General surface no lines I've been studying surfaces recently and I came across the following statement: A general surface $S \subset \mathbb{P}^{3}$ of degree $m \geq4$ contains no lines. Does anyone have any idea how to prove this? or at least a sketch proof?
 A: Let $G$ be the (($2n-2$)-dimensional) Grassmannian $\mathbb G(1,n)$ of lines in $\mathbb P^n$, and let $\mathbb P^N$ be the space of all hypersurfaces $S\subset \mathbb P^n$ of degree $d$. Here, $N=\binom{n+d}{d}-1$. Look at the incidence correspondence
$$\mathcal Y=\{([S],[\ell])\in \mathbb P^N\times G\,|\,\ell\subset S\}\subset \mathbb P^N\times G,$$ and at the two projections
$$p:\mathcal Y\to \mathbb P^N\,\,\,\,\,\,\,\,\,\,\textrm{and}\,\,\,\,\,\,\,\,\,\,q:\mathcal Y\to G.$$
For every $[\ell]\in G$, we have $$q^{-1}([\ell])=\{[S]\in\mathbb P^N\,|\,\ell\subset S\}\cong \mathbb P^{N-d-1}.$$ Thus, $\mathcal Y$ is irreducible of dimension $$(2n-2)+(N-d-1)=N+2n-d-3.$$ Now, if $d>2n-3$, then $\dim \mathcal Y<N$, and then the general hypersurface $S\subset \mathbb P^n$ contains no lines.
In your situation, $n=3,d=4$, so $4=d>2n-3=3$.
Added: the reason why $\dim q^{-1}([\ell])=N-d-1$ is that the condition "a fixed line $\ell\cong\mathbb P^1$ lies on some surface $S$" is determined by $d+1$ linear conditions, where $d+1=h^0(\ell,\mathcal O_\ell(d))$. More precisely: there exists $S$ such that $\ell\subset S$ if and only if there exists a polynomial $f:=f_S\in H^0(\mathbb P^n,\mathcal O_{\mathbb P^n}(d))$ such that $f|_\ell=0$, where by $f|_\ell$ I mean the image of $f$ under the restriction map $\rho_\ell:H^0(\mathbb P^n,\mathcal O_{\mathbb P^n}(d))\to H^0(\ell,\mathcal O_\ell(d))$. We have $$\dim\ker \rho_\ell=(N+1)-(d+1),$$ but of course we consider that multiple polynomials give the same surface, so a $+1$ disappears in the displayed equation. So finally $$q^{-1}([\ell])=\mathbb P(\ker \rho_\ell)\cong \mathbb P^{N-d-1}.$$
