## Exhibit a residue system for a given quotient of the Gaussian integers

Exhibit a complete residue system for $\mathbb{Z}[i]/(2+i)$. Plot the points $\{a+bi \ |\ 0 \leq |a|,|b| \leq 4 \}$ in the complex plane, coloring each point according to which coset of $(2+i)$ it is in.

Note that $N(2+i) = 5$. Now each residue class mod $2+i$ has a representative whose norm is strictly less than 5 by the division algorithm. It is easy to see then that the distinct residue classes have representatives in the set $\{(0,0), (\pm 1,0),$ $(\pm 2,0), (\pm 1, \pm 1),$ $(0, \pm 1), (0,\pm 2) \}$. We claim that all but 5 of these are redundant. Indeed:

• $(2) - (-i) = 2+i$ and $(-i) - (-1+i) = (2+i)(-i)$.
• $(-2) - (i) = -(2+i)$ and $(i) - (1-i) = (2+i)i$
• $(-1) - (1+i) = -(2+i)$ and $(-2i) - (-i) = (2+i)(-1)$
• $(1) - (-1-i) = 2+i$ and $(2i) - (1) = (2+i)i$

Thus a complete residue system is contained in the set $\{0, 1, -1, i, -i\}$. We claim that this set is itself a residue system. To see this, we need to verify that no two are congruent mod $2+i$. First, since $2+i$ is not a unit, none of $1$, $-1$, $i$, and $-i$ is congruent to 0 mod $2+i$. Now $(1) - (i) = 1-i$ has norm 2, and thus cannot be divisible by $2+i$ which has norm 5. Similarly, $(1) - (-1) = 2$ has norm 4, $(1) - (-i) = 1+i$ has norm 2, $(-1) - (i) = -1-i$ has norm 2, $(-1) - (-i) = -1+i$ has norm 2, and $(i) - (-i) = 2i$ has norm 4. Thus these representatives are distinct mod $2+i$. Hence $\{0,1,-1,i,-i\}$ is a complete residue system mod $2+i$.

Note that $\alpha \equiv \alpha + (2+i) \mod (2+i)$. Likewise, we may add any associate to $\alpha$ and remain in the same congruence class. If we color the elements congruent to 0 green, to 1 purple, to $i$ red, to $-1$ blue, and to $-i$ orange, then a small region around the origin in the complex plane appears as follows.