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I've been trying to solve this problem but I got stuck on it:

Given is the ideal $I=(2+i)$ in the ring of Gaussian integers $\mathbb{Z}[i]$. How many elements does the quotient ring $\mathbb{Z}[i]/I$ contain?

I'm just a dummy but I will show you what I have been trying to do.

My first observation is that we need to find out how many cosets the subgroup I has. In order to find that out I wrote: $I=\{(a+bi)(2+i):a,b \in \mathbb{Z}\}=\{(2a-b)+(2b+a)i:a,b \in \mathbb{Z} \}$.

Now we see that for all $z \in I, \ 5\mid 2 \text{Re}(z)+\text{Im(z)}$, because $ \ 2(2a-b)+(2b+a)= 5a$. Now denote the set $A = \{z \in \mathbb{Z}[i]: 5\mid 2 \text{Re}(z)+\text{Im(z)}\}$. I don't know if $A$ is a subring or an ideal or neither, but it's a subgroup of $\mathbb{Z}[i]$ with regard to "$+$", containing 4 other cosets, right?

As I noticed $I \subseteq A$, although I failed to proof that $A \subseteq I$, so all I've got now is that $|\mathbb{Z}[i]/I|\geq5$, without the knowledge if even this is bounded from above.

Could some of you please tell me if this makes sense, and give an hint? Thank you!

Koenraad van Duin
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    Hint: Every coset can be written as $(a+0i)+I$ for some $a$. Then you just need to realize which distinct $a$ give distinct cosets. – Thomas Andrews Jul 29 '13 at 21:08
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    I don't think this post should have been closed as a duplicate; it is asking for confirmation of a certain approach to solving the question, an approach which is not presented in any of the answers to the question of which it is allegedly a duplicate. – Matt E Jul 29 '13 at 22:34
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    Incidentally, it mystifies me that this post was down-voted. It states the problem clearly, and gives an excellent presentation of non-trivial work on the problem. What more does one want in a question? – Matt E Jul 29 '13 at 23:18

2 Answers2

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Your approach is a good one. (There are others, as in the linked material at the end of this answer, but there is no reason not to continue pursuing your approach.)

Why don't you try to show that $A = I$?

Given $x$ and $y$ (integers) such that $5 | 2x + y$, try to find $a$ and $b$ so that $x = 2 a - b$ and $y = a + 2b$. (You can just try to solve for $a$ and $b$; when you do, your divisibility condition will come in useful to ensure that $a$ and $b$ are integers.)

Once you do this, you'll be finished, since, as you observed, $A$ has index $5$ in $\mathbb Z[i]$.

For a different perspective on this kind of question, you could look at this answer (and the other answers posted there). But I would encourage you to push your original approach all the way through!

Community
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Matt E
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  • Dear Matt, his idea is clearly explained in the answer posted by Brandon Carter here, especially the surjectivity part. I don't think Brandon's approach is a "different perspective". Maybe just a better one. –  Jul 30 '13 at 07:09
  • @YACP: Dear YACP, Maybe I'm wrong, but based on the way the OP expresses themself in the question, I'm not sure that they would recognize their approach in Brandon Carter's answer. It is phrased a lot more formally (introducing the map $\phi$ and so on), and the explicit discussion of $I$ and $A$ as in the OP's post is suppressed in favour of a more formalistic approach. Certainly Brandon's approach is closer to how I would approach such a question, but I'm not convinced it is the right way to help the OP. As I wrote, I think it's better to help the OP complete their own approach. Regards, – Matt E Jul 30 '13 at 14:00
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Exercise: Show, in general, that the number of Gaussian integer equivalence classes, modulo a Gaussian integer $z$ is exactly $N(z)$, where $N$ is the norm function.

Ahaan S. Rungta
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