Find all idempotent elements in the group algebra $\mathbb CC_3$ Sorry but I'm quite new to group algebras and even Latex so if this is all wrong I apologize.
By $\mathbb CC_3$ I mean the group algebra of the cyclic group of order 3 in the complex numbers
A group algebra is a ring so is it best to say $C_3$ has elements $e$, $a$ and $b$ instead of thinking of C3 having a generator g because the group action would be written + and if we said $C_3$ had elements $e$ $g$ and $2g$ that's pretty ambiguous because in $\mathbb CC_3$ $2g$ means something quite different right? Similarly if we wrote the group binary relation multiplicatively and said $C_3$ had elements $e$, $g$ and $g^2$ that would also be troublesome because we're working in a ring with + and *, right? (although in this specific example it would work out okay since $2*2=2+2$)
but I mean the point is if the elements are $e$ (the identity), $a$ and $b$, we think of them as e=0, a=1 and b=2 with addition and multiplication mod 3 yes? I'm just reluctant to actually call the elements 0,1,2 because it would look somewhat confusing in the group algebra, right?
When we say that $\mathbb CC_3$ is a group algebra we mean it's just a set of finite linear combinations of the form $Ae + Ba + Cb$ where $A,B,C$ belong to the set of complex numbers, right? The coefficients $A,B,C$ in no way at all "mean" anything with respect to the group, right? (eg, there's no way that $Ba=b$ for some $B in C$, like say B=2) right? $A. B and C$ are just symbols from the field of complex numbers and that's it, right?
How would one go about finding all the idempotent elements of this group algebra? How would you know you've found them all? Does one just solve:
$$(Ae + Ba + Cb)(Ae + Ba + Cb)=(Ae + Ba + Cb)$$
which implies
B=1 or 0 by looking at a coefficient, etc?

I took the approach you mention.  If $a$ is a generator of $C_3$, and $x,y,$ and $z$ are complex numbers, then the equation $(x\cdot e+y\cdot a+z\cdot a^2)^2=x\cdot e+y\cdot a+z\cdot a^2$ gives the following system:
$$x^2+2yz=x\\y^2+2xz=z\\z^2+2xy=y$$
You can solve this by hand if you want, but Wolfram|Alpha gives $8$ solutions.
I imagine there is a better way to do this.
