Invertible elements of $\mathbb{Z}_3[x] / (x^4+x^3-1)^3$

Question

Let $F=\mathbb{Z}/3\mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.

I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$

I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):

$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)

$|R|$ = $\mathrm{3}^{12}$

$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)

Then:

$\mathrm{3}^{12} = 3^4 \cdot |M|$ so $|M| = 3^8$

But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?

Answer 1

Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis $\{1,x,x^2,x^3\}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.