# A problem regarding the defining polynomial of a projective hypersurface

I was actually trying to find out how the polynomial defining a degree $$5$$ projective hypersurface looks like or atleast what can be said about the polynomial provided the hypersurface satisfies some conditions.After reducing algebraic geometry conditions on hypersurface to a statement regarding polynomials it turns out as follows.

Let, $$f(x,y,z,w)$$ be an irreducible homogeneous degree $$5$$ polynomial in $$K[x,y,z,w]$$, such that,

$$1)$$ $$f(x,y,z,$$0$$)=g(x,y,z).h^2(x,y,z)$$ ,Where $$g$$ is a degree $$1$$ homogeneous polynomial in $$K[x,y,z]$$ and $$h$$ is a degree $$2$$ homogeneous irreducible polynomial in $$K[x,y,z]$$ . AND

$$2)$$ If we set $$Z$$={[$$a_0$$:$$a_1$$:$$a_2$$]$$\in P^2$$|$$h^2$$($$a_0$$,$$a_1$$,$$a_2$$)=$$0$$} ,then all four partial derivatives of $$f(x,y,z,w)$$ evaluated at all points of $$Z$$ vanishes.

Then how will $$f(x,y,z,w)$$ will look in $$K[x,y,z,w]$$ ?

I guess that it will look like $$f(x,y,z,w)=G(x,y,z,w).H^2(x,y,z,w)+T(x,y,z,w)$$, where $$G(x,y,z,0)=g(x,y,z)$$ and $$H(x,y,z,0)=h(x,y,z)$$ and $$T(x,y,z,w)$$ is a homogeneous irreducible polynomial of degree $$5$$ where presence of the variable $$w$$ cannot be denied.

But,I believe that more concrete expression can be written down for $$f$$ satisfying this two conditions.I do not know what techniques to use to find more concrete answer.

Any help from anyone is welcome.

• I find it to be a bit contradictory if we say that for every open set inside $z$ all partial derivative vanishes means it is constant there .But then it cannot be irreducible. – HARRY Nov 2 '18 at 14:08