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Following Quotient ring of Gaussian integers, their extended conclusion is $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$. However it does not convince me, at least, one example below:

Let $a=2,b=0$, I cannot find explicit isomorphism between $\mathbb{Z}[i]/2\mathbb{Z}[i]$ and $\mathbb{Z}/4\mathbb{Z}$.

The coset leaders of $\mathbb{Z}[i]/2\mathbb{Z}[i]=\{0,1,i,1+i\}$, and the coset leaders of $\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$.

I appreciate if anyone could give a bijection mapping between the two quotient rings.

I am not sure I am right or not but will be happy to discuss with anyone who is interested in.

From my point of view, $\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$ (I know it is not good to give this expression, just for convinence) has 2 units $1,3$ where the sum is $0$, but $\mathbb{Z}[i]/2\mathbb{Z}[i]$ has two units, say, $i,1$ and their sum is not zero. The structure of the two quotient rings are different, and hence the conclusion $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$ is not correct.

Anyone here could make things clearer. Thanks a lot.

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    $\begingroup$ That's only true if $(a,b)=1$... $\endgroup$ – Thomas Andrews Jul 21 '14 at 23:31
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You'll never find such an isomorphism, because $x+x=0$ for all $x\in\mathbb Z[i]/2\mathbb Z[i]$, but not for all $x\in\mathbb Z/4\mathbb Z$.

The theorem you state is only true if $\gcd(a,b)=1$.

See this answer from the same question.

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  • $\begingroup$ Thanks Thomas, it is clear now. $\endgroup$ – Yi W Jul 22 '14 at 0:21
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The following holds for any nonzero Gaussian integer $a+bi$.

If $d=gcd(a,b)$, then $\mathbb{Z}[i]/(a+bi)$ is isomorphic (as a group) to $$\mathbb{Z}/d\mathbb{Z} \oplus \mathbb{Z}/ f\mathbb{Z},$$

where $f=\frac{a^2+b^2}{d}=aa'+bb',$ and $a=a'd, b=b'd.$

There might be a more explicit link, but here one sees how the quotient group is classified through the use of matrices in $GL_n(\mathbb{Z})$. http://www.cs.uleth.ca/~holzmann/notes/abelian.pdf

Take the vectors $u_1=(a,b), u_2=(-b,a)$ that form a basis of the lattice $(a+bi)$. The Bézout identity reads $$ma'+nb'=1,$$

and suggests a $2\times 2$ invertible matrix (over $\mathbb{Z}$); a new basis of our lattice is now given by $$v_1=b' u_1+a'nu_2=(0,aa'+bb')=(0,f),$$ and the vector $v_2= m u_1-n u_2=(d, d(b'm-a'n)).$ Changing now the basis in $\mathbb{Z}\oplus\mathbb{Z}$ to $a_1=(1,0), a_2=(1,s)$ where $s=b'm-a'n$ one has:

$$(a+bi)= \mathbb{Z} fi \oplus \mathbb{Z} d.(1+si) \subset \mathbb{Z}i \oplus \mathbb{Z} (1+si)=\mathbb{Z}[i],$$ which yields the desired result.

Note that the cardinal of the group $=df$ equals the determinant of the initial matrix, which is $a^2+b^2$.

\paragraph{DECOMPOSITION AS A RING}

In order to study the decomposition as a ring, one need factor the number $$a+bi=\prod \pi_i^{e_i},$$ then apply the Chinese number theorem.

Primes in this ring are $1+i, p$ (where $p$ is an odd prime which is $3$ modulo $4$), and prime factors of prime integers $q$ which are $\equiv 1 (\mbox{ mod } 4)$; this last type are of the form $m+ni$ where $m^2+n^2=q.$ This is done as follows, for any prime $p$:

$$\mathbb{Z}[i]/(p)=\mathbb{Z}[x]/(p,x^2+1)=\mathbb{F}_p[x]/(x^2+1),$$ and this ring is nonreduced when $p=2.$ If $p$ is odd, the ring is a domain (hence a field) exactly when $x^2+1$ has no zeros in $\mathbb{F}_p$, which is precisely when $p\equiv 1 (\mbox{ mod }4)$ (look up the Legendre symbol).

Finally, if $4|p-1$ then the above ring has two different primes, which translates into saying that $(p)=(\pi_1)\cap (\pi_2)$. After working a bit more, one derives $$p=\pi_1 \pi_2$ and $N(\pi_i)=p$, which yields $\pi_1=\overline{\pi_2}.$ The result follows.

A more detailed discussion would considerably extend this paragraph.

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  • $\begingroup$ Thanks Theon, I will have a look of the link you suggested. $\endgroup$ – Yi W Jul 27 '14 at 23:24
  • $\begingroup$ what about as a ring? $\endgroup$ – M. Van Feb 24 '17 at 15:57
  • $\begingroup$ As a ring, it fully depends on the decomposition into prime factors, which it surely has, since it is an Euclidean domain, using the (square of the) norm as the corresponding function. I added a paragraph on that. $\endgroup$ – Theon Alexander Mar 2 '17 at 18:31

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