Determining cardinality and units of $\Bbb Z_3[i]$ Let $\Bbb Z_3[i] = \{a + bi \mid a, b ∈ \Bbb Z_3, i^2 = −1\}$ and denote $·_3, +_3, −_3$ the multiplication, the addition and the subtraction $mod$ $3$. With these notations, the addition and the multiplication in $\Bbb Z_3[i]$
are defined as follows. For any $x = a + bi$ and $y = c + di$ with $a, b, c, d, ∈ \Bbb Z_3$ we have:
$x + y = (a +_3 c) + (b +_3 d)i$
$x · y = (a ·_3 c −_3 b ·_3 d) + (a ·_3 d +_3 b ·_3 c)i$
a). How many elements are in $\Bbb Z_3[i]$? Explain.
b). Find the units in $\Bbb Z_3[i]$ and compute their inverses
This is really confusing me but my attempt thus far:  for $x = a+bi$, $a, b$ can assume values from $\Bbb Z_3: 0, 1, 2$. So there are $9$ possible values for $\forall x \in\Bbb Z_3[i] $. Under the operations, there are $81$ possible ways to perform $x+y$, $x-y$, and $xy$ each respectively. I'm sure there is some repetition among each of these sets which would reduce the number of elements in the set. Aside from listing the elements out, I don't really know where to begin. 
Any help would be much appreciated.
 A: (a) is solved correctly by OP.
For (b), you first need to know what the multiplicative identity is, which is $1+0i$.  Hence, you want to determine which of the nine choices for $a,b$ have a corresponding $c,d$ so that $x\cdot y=1+0i$.  There are nine mini-problems.
Sample: $x=1+i$, i.e. $a=b=1$.  Now we need $ac-bd=c-d=1$ and $ad+bc=c+d=0$.  Adding, we get $2c=0$, so $c=0$ since we are in $\mathbb{Z}_3$.  But now $c+d=0$ tells us $d=0$, and $c-d\neq 1$.  Hence there is no choice of $y=c+di$ so that $x\cdot y=1+0i$.
Now, set $x=1+2i$ and do it all over again, then for the other seven elements.
A: Once you get more experience with these matters, you’ll see that there are many ways of finding reciprocals of elements in finite fields.
In this case, since your field is quadratic over the field with three elements, which I’ll denote $\Bbb F_3$, you can do the same thing as you did in High School to find the reciprocal of, for instance $1+i$:
$$
\frac1{1+i}=\frac1{1+i}\cdot\frac{1-i}{1-i}=\frac{1-i}{1-i^2}=\frac{1-i}2=-1+i\,,
$$
since $2=-1$ in $\Bbb F_3$.
Another much better way, once you know the theorem that the set of nonzero elements in a finite field is a cyclic group, that is, all are powers of one well-chosen element, you can blunder about till you find one of these “primitive elements”, and then write down its powers in order. One choice here is $R=1-i$, and you get $R^2=i$, $R^3=1+i$, $R^4=-1$, and I’ll let you complete the list, going all the way to $R^8=1$. Then you see that since $1+i=R^3$, its reciprocal is $R^5$, which you have on your list.
