Does every hypersurface in the projective plane contain a projective line?

Question

Consider $P^{2}(\mathbb{C})$, the space of all lines through the origin in $\mathbb{C}^{3}$ (or $\mathbb{R}^3$ if that works better). Let $X\subset P^{2}(\mathbb{C})$ be a (nonempty) hypersurface (algebraic in some coordinates). Must it be the case that for some $2$-dimensional subspace $V$ of $\mathbb{C}^3$, $P^1(V)\subset X$?

More generally, given $1<m<n$, must $n-1$-dimensional hypersurfaces in $P^{n}(\mathbb{C})$ (for $n\geq 2$) contain a $P^{m-1}(V)$ for some $m$-dimensional subspace $V\subset \mathbb{C}^{n+1}$?

I'm not a geometer (set theorist in unfamiliar waters here), so I don't have much of a feeling for these questions or the techniques involved, but my understanding is that one can show (using cohomology or Schubert calculus) that every hypersurface intersects every projective line.

Answer 1

Q1: No. Indeed it is easy to see that each two-dimensional subspace is defined by some linear equation, so any homogeneous polynomial which has no linear factors provides a counterexample, e.g. $x^2+y^2+z^2$.

Q2: No. You can easily count the dimension of the space of all hypersurfaces and the dimension of the space of hypersurfaces that contain a given line, as functions of the degree $d$. The dimension of the hyper surfaces that contain any line is the second number, plus the dimension of the space of lines. But the difference between the first two numbers can easily be seen to equal the dimension of the space of degree $d$ polynomials in two variables, so for $d$ large enough it is larger than the dimension of the space of lines. Because there are hyper surfaces with no lines, they can't contain any higher-dimensional subspaces.

You don't need to use cohomology or Schubert calculus to show this, it follow from the fundamental theorem of algebra. You have a polynomial of degree $d$ in $n+1$ variables. Intersecting with a line is just throwing away all but $2$ variables. It is still a polynomial of degree $d$, so it has roots.