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.
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just multiply in $\mathbb{Z}_3[x]/I$ – JJR May 08 '17 at 21:00
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What don't you understand? There are at least three different statements bundled up here: that those elements are units, that the listed pairs are multiplicative inverses, and that there are no other units beseides these. Do you know why any of these statements are true? Do you know how you could check any of them? – Eric Wofsey May 08 '17 at 21:50
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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 $u$.
lhf
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See also https://math.stackexchange.com/questions/60969/every-nonzero-element-in-a-finite-ring-is-either-a-unit-or-a-zero-divisor. – lhf May 08 '17 at 22:47