Determine all ideals of $ \ \large \frac{\mathbb{Z}_3[x]}{(x^3+x+2)} \ $ Determine all ideals of $ \ \large \frac{\mathbb{Z}_3[x]}{(x^3+x+2)} \ $.
Answer:
The polynomial  $ \ x^3+x+2 \ $ is not irreducible in $ \ \mathbb{Z}_3 \ $ .
So we have to now factorise $ \ x^3+x+2 \ $ in order to get ideals.
But I can't factorise the polynomials andcan not find the ideals .
Please help me out.
 A: The ideals of any quotient ring $A/I$ coresponds to the ideals of $A$ that contain $I$. The ideals of $\Bbb Z_3[x]$ that contain $(x^3+x+2)$ are $(x^3+x+2), (x-2),(x^2-x-1)$ and $(x^2-x-1, x-2) = \Bbb Z_3[x]$. Divide each of these out by $(x^3+x+2)$, and you have your four ideals.
As for the Chinese reminder theorem (which, as it turns out, was unnecessary here), it says that the canonical ring homomorphism
$$
\Bbb Z_3[x]/(x^3+x+2)\to \Bbb Z_3[x]/(x-2)\times \Bbb Z_3[x]/(x^2-x-1)
$$
is an isomorphism. The ring on the right is a product of two fields (the first with three elements, the second with $9$), which makes looking for ideals easy (they are $(0), ((1,0)), ((0,1))$ and the whole ring)
A: The ideals of $\dfrac{\mathbb{Z}_3[x]}{(x^3+x+2)}$ correspond to the ideals of $\mathbb{Z}_3[x]$ that contain $(x^3+x+2)$.
Since $\mathbb{Z}_3[x]$ is a PID, the ideals of $\mathbb{Z}_3[x]$ that contain $(x^3+x+2)$ are the ideals generated by the factors of $x^3+x+2$.
We have $x^3+x+2= (x + 1) (x^2 + 2 x + 2)$ in $\mathbb{Z}_3[x]$. These factors are irreducible. Therefore, we have four the ideals of $\mathbb{Z}_3[x]$ that contain $(x^3+x+2)$:


*

*$(1)$

*$(x+1)$

*$(x^2 + 2 x + 2)$

*$(x^3+x+2)$
