On the order of $\mathbb{Z}[X]/(f,g)$ and the resultant $R(f,g)$. I suspect that $\#\mathbb{Z}[X]/(f,g)=|R(f,g)|$ holds for any two non-constant polynomials $f,g\in\mathbb{Z}[X]$, where $R(f,g)$ is the resultant of $f$ and $g$. I am however unable to prove it. I'd like to know whether or not this is true, and if so, a hint as to how to prove it.
For $f,g\in\mathbb{Z}[X]$ splitting as $f=a\prod_{i=0}^m(X-\alpha_i)$ and $g=b\prod_{j=0}^n(X-\beta_j)$ over $\overline{\mathbb{Q}}$, I understand the resultant of $f$ and $g$ to be
$$R(f,g):=a^nb^m\prod_{i=0}^m\prod_{j=0}^n(\alpha_i-\beta_j).$$
Edit: From the first few replies it is clear that I rushed this post.
I should mention that for my own purposes I only require a proof for distinct $f$ and $g$, both monic and irreducible over $\mathbb{R}$. More explicitly; $f$ and $g$ are both either linear, or quadratic with negative discriminant.
It seemed to me that the desired result would generalise, though clearly some assumptions must be made. Perhaps it is enough to assume both $f$ and $g$ monic, and the $\alpha_i$ and $\beta_j$ all distinct?
 A: Let $f,g \in \mathbb{Z}[X]$ be two non-constant polynomials. Assume that $f$ is monic. Then:

*

*$\mathrm{Res}(f,g) \neq 0$ if and only if $f$ and $g$ are coprime;

*if $\mathrm{Res}(f,g) \neq 0$, then $ \mathbb{Z}[x]/(f,g)$ is a finite abelian group of order $\vert \mathrm{Res}(f,g) \vert$.
Proof of (1): this is well known when we work over a field. In our case we need to use the Gauss lemma because $\mathbb{Z}$ is a unique factorization domain.
Proof of (2) (following Gerry Myerson): the $\mathbb{Z}$-module $M = \mathbb{Z}[x]/(f)$ is free of finite rank because $f$ is monic. The endomorphism $\varphi \in \mathrm{End}_\mathbb{Z}(M)$ given by the multiplication by $g$ has determinant equal to $\mathrm{Res}(f,g)$. Moreover the kernel of $\varphi$ is zero because $f$ and $g$ are coprime by (1). If one diagonalises $\varphi$ via the Smith normal form, one can see that $\operatorname{coker} \varphi = \mathbb{Z}[x]/(f,g)$ is isomorphic to a direct sum of finite cyclic groups and the cardinality of $\operatorname{coker} \varphi$ is exactly $\vert \det \varphi \vert$.
