Let $f,g\in\mathbb{C}[x_1,\ldots,x_n]$ be homogeneous of degree $d>1$ such that $f$ is irreducible and no polynomial in the linear span of $f$ and $g$ is a square of another polynomial. Is it true that the variety $V(f,y^2-g)$ is irreducible (where $y$ is another variable)?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
No, take $f=b^4-a^3c$ and $g=ac^3.$ Now the variety $V(b^4-a^3c,y^2-ac^3)$ being reducible is equivalent to the ideal $(b^4-a^3c,y^2-ac^3)$ being prime. But we have $(ya-b^2c)(ya+b^2c)=y^2a^2-b^4c^2=a^2(y^2-ac^3)+c^2(a^3c-b^4),$ and $ya-b^2c,ya+b^2c\notin(b^4-a^3c,y^2-ac^3).$
On the other hand, if the variety defined by $f$ is normal, then the statement holds. This is because $V(f,y^2-g)$ is a double cover of $V(f)$, and a reducible double cover of an irreducible normal variety must have a section (because any birational finite morphism to an irreducible normal variety is the identity.) Pulling back the function $y^2$ along this section gives a perfect square which is a linear combination of $f$ and $g$.
The above example was produced by picking a non-normal $f$ and choosing $y$ to correspond to a function on the normalization that does not descend to $V(f).$ Note that the above surface can be parametrized by $(x,y,z)=(s^4,s^3t,t^4),$ and so corresponds to the ring $k[s^4,s^3t,t^4].$ This ring has normalization $k[s^4,s^3t,s^2t^2,st^3,t^4]$, and $y$ corresponds to the element $s^2t^6.$
