Abstract Algebra - Monoid

Here's a question.

I understand how to prove that it's associative but I don't understand how it's proved to have an identity element.

Also, I understand that for identity, $(x,y,z) * (e,f,g) = (e,f,g) * (x,y,z) = (x,y,z)$

Here's how it's proved.

Where did the $(0,0,1)$ come from?

Also the $(1,1,1)$ used to prove that it's not invertible, where did the that come from?

-

This is the multiplication of the ring $\mathbb{R}[x]/(x^3-1)$, written in the basis $x^2,x,1$. Thus it is a commutative monoid, but not a group (there is an absorbing element, the zero). No fiddly calculations are needed.

-

See the definition of the binary operation. Remember * is not the usual multiplication.

In the first case, you have $b_1, b_2=0$ and $b_3=1$, then $(a_1,a_2,a_3)*(0,0,1)=(a_1,a_2,a_3)$.

You have to pay attention of the definition of the binary operation *

Now you have to substitute $b_1,b_2=0$ and $b_3=1$ in the expression: $(a_1b_3+a_2b_2+a_3b_1,a_1b_1+a_2b_3+a_3b_2,a_1b_2+a_2b_1+a_3b_3)$

See that so many terms will be canceled because $b_1,b_2=0$ and you will have just $(a_1,a_2,a_3)$

-
I don't understand how $b1, b2=0$ and $b3=1$. I don't see how the definition tells me about the identity element. –  Adegoke A Apr 22 '13 at 10:35
@AdegokeA I will enlarge my answer –  user42912 Apr 22 '13 at 10:38
@AdegokeA Did you understand know? –  user42912 Apr 22 '13 at 10:49
@AdegokeA Now you have to substitute $b_1,b_2=0$ and $b_3=1$ in the expression: $(a_1b_3+a_2b_2+a_3b_1,a_1b_1+a_2b_3+a_3b_2,a_1b_2+a_2b_1+a_3b_3)$ –  user42912 Apr 22 '13 at 10:58
@AdegokeA Because when we substitute in the same way we did with $(0,0,1)$ we have $(a_1,a_2,a_3)*(1,0,1)=(a_1+a_3,a_1+a_2,a_2+a_3)$ and remember you want to have a triple $(b_1,b_2,b_3)$ such that $(a_1,a_2,a_3)*(b_1,b_2,b_3)=(a_1,a_2,a_3)$ and (0,0,1) is the right triple having this property –  user42912 Apr 23 '13 at 10:52