What are the ideals of the ring $\mathbb{Z}_3[x]/\langle x^4+x^3+x+1\rangle$?
I know that $\mathbb{Z}_3[x]$ is a PID. In addition, $(x+1)^4=x^4+x^3+x+1$ in $\mathbb{Z}_3[x]$. Therefore, our ring is nothing but $\mathbb{Z}_3[x]/\langle (x+1)^4\rangle$. But, now how to proceed further? Is the ring isomorphic to $\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3$? Whence its ideals would be four tuples of the form $(a,b,c,d)$ where each of $a,b,c,d$ is an ideal of $\mathbb{Z}_3$, which is either $(0)$ or $\mathbb{Z}_3$? Any hints? Thanks beforehand.
This is where the correspondence theorem is useful: there is a one-to-one order preserving correspondence between the of $\Bbb Z_3[x]/((x+1)^4)$ and ideals of $\Bbb Z_3[x]$ which contain $((x^4+1))$. Further, as you stated $\Bbb Z_3[x]$ is a PID, therefore ideals are of the form $I = (f)$ where $f \in \Bbb Z_3[x]$. So if an ideal $(f) =I \subset \Bbb Z_3[x]$ contains the ideal, $((x+1)^4)$, then $f$ must divide $(x^4+1)$. Therefore the ideals of $\Bbb Z_3[x]/((x+1)^4)$ are of the form $\pi(I)$, where $I = (f)$ is a polynomial(in $\Bbb Z_3[x]$ of course) which divides $(x+1)^4$ and $\pi: \Bbb Z_3[x] \to \Bbb Z_3[x]/((x+1)^4)$ is the quotient map.