Elementary question on quotient group construction I am going though an introductory proof text book and was hoping if someone could verify/correct my attempt at the following problem :

Let $(\Bbb{Z}[\sqrt3],+,.)$ where $\Bbb{Z}[\sqrt3] \subset\Bbb{R}$ and
  $\Bbb{Z}[\sqrt3]=\{i+j\sqrt3 :i,j \in \Bbb{Z} \}$. Consider the Ideal
  of the ring given by $I=\{a+b\sqrt3 : 3|a \}$.
Construct and Evaluate the quotient ring $\Bbb{Z}[\sqrt3]\text{\I}$.

Attempt
(i) The equivalence relation ~$_{I}$ is defined by x ~$_{I}$ y iff $(x-y)\in I$ for all $x,y \in \Bbb{Z}[\sqrt3]$.
(ii) x ~$_{I}$ y


*

*iff $(x-y)\in I$

*iff $x-y = a+b\sqrt3$

*iff $x-y = 3k + b\sqrt3 \text{ for some } k \in \Bbb{Z}$

*iff $y=x-3k - b\sqrt3 $
(ii)By theorem, x ~$_{I}$ y implies that $[x]=[y]$.
With $y=x-3k - b\sqrt3 $, then
$\Bbb{Z}[\sqrt3]\text{\I}$


*

*$=\{[x] : x \in \Bbb{Z}[\sqrt3] \}$

*$=\{[x-3k - b\sqrt3] : x \in \Bbb{Z}[\sqrt3] \}$
(iii)By theorem,since I is an ideal of $\Bbb{Z}[\sqrt3]$, then $(\Bbb{Z}[\sqrt3]\text{\I}),\bigoplus,\bigodot)$ is also a ring with well-defined binary operations.
Let $h,g \in \Bbb{Z}[\sqrt3]$ ,then $[h]\bigoplus[g]$ 


*

*= $[h-3k - b\sqrt3] \bigoplus [g-3l - c\sqrt3]$

*= $[h-3k - b\sqrt3 + g-3l - c\sqrt3]$

*= $[h+g - \sqrt3 (\sqrt3(k+l) + (b+c))]$
..similar procedure for $\bigodot$
Is this what is meant by construction? 
Any help appreciated.
 A: I would say you need to get a better idea what the elements of $\Bbb Z\bigl[\sqrt{3}\bigr]/I$ are, before you have a feel for what it looks like as a ring. By definition, the elements of $\Bbb Z\bigl[\sqrt{3}\bigr]/I$ are all equivalency classes of elements of $\Bbb Z\bigl[\sqrt{3}\bigr]$ under the relation $\sim_I$ you mentioned above. However, that doesn't really paint a clear picture, yet.
You're right that for any given $x,y\in\Bbb Z\bigl[\sqrt{3}\bigr],$ we will have $[x]=[y]$ if and only if $x-y=3k+b\sqrt{3}$ for some $k,b\in\Bbb Z,$ but this doesn't use the structure of $\Bbb Z\bigl[\sqrt{3}\bigr],$ at all, so doesn't clarify enough. We can do better! In particular, we know that $x=i_1+j_1\sqrt{3}$ and $y=i_2+j_2\sqrt{3},$ where $i_1,i_2,j_1,j_2\in\Bbb Z.$ Thus, we can conclude that $x\sim_I y$ if and only if the following (equivalent) conditions hold: $$x-y=3k+b\sqrt{3}\textrm{ for some }b,k\in\Bbb Z\\(i_1-i_2)+(j_1-j_2)\sqrt{3}=3k+b\sqrt{3}\textrm{ for some }b,k\in\Bbb Z\\i_1-i_2=3k\textrm{ and }j_1-j_2=b\textrm{ for some }b,k\in\Bbb Z$$ However, since $j_1$ and $j_2$ are integers, then this last one is equivalent to $$i_1-i_2=3k\textrm{ for some }k\in\Bbb Z,$$ meaning that $i_1\sim_{(3)}i_2.$ Since the $j$-values had no effect on the equivalency class, then note in particular that $x\sim_I i_1$ which means that we can choose an integer representative for each equivalency class. The equivalence $\sim_{(3)}$ gives us even more, though!
Given any $i\in\Bbb Z,$ there exists a unique $m\in\{0,1,2\}$ and $n\in\Bbb Z$ such that $i=m+3n.$ In particular, say $i_1=m_1+3n_1,$ so since $$x-m_1=i_1+j_1\sqrt{3}-m_1=3n_1+j_1\sqrt{3}\in I,$$ then $[x]\in\bigl\{[0],[1],[2]\bigr\}.$ Thus, $\Bbb Z\bigl[\sqrt{3}\bigr]/I=\bigl\{[0],[1],[2]\bigr\}.$ At that point, it is much easier to see and talk about what $\oplus$ and $\odot$ look like.
