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Does $\mathfrak m$ divide the discriminant of the cubic polynomial $f(x)$?

Let $f(x)$ be a cubic polynomial with discriminant $\Delta$, and let $\mathfrak m$ be a prime ideal such that $f(x)\equiv0\pmod{\mathfrak m}$. Can it be shown that $\mathfrak m$ divides $\Delta$?
In particular, does the cubic $x^3+x^2-3x+2$ have a prime divisor $\mathfrak m\lhd\mathbb Z[\sqrt{ -3}]$ such that $f(x)\equiv0\pmod{\mathfrak m}$ for every $x\in\mathbb Z[\sqrt{ -3}]$?

A:

In the case when $\mathfrak m$ is principal, the discriminant is given by $$\Delta = \frac{4d^3}{27}$$ (see here). From there, the result follows by dividing by $4$ and noting that $d$ is a unit in $\mathbb Z[\sqrt{ -3}]$.
Alternatively, you can use the following generalisation of the result you mentioned, which shows that there are quadratic forms $Q(x,y)$ with integer coefficients and such that $Q(x,y)\equiv0\pmod{\mathfrak m}$ for all $x,y$.

Theorem. Let $Q(x,y)$ be a quadratic form with integer coefficients, and let $\mathfrak m$ be a prime ideal of $\mathbb Z[x,y]$ such that the matrix $$\left(\begin{array}{cc} x &y \\ Q(x,y) & -Q(x,y) \end{array}\right)$$
is $\mathfrak m$-integral for all $x,y\in\mathbb Z$. Then $Q(x,y)\equiv0