$\#ℤ[i]/\alpha\betaℤ[i] = \#ℤ[i]/\alphaℤ[i] · \#ℤ[i]/ \beta ℤ[i]$

Question

For arbitrary $\alpha = a+bi, \beta = c+di \in ℤ[i] \setminus \{0\}$, I'm trying to prove the title: $$\#ℤ[i]/\alpha\betaℤ[i] = \#ℤ[i]/\alphaℤ[i] · \#ℤ[i]/ \beta ℤ[i]$$

I think this is quite easy in the case where $(\alpha) := \alpha ℤ[i]$ and $(\beta)$ are coprime, because then you can use the Chinese Remainder Theorem.

However, we do not restrict ourselves to that case, so I'm afraid a whole other method is needed. What I have shown earlier is that for any ring $R$ with ideals $I, J \trianglelefteq R$,

$$ R / (I+J) \cong (R/I)/\pi[J],$$

where π is the canonical projection. This gets you to

\begin{align*} ℤ[i]/\alpha\betaℤ[i] &= ℤ[i]/(\alpha\beta) = ℤ[i]/(ac-bd + i(ad+bc))) \\ &\cong \left(ℤ[i]/(ac-bd)\right) / \,π[(ad+bc)] \end{align*}

But still I don't see how that gets us where we want to be.

Any hints will be greatly appreciated.

Answer 1

I'll write $R=\Bbb Z[i]$, and use the usual index notation, so $|R:\alpha R|$ is your $\# R/\alpha R$. Then $$|R:\alpha\beta R|=|R:\alpha R||\alpha R:\alpha\beta R|$$ so we just need to prove that $$|\alpha R:\alpha\beta R|=|R:\beta R|.$$ But this follows from multiplication by $\alpha$ being a group isomorphism $R\to\alpha R$ which takes $\beta R$ to $\alpha\beta R$.