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I hope to better understand the notion of a quotient ring through this example:

I am given $R=\mathbb{Z}[i]=\{a+bi:a,b\in \mathbb{Z}\}$ and $M=\{a+bi: 3|a,3|b\}$. I have already shown that $M$ is a maximal ideal of $R$, but I am also asked to show that $R/M$ is a field with $9$ elements. I understand that $R/M$ is the set of cosets of $M$ in $R$, but for rings I am confused as to what these look like. For instance, do I look at $k+M$ or $kM$ for $k\in R$? How am I to show that there are $9$ cosets?

$\mathbf{Note:}$ If you can clarify what $R/M$ looks like for me, I think I would be able to find the answer on my own.

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Hint: You could write out the multiplication table, but there's an easier argument: What happens when you mod out by a maximal ideal?

Quotient rings are made up of additive cosets, i.e. the elements of your quotient ring $R/M$ look like $k+M$ with $k\in R$.

Here are a few cosets, all distinct from eachother: $M$, $1+M$, $1+i+M$. I also claim that $2+M=5+3i+M=2+6i+M$ (why?). Hopefully this is enough for you to see what's going on. Feel free to comment if you still have questions.

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  • $\begingroup$ I think I see where this is headed. To be sure, can I say that $2+M=8+3i+M$? $\endgroup$ – user113525 Apr 15 '14 at 3:00
  • $\begingroup$ Indeed you can! $\endgroup$ – Brett Frankel Apr 15 '14 at 3:10
  • $\begingroup$ Now try and write down representatives for each of the $9$ cosets. $\endgroup$ – Brett Frankel Apr 15 '14 at 3:11
  • $\begingroup$ In the comment section for the answer above, I gave my best understanding for why this is: $a \equiv i \mod 3$ and similarly for $b$ which will mean there are $3^2=9$ cosets? $\endgroup$ – user113525 Apr 15 '14 at 3:12
  • $\begingroup$ I think you're getting after the right thing, but to make sure, just write them all down. There's only $9$, so it won't take long. $\endgroup$ – Brett Frankel Apr 15 '14 at 3:15
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This used to confuse me as well. Generally, the line of thinking that helps me is that two cosets $k$ and $k^{'}$ are equivalent mod $M$ if and only if $k-k^{'}$ is an element of $M$. In your example that means two cosets $\overline{a+bi}$, $\overline{c+di}$ are equivalent if and only if $3 \mid a-c$ and $3 \mid b-d$. Think about which kinds of elements satisfy this. You should be able to think of 9 natural coset representatives.

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    $\begingroup$ In particular, you can ignore the ring structure, and look at this as simply a question of a quotient of abelian groups, and the abelian group structure is pretty simple! $\endgroup$ – user14972 Apr 15 '14 at 3:02
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$M$ is $(3)$, the principal ideal generated by $3$ in $\mathbb{Z}[i]$. Can you think of a good set of representatives for all the cosets of $(3)$? What values of $a$ and $b$ might you pick for $(a + bi) + (3)$?

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    $\begingroup$ Is it, $a \equiv i \mod 3$ for $i\in \{0,1,2\}$, and similarly for $b$? This will mean there are $3^2=9$ cosets. Is this right? $\endgroup$ – user113525 Apr 15 '14 at 3:05
  • $\begingroup$ Sounds good to me! $\endgroup$ – André 3000 Apr 15 '14 at 3:56

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