# Show quotient ring of Gaussian integers is indeed a ring

Let $R=\mathbb{Z}[i]/(1+2i)$.

I want to show that this is a ring and also find it's order.

I've had a look at some similar questions and worked out that the elements of $R$ are $0,1,2,1+i$ and $2+i$. Is there a formal way to prove this so that I can show that the order of $R$ is $5$?

Also, would it be enough to show that $R$ is a subring of $\mathbb Z[i]$? If this is the case I believe I would be correct in saying that I all I need to show is the following:

$1.$ $R$ is closed under addition

$2.$ $1_{\mathbb{Z}[i]}=0 \in R$

Is that correct?

I could show the second part easily enough but how would I go about proving closure?

• I agree with Ferra that you should try and write down an isomorphism to extend your understanding. A good way to start it to show that the element 5 is actually contained in the ideal (2i+1) (do this for yourself). After this you should be able to show that also (i-2) is contained in the ideal. That should help. – Joachim Feb 21 '15 at 12:05

There is a natural ring image of $\,\Bbb Z\,$ in $R = \Bbb Z[i]/(1\!+\!2i)\,$ by mapping integer $\,n\,$ to $\ n \pmod{1\!+\!2i}$ by composing the natural maps $\,\Bbb Z\to \Bbb Z[i]\to \Bbb Z[i]/(1\!+\!2i).\,$ This map $\, h\color{#0a0}{ \ {\rm is\ surjective\ (onto)}}$ by
$\quad {\rm mod}\ \alpha =1\!+\!2i\!:\ \, 5\equiv \alpha\bar\alpha \equiv 0\$ so $\ 2i\equiv -1\overset{\large \times 3^{\phantom{I}}\!}\Rightarrow\, \color{#c0f}i \equiv \color{#c0f}{-3}\$ so $\ a+b\,\color{#c0f}i\equiv a\color{#c0f}{-3}b\in\Bbb Z\,$
Next, by $\rm\color{blue}{RD = \,Rationalizing\,}$ a Denominator, we compute that $\,I = \color{#c00}{\ker h = 5\,\Bbb Z}\,$ as follows
$\quad n\in I\iff 1\!+\!2i\mid n\ \ {\rm in}\ \ \Bbb Z[i]\iff \dfrac{n}{1\!+\!2i}\in \Bbb Z[i]\color{blue}{\overset{\rm\ \ RD}\iff} \dfrac{n(1\!-\!2i)}{5}\in\Bbb Z[i]\iff \color{#c00}{5\mid n}$
Therefore, applying the First Isomorphism Theorem, $\, \color{#0a0}{R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{5\,\Bbb Z}.$
$R$ cannot be a subring of $\mathbb Z[i]$ because it is a torsion abelian group while $\mathbb Z[i]$ is torsion free. The quotient of a commutative ring by an ideal is always a ring... To show that $R$ has order $5$ a good way is writing down an isomorphism $R\to \mathbb F_5$
• @23574579686, the bracket notation means "the ideal generated by", i.e. $(1 + 2i)$ is the ideal generated by $1+2i$. If that does not sound familiar you might want to look that up. – Joachim Feb 21 '15 at 12:04