What is the order of quotient rings? I have a question related to determine the order of a quotient ring in the following:


Consider the quadratic ring $\mathbb{Z}[\sqrt{-n}]$, and $\alpha$ and $\alpha'$ are 2 irreducible elements. Can we determine the order of the quotient ring $(\alpha, \alpha')/(\alpha)$, here $(\alpha, \alpha')$ is the ideal generated by $\alpha, \alpha'$?


Thanks so much. I really appreciate it.
 A: It’s not quotient rings you’re dealing with, but quotient modules over the base ring $\mathbb Z[\sqrt{-n}\,]$. Consider the multiple inclusion
$$
(\alpha)\subset(\alpha,\alpha')\subset\mathbb Z[\sqrt{-n}\,]\,.
$$
Now one of the Isomorphism Theorems tells you that $\mathbb Z[\sqrt{-n}\,]/(\alpha)\supset(\alpha,\alpha')/(\alpha)$, and the quotient is isomorphic to the object whose cardinality you’re trying to determine. The two groups (modules over your number-ring, actually) are finite, and in fact the cardinality of $R/(\alpha)$ is, according to a well-known theorem of algebraic number theory, equal to $|\mathcal N(\alpha)|$, where $\mathcal N$ is the field-theoretic norm from $\mathbb Q(\sqrt{-N}\,)$ down to $\mathbb Q$, and $R$ is the full ring of algebraic integers in $\mathbb Q(\sqrt{-N}\,)$. (This ring will be $\mathbb Z[\sqrt{-N}\,]$ in the good cases that $N$ is square-free and not congruent to $3$ modulo $4$.) In fact, since you’re presumably dealing with an imaginary field, the norm is always positive, so we can dispense with the absolute-value bars.
At this point we can’t say much more without further information about $N$, $\alpha$, and $\alpha'$. As an interesting example, consider $N=6$, $\alpha=\sqrt{-6}$, and $\alpha'=3$. Then $(3,\sqrt{-6}\,)$ is the (unique) prime ideal of the integer-ring above the rational prime $3$, and its index in the whole is $3$, while the index of $(\sqrt{-6}\,)$ is $6$, and of $3$ is $9$. If $N$ is not square-free, alternatively if $N\equiv3\pmod4$, then things get rather messy.
