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For the number $2+i$, since $2^2+1^2=5$ is a prime number, so Gaussian integer $z = 2 + i$ is a prime. But this number is a complex number so

  1. Does it make sense to find the addition and multiplication table modulo $2+i$?
  2. If it does, is there a way that I find it? Just like finding the addition and multiplication table for a normal prime number $3$ or $5$

Thanks in advance. (I don't know too much about abstract algebra, or things like groups, rings, and fields. I only understand some elementary number theory. At present, I know how to find the addition and multiplication tables of real integers. For example, I can write the table of module 3 on $\mathbb Z$ or module 5 on $\mathbb Z$)

Bill Dubuque
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M_k
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  • $\mathbb{Z}[i] / (2+i) = \mathbb{Z}[x] / (x^2+1, 2+x) =(\mathbb{Z}[x] / (2+x)) / (x^2+1) = \mathbb{Z} / (5)$. – David Lui Jan 22 '23 at 21:30
  • Do you mean that the table for modulo $2+i$ is just the table for modulo 5 on the real integer $\mathbb Z$? I don't quite understand your processes. Could you please them in detail? Thanks. – M_k Jan 22 '23 at 21:34
  • As in the linked dupe $,\Bbb Z[i]/(2+i)\cong \Bbb Z/5,$ via $,a+bi\to a-2b\bmod 5\ \ $ – Bill Dubuque Jan 22 '23 at 21:38
  • So the operation tables are the same as those for arithmetic $!\bmod 5,,$ e.g. $$\begin{align} 1+i&\equiv 1-2\equiv 4\ 1-i&\equiv 1+2\equiv 3 \end{align}\Rightarrow (1+i)(1-i)\equiv 4(3)\equiv 2\qquad $$ – Bill Dubuque Jan 22 '23 at 22:49

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