This is a proof from Section 7.1 of Undergraduate Algebraic Geometry by Reid.
Suppose $S\subset \mathbb{P}^3$ is a nonsingular cubic surface, given by a homogeneous cubic $f=f(x,y,z,t)$. Consider the intersection of a plane $\Pi\subset \mathbb{P}^3$ with $S$.
The author shows that a multiple line is impossible.
Let $\Pi: t=0$ and a line $l: z=0\subset \Pi$, then to say $l$ is a multiple line of $S\cap \Pi$ means that $f$ is of the form $$f=z^2\cdot A(x,y,z,t)+t\cdot B(x,y,z,t)$$
Then $S$ is singular at a point where $z=t=B=0$. I understand the proof so far. But then he states that this is a nonempty set, since it is the set of roots of $B$ on the line $l: z=t=0$.
I don't understand why it has to be nonempty. Is it necessary to assume the underlying field $k$ is algebraically closed? Otherwise why is it impossible that the quadratic form $B$ lies just above or below the $xy$-plane, so its intersection with $l$ is the origin which does not exist in $\mathbb{P}^3$?
Thank you for your help!
Yes, we need the field to be algebraically closed. Then it is clear. $B(x,y,0,0)$ is a homogenous polynomial in $x,y$. Such a polynomial certainly admits a root other than $(x=0,y=0)$.