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I want to show that $\Bbb Z[i]/(2+3i)$ is a finite field and count it's elements.

I don't really know how to show that this field is finite. I start by trying to understand that definition of this ring, so I write

$\Bbb Z[i]/(2+3i)=${$a+bi+2+3i:a,b \in \Bbb Z$}

I don't really know what to do from here. The knowledge needed to solve this shouldn't be much farther than the Chinese Remainder theorem, so I think it should be relatively simple but I don't see how to approach it. Any help is appreciated, thanks

mweiss
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Jeff
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  • You say "the Chinese Remainder Theorem," but there are many forms of this theorem. To which do you refer? – Cameron Buie Dec 17 '15 at 03:36
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    It may be that you will find this helpful: note that $(2+3i)(2-3i)=13$, so that $(2+3i)|13$ and consequently $(2+3i)\supset(13)$. This means, in turn, that there is a surjective homomorphism of rings, $\Bbb Z[i]/(13)\to\Bbb Z[i]/(2+3i)$. – Lubin Dec 17 '15 at 04:20

2 Answers2

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Here are a few hints:

  1. In this ring, $2+3i = 0$. That means that any time you encounter $3i$ you can replace it with $-2$. Using this technique you can rewrite any element in the form $a+bi$ where $b$ is either $0$, $1$, or $2$.
  2. Also, since $2+3i = 0$ in this ring, we also have $4+6i=0$. But also we have $(2+3i)^2=0$. Combining those identities you should be able to find a positive integer $n$ with the property that $n=0$ in this ring.
mweiss
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Your "definition of the ring" is already wrong, which is why you're not understanding it at all. $\{ a+bi+2+3i : a,b\in\mathbb{Z} \}$ is exactly the same as $\{ a+bi : a,b\in\mathbb{Z} \}$! Whenever you form equivalence classes, each element is a class of elements that are equivalent under the equivalence relation, which in this case is modulo $(2+3i)$. Once you get that right, it should be obvious why there are only finitely many equivalence classes of $\mathbb{Z}[i]$ mod $(2+3i)$.

user21820
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  • Good point on showing that it's equivalent to that definition which is nonsensical! So I'm modulating by (2+3i), so it should be a+bi (mod 2+3i), is that it? Then it's obvious, it's like showing $\Bbb Z/3\Bbb Z$ is a finite field? – Jeff Dec 17 '15 at 03:44
  • @Jeff Don't forget that you have to show it's a field, not just that it's finite. – Andrew Dudzik Dec 17 '15 at 03:46
  • Yes but once the elements are counted showing its a field is relatively simple, I think understanding the concept of it was the important part, thank you – Jeff Dec 17 '15 at 03:48
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    @Jeff: Sort of. In general it would be best to fully understand the concept of equivalence classes. The quotient ring you are talking about here is ${ { x+(2+3i)y } : x,y \in \mathbb{Z}[i] }$. You can see that any equivalence class containing $x$ contains all the elements of $\mathbb{Z}[i]$ that are equivalent to $x$ mod $(2+3i)$. – user21820 Dec 17 '15 at 04:16