List all elements in the residue field $Z[i]/(q)$

Question

Consider a Gaußian prime $q$.

How to list all elements in the residue field $Z[i]/(q)$? Is there any formulas or criteria? Here I'm looking for the case $q$ is a complex number, as I can do the real $q$ case.

Thanks.

Answer 1

Well if $q$ is also rational, i.e. $q\equiv 3\mod 4$ is a rational prime, then it's really easy, you just have

$$\Bbb Z[i]/(q)=\{a+bi : 0\le a,b<q\}.$$

You already knew that, but I like to be complete.

If $q=1+i$, then note that $(1+i)(1-i)=2$ so $2\equiv 0\mod (1+i)$ so

$$\Bbb Z[i]/(1+i)\subseteq \{a+bi : 0\le a,b\le 1\}$$

Since $1= (1+i)+i$ we see $1\equiv i\mod (1+i)$ hence the representatives are $0,1$, and no smaller set will do.

Finally if $q=a_0+b_0i$ is such that $a_0^2+b_0^2=p$ is a rational prime, then--as in the case of $2$ we note that $p\equiv 0\mod q$ so

$$\Bbb Z[i]/(q)\subseteq \{a+bi: 0\le a,b\le p-1\}$$

For this we note that $a+bi\equiv a+a_0+(b+b_0)i$, so you knock it out by starting with $0,1,2,3,4,\ldots, p-1$ and noting that these represent $p$ equivalence classes. Then by adding $q$ which has $a<p$ clearly, so the residue classes of $a$ and $b$ mod $p$ generate $\Bbb Z/p$, i.e. each other $a+bi$ in our set on the RHS above with a non-zero imaginary part is equivalent to one of the $0,1,\ldots, p-1$, indeed choose $0\le x\le p-1$ for $c+di$ by choosing $y$ so that $yq+c+di\in\Bbb R$. By definition $yq+c+di\equiv x\mod q$. So you can use the set

$$S_q=\{x: 0\le x\le p-1=|q|^2-1\}$$

to represent $\Bbb Z[i]/(q)$.

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Addendum (for the well-informed reader)

In case you (the op) or anyone are reading this who have a higher-tech view of algebraic number theory, you can justify my last structure theorem even more easily: The norm of an ideal is its index, hence it is sufficient to show each $0\le x\le p-1$ are inequivalent modulo $q$. However, as previously noted $b$ generates $\Bbb Z/p$, hence if $a\equiv b\mod q$ and $0\le a,b\le p-1$, then $a=b+kq$ which only has zero imaginary part from among the candidates if $k\equiv 0\mod p$, i.e. $a=b$.