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.

  • $\begingroup$ 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. $\endgroup$ – HARRY Nov 2 '18 at 14:08

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