Units in $\Bbb Z / 3 \Bbb Z[x] / I$ where $I=(x^3+2x^2)$

Question

I'd like to know how to find all the units in the quotient ring $R/I$ where $I=(x^3+2x^2)$ and $R=\Bbb Z / 3 \Bbb Z[x]$.

I know $$R/I = \{a_0+a_1x+a_2x^2 + I \; / \; a_0+a_1x+a_2x^2 \in R\},$$ so $R / I$ has $3^3=27$ elements.

Now, as $x^3+2x^2 = x^2(x+2)$ and $\Bbb Z / 3 \Bbb Z[x]$ is a PID, $I$ is not maximal. Thus, $R/I$ does not has $26$ units (it would have been so easy). How can I find all the units in $R/I$?

Answer 1

By Bézout's theorem, the units in $R/I$ correspond bijectively to polynomials of degree at most $2$ in $R$ that are coprime with $x^3+2x^2=x^3-x^2=x^2(x-1)$. So look for polynomials $ax^2+bx+c$ that don't have $0$ or $1$ as roots, that is, you need $c\ne0$ and $a+b+c\ne0$. Choosing $a=0$ is allowed.