[I cross-posted this question as https://mathoverflow.net/questions/431454. (That version is also slightly updated.)]
Let $f = \sum_{i=m}^n f_iy^i \in \mathbb{C}[x,y]$ be a polynomial (where $f_i \in \mathbb{C}[x]$ with $f_m,f_n \ne 0$ and $n > m$), and let $\operatorname{Disc}_y(x) \in \mathbb{C}[x]$ be the discriminant of $f$ with respect to $y$ (i.e., for every $a \in \mathbb{C}$, $\operatorname{Disc}_y(a)$ is the discriminant of $f(a,y)$). I got the impression that $f$ is weighted homogeneous (with possibly negative weights) if and only if the following three conditions hold:
- $f_m$ is a (non-zero) monomial in $x$
- $f_n$ is a (non-zero) monomial in $x$
- $\operatorname{Disc}_y$ is a non-zero monomial in $x$
The direction "$\Rightarrow$" should not be too hard to prove: 1 and 2 are clear, and I think that 3 follows from the formula expressing the discriminant in terms of the coefficients of a polynomial. My question is: Is "$\Leftarrow$" also true? This seems more surprising, given that it yields strong conditions on $f$; since the question seems rather elementary, I am hoping that this might be known.
Some thoughts/remarks (which didn't yield a solution yet):
- If one thinks of $f$ as defining a variety $V \subset \mathbb{C}^\times \times \mathbb{P}^1\mathbb{C}$, then 1 and 2 say that $V \subset \mathbb{C}^\times \times \mathbb{C}^\times$, and 3 says that the projection to the first coordinate is unramified.
- Consider $g := y \cdot \partial f/\partial y$. It is not hard to verify that 1, 2 and 3 together are equivalent to: the subvarieties of $\mathbb{C}^\times \times \mathbb{P}^1\mathbb{C}$ defined by $f$ and $g$ have empty intersection. Applying a version of Bezout's Theorem for subvarieties of $\mathbb{P}^1\mathbb{C} \times \mathbb{P}^1\mathbb{C}$ thus yields a lot of intersections above $0$ and $\infty$.
- If $n$ (the degree in $y$) is 2, then one can easily prove the claim using the formula for the discriminant. I was also able to prove the claim when $f$ consists of only three monomials.