Find the congruence classes (mod $(3 + \sqrt{ −3})/2)$ in $Q[ \sqrt{ −3}]$. Find the congruence classes (mod $(3 + \sqrt{ −3})/2)$ in $Q[ \sqrt{ −3}]$.
This one really has me stumped. How do I find a clear, number-theoretic approach towards a solution to this problem?
EDIT: I have solved this problem, but am looking for a solution that preferably doesn't really use the notions of rings and specifically fields/field extensions
Since -3 = 1 mod 4 the ring of integers in Q(√(-3)) is Z[ (-1+√(-3))/2 ]. (-1 + √(-3))/2 is a root of the irreducible polynomial f(X) = X^2 + X + 1 over Z, and is a cube root of 1, X^3 =1. Z[ (-1+√(-3))/2 ] is a Euclidean domain, and so a UFD. 
Norm ((3 + √(-3))/2) = 3, so (3 + √(-3))/2 must be a prime { else if = QR then Norm(Q) Norm(R) = 3 --- doesn't make sense unless Q or R is unit}. 
Since (3 + √(-3))/2 is a prime then Z[ (-1+√(-3))/2 ] mod (3 + √(-3))/2 is a field {a general fact}, containing Z mod (3 + √(-3))/2. Since (3 + √(-3))/2 divides 3, then Z mod (3 + √(-3))/2 is really the subfield Z mod 3. More precisely if A and B are in Z and A = B mod (3 + √(-3))/2 then A-B = (3 + √(-3))/2*Q where Q is an integer in Z[(-1+√(-3))/2]. Taking norms you have (A - B)^2 = 3*Norm(Q), so 3 divides A-B and A = B mod 3. 
So Z[ (-1 + √(-3))/2] mod (3 + √(-3))/2 is the field extension of Z mod 3 gotten by adjoining a root of f(X) = X^2 + X + 1 to Z mod 3. But X^2 + X + 1 = (X-1)^2 mod 3, so f(X) splits completely mod 3, which means that there is no extension of Z mod 3, ie,         Z[ (-1 + √(-3))/2 ] mod (3 + √(-3))/2 = Z mod 3. 
So the residue class representatives are just 0, 1 and 2. 
 A: Here’s a hint only: recall that $\frac{1-\sqrt{-3}}2=\zeta$ is a primitive sixth root of unity in your field — and if you didn’t know that, you’d better check this fact right now. But when you multiply out $\zeta\sqrt{-3}$, you get your modulus $\frac{3+\sqrt{-3}}2$.
EDIT (expansion). In doing a problem like this, you need to have a precise idea of what the integers of your field $K=\Bbb Q(\sqrt{-3}\,)$ are. Recall that they’re the numbers that are in $K$ and are roots of a monic polynomial with integer ($\Bbb Z$) coefficients. As it happens, in this field, the integers are precisely the numbers $$\frac{a+b\sqrt{-3}}2$$ where $a$ and $b$ are in $\Bbb Z$ and have the same parity: either both odd or both even. So your job is to find a complete set of representatives for the congruence classes modulo $\frac{3+\sqrt{-3}}2$, equivalently modulo $\sqrt{-3}$. Remember that for $z$ and $w$ in your ring of algebraic integers of $K$, $z\equiv w\pmod m$ if and only if there’s a $K$-integer u for which $z-w=mu$. (Here, $m$ stands for your modulus, whatever it is.)
Now go forth and prove.
SECOND EDIT: You have asked for my proof that a complete set of representatives of the congruence classes modulo $\frac{3+\sqrt{-3}}2$ may be taken to be $\{0,1,2\}$.
I’ve already pointed out that congruence modulo $\frac{3+\sqrt{-3}}2$ is the same as congruence modulo $\sqrt{-3}$, and I’ve asked you to believe that the algebraic integers of $K=\Bbb Q(\sqrt{-3}\,)$ are the numbers $\frac{a+b\sqrt{-3}}2$ for which $a$ and $b$ are natural integers of the same parity. First I want to show that $0\not\equiv1\pmod{\sqrt{-3}}$, similarly that $0\not\equiv 2$. Well, for $1$ to be congruent to $0$, it’d have to be a multiple of $\sqrt{-3}$, clearly not so, same for $2$. So that disposes of that question: there are at least three congruence classes.
Next I want to show that every algebraic integer $\frac{a+b\sqrt{-3}}2$ is congruent to a natural integer $n$, indeed it’s congruent to $-a$, since
$$
\frac{a+b\sqrt{-3}}2-(-a)=\frac{3a+b\sqrt{-3}}2=\frac{b-a\sqrt{-3}}2\cdot\sqrt{-3}\,,
$$
in which $a$ and $b$ started out with the same parity, so that the multiplier of $\sqrt{-3}$ is still an algebraic integer.
Finally, every natural integer is congruent to $0$, $1$, or $2$ modulo $3$, and thus a fortiori modulo $\sqrt{-3}$.
What have I shown? Everything is congruent to one of the three first natural numbers, and these are incongruent to each other. So there are just three congruence classes.
A: I know this question has already been answered, but I'd like to offer a graphical solution too for those you come in the future.
If you are familiar with the concept of a lattice, this problem becomes very easy.
In case you don't know what a lattice is, here is a quick description:
To create the lattice, first you have to graph the multiples of the integer in question. The x-axis is the $a$ value of $\alpha$ and the y-axis is the $b\sqrt{d}$ value of $\alpha$. To generate multiples of $\alpha$, you would multiply it by $x+yi$ where x and y are integers. If you connect the grid points or multiples you create a lattice.
For this problem, you only need 1 lattice grid, ie any 4 points or multiples that creates a grid. The next step is to find all quadratic integers in the field $Q[\sqrt{d}]$ (where d=-3 in our case) that are "inside" this grid or box (but not on it). The final step is to subtract the value of the nearest gridpoint from each of these quadratic integers, and you get your congruence classes.
In our case, $\alpha=\frac{3+1\sqrt{-3}}{2}$, so 4 easy points/multiples would be $(0,0)$, $(1.5,.5\sqrt{-3})$, $(1.5,-.5\sqrt{-3})$, and $(3,0)$. We can graph all quadratic integers of $Q[\sqrt{-3}]$ in that area and we see that there are only 2 eligible points, being $(1, 0\sqrt{-3})$ and $(2, 0\sqrt{-3})$. Subtracting this from the nearest lattice values (0 for the first one and 3 for the second) we get the values 1 and 2. So your final answer would be 0,1, and 2.
Here is a picture of a lattice that I found off the internet. The yellow points are multiples of your $\alpha$ (in the picture, its just a+bi), and the red points represent quadratic integers inside the lattice of that specific field

