it asThe defining equation would give the polynomial identityThe only units of $k[t]$ are nonzero constants. Since a product is a unit only when every factor is a unit, all three linear polynomials must be constant, so $b_1=b_2=b_3=0$. The alleged line is a point, a contradiction. Therefore