Units in a quotient ring

Question

Let $S=\mathbb{Z}_3[x]/I$, where $I=\mathbb{Z}_3[x](x^2-x+1)$. The units and their corresponding multiplicative inverses are

$[x]$, $[-x+1]$

$[-x]$,$[x-1]$

$[-1]$, $[1]$

I don't really understand why this is the case.

Answer 1

Let $u=x+1 \bmod I$. Then $S=\mathbb Z_3 [u]$ with $u^2=0$, which is easier to work with.

The elements of $S$ are of the form $a+bu$, with $a,b \in \mathbb Z_3$ and so $S$ has nine elements.

No element of the form $bu$ is a unit because $u^2=0$.

The other six elements are units:

• $1 \pm u$ are mutual inverses.

• $-1 \pm u$ are mutual inverses.

• $\pm 1$ are mutual inverses.

which is your list in terms of $u$.