Consider the homogeneus polynomial \begin{equation}F=-4y_0y_2^3+18y_0y_1y_2y_3 +y_1^2y_2^2-4y_1^3y_3-27y_0^2y_3^2\in\mathbb{C}[y_0,y_1,y_2,y_3]\end{equation} which is the discriminant of a general cubic polynomial. It defines a hypersurface $D:=V(F)\subset\mathbb{P}^3$ that contains the curve $C$ given parametrically as the image of the morphism \begin{align*} \mathbb{P}^1 &\to\mathbb{P}^3 \\ [x:y] &\mapsto [x^3:3x^2y:3xy^2:y^3]. \end{align*} One can check that $C$ is precisely the set of singular points of $D$. In particular, $D$ is not isomorphic to $\mathbb{P}^2$. My question is:

Does $D$ have a normalization (a normal variety $X$ and a finite birational isomorphism $X\to D$)? Is this normalization smooth? Is it isomorphic to $\mathbb{P}^2$?


Every variety has a normalization. In the particular case of $D$ a normalization can be described as follows.

Consider the space $D'$ of pairs $(f,x)$, where $f \in \mathbb{P}^3$ is a cubic polynomial (up to a scalar) and $x \in \mathbb{P}^1$ is a point, such that $x$ is a root of $f$ of multiplicity at least 2. Clearly, $D'$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$ (one can associate to $(f,x)$ the point $x$ and the linear polynomial $f/l_x^2$, where $l_x$ is a linear polynomial with root at $x$).

On the other hand, the projection $D' \to \mathbb{P}^3$, $(f,x) \mapsto f$, gives a finite birational map $D' \to D$. Since $D'$ is smooth, it is a normalization of $D$.

  • $\begingroup$ Dear @Sasha, isn't your map $D'\to D$ one-to-one? Because given $f\in D$ there is only one point $x\in\mathbb{P}^1$ such that $x$ is a root of $f$ of multiplicity at least $2$. $\endgroup$ – Marco Flores Dec 2 '16 at 17:13
  • $\begingroup$ It is indeed one-to-one, like the normalization of a cusp. $\endgroup$ – Sasha Dec 2 '16 at 18:58

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