Question: Let $d \geq 4$. I'm interested by know if there is a surface $S$ of degree $d$ in $\mathbb P^3\_{\mathbb C}$ such that $S$ does not contains a line. I know 
I have no idea how to do it.

Answer: 
This is the famous Noether-Lefschetz theorem. The answer is that for a "very general" such surface -- meaning away from a countable union of proper closed subsets in the parameter space of all degree-d surfaces -- the only algebraic curves are complete intersections with other surfaces. In particular, there are no lines, no conics, etc. on such a surface. While this is true in general, Mumford in the 60s or maybe 70s gave a challenge to come up with a specific example of even one quartic surface not containing a line, a challenge that was not met until just a few years ago.
I think that

$$f(x,y,z) = x^4-3 x y+x^2 z+z^3+y^2+5$$

provides an example of a surface in ${\mathbb A}^3$ that does not contain a line in in ${\mathbb A}^3$.

Substitute $x=u\_0 + t p\_0$, $y = v\_0 + t q\_0$ and $z = w\_0 + t r\_0$ in $f(x,y,z)$.
Collect coefficients of powers of $t$:

$$C\_4 t^4 + C\_3 t^3 + C\_2 t^2 + C\_1 t + C\_0$$

where the $C\_i \in {\mathbb Q}[u\_0,v\_0,w\_0,p\_0,q\_0,r\_0]$

Now calculate three grobner bases of $\{C\_i\}\_{i=0..4} \cup \{h\}$ where $h$ is either $p\_0 - 1$, $q\_0 - 1$ or $r\_0 -1$. All three reduce to the unit ideal, proving that $V(f(x,y,z))$ does not contain a line in ${\mathbb A}^3$.