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I hope to show that $$\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$$ for $(a,b)=1$.

I made an effort to find a homomorphism but I failed.

Can you give a hint?

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Hint: $(a,b)=1$ if and only if there exist integer $m,n$ such that $an+bm=-1$.

Incidentally, the result should be that $$\Bbb Z[i]/(a+bi)\cong\Bbb Z_{a^2+b^2}.$$

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  • $\begingroup$ Could you explain me more precisely? I can't get the homomorphism with an+bm=-1 $\endgroup$
    – Pearl
    Jun 10, 2016 at 5:25
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    $\begingroup$ I don't have time to get into the details, I'm afraid, but specific instances of this sort of question can be found here, here, and here. Hopefully that's enough to give you the idea of what such a homomorphism should look like. $\endgroup$
    – Cameron Buie
    Jun 10, 2016 at 12:48
  • $\begingroup$ @Pearl: Let me know what you're able to do with those, and if you're still stuck, I'll try to explain it further when I have the time. $\endgroup$
    – Cameron Buie
    Jun 10, 2016 at 16:13
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    $\begingroup$ Thank you for link. It is easy to solve with starting from $Z$ to $Z[i]/(a+bi)$. I first thought from $Z[i]$ to $Z_{a^2+b^2}$ ^0^ $\endgroup$
    – Pearl
    Jun 11, 2016 at 2:21
  • $\begingroup$ You're very welcome! $\endgroup$
    – Cameron Buie
    Jun 11, 2016 at 13:47

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