i just wanted to know some necessary and sufficient conditions



condition for the existence of lines on degree four hypersurface in P^3..
Here's how to obtain such a condition in principle. Doing it in practice will require a computer algebra system, and may need a lot of computer power:


The generic line in $\mathbb{P}^3$ can be parameterized as $(x:y:ax+by:cx+dy)$ for some $(a,b,c,d)$. More specifically, this is the dense Schubert cell in $G(2,4)$. Let your degree $4$ polynomial be $F$. Write
$$F(x,y,ax+by,cx+dy) = A(a,b,c,d) x^4 + B(a,b,c,d) x^3 y + \cdots + E(a,b,c,d) y^4.$$


So $A$, $B$, $C$, $D$ and $E$ are $5$ polynomials in $4$ variables, each non-homogenous with maximal degree $4$. Let their resultant be $R$. (If you haven't heard of multivariate resultants, skip to the end.) Then $R=0$ is a necessary condition for $F$ to contain a line. There will most likely be some factor $Q$ of $R$ which is a necessary and sufficient condition, but I don't immediately see a better way to obtain $Q$ than by factoring $R$ and trying points on each irreducible component.


Why does this happen? Well, if there is a line $\ell$ in $F=0$, and $\ell$ lies in the dense cell of $G(2,4)$, then $(A,B,C,D,E)$ have a common root so $R$ vanishes. If we have a pair $(F, \ell)$, but $\ell$ is not in the dense cell, then let $g(t)$ be a generic path in $PGL\_4$, with $g(0) = \mathrm{Id}$. Then $g(t) \ell$ is in the dense cell for $t \neq 0$, and the preceeding argument shows that $R$ vanishes on $g(t)^\* F$. By continuity, $R$ vanishes on $F$.


Why isn't this argument invertible? The vanishing of $R$ is equivalent to the homogenized versions of $(A,B,C,D,E)$ having a common root in $\mathbb{P}^4$. But what we want to know is whether the hypersurfaces they define have a common root in $G(2,4)$. There is a birational map $G(2,4)$ to $\mathbb{P}^4$, but it is not an isomorphism, so the hypersurfaces might meet in $\mathbb{P}^4$ but become disjoint in $G(2,4)$. I don't know for sure whether this happens in this case, but it often does. The good thing is that we suspect that the actual locus we are hunting for is codimension $1$, so we can find it as a factor of the polynomial $R$.


There is a computation we should do before this one, because it will help check for errors and tell us what to expect. We should compute what we expect the degrees of $Q$ and $R$ to be. **EDIT:** In my pother answer, I show that $Q$ should have degree $320$.


Finally, I wanted to point you to a quick online reference for computing resultants of multi-variate polynomials, but all the sources I checked only had the one variable case (or, equivalently, two variables with homogenous polynomials). The famous reference here is the book of Gelfand, Kapranov and Zelevinsky, but that is quite an undertaking. The one thing I will do is alert you to the Cayley trick: The resultant of $(A,B,C,D,E)$ is the same as the discriminant of 
$$A(a,b,c,d) + B(a,b,c,d) w + \cdots + E(a,b,c,d) z$$
where $(w,x,y,z)$ are auxilliary variables. So you can compute a disciminant in $8$ variables instead of a resultant in $4$, if your software prefers that.


