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.

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