Which ideal gives a non trivial quotient ring?

Question

This was an assignment problem and my answer turned out to be correct but I want to know if my concepts are accurate or if I got lucky since this was an MCQ. This question has been asked here and answered using homomorphism theorems but the professor hasn't covered it and I can't understand homomorphism theorems with respect to quotient rings.

For which ideal does the quotient ring become a non-trivial ring?

Options are $(1),(x),(x^2),(x+1)$

The ring consists of $1+(x^3+1), x+(x^3+1),x^2+(x^3+1), 1+x+ (x^3+1), 1+x^2+(x^3+1), x+x^2+(x^3+1),1+x+x^2+(x^3+1), (x^3+1) $

Now I've to consider the quotient ring of this ring with respect to other ideals.

The elements of the new quotient ring will be of the form $r+I$ where $I$ is the ideal we are considering. For $(1)$, it's trivial to see all elements are the same.

For $(x), (x^3+1)/(x)\equiv 1+(x^3+1)+(x)\equiv 1+(x)$ and all elements of $\mathbb{Z}_2[x]/(x^3+1)/(x)$ become equal to $1+(x)$.

Not really sure what I'm doing here but it's something similar to congruence. The only problem is I'm finding congruence class of a congruence class

For $(x^2), (x^3+1)/(x^2) \equiv 1+(x^2)$ and $1+(x^3+1)/(x^2)$ become equal to $1 + (x^2), x+(x^3+1)/(x^2)$ becomes equal to $x+x^4 - x^4+(x^3+1)/(x^2) \equiv -x^4 + (x^3+1)/(x^2) \equiv 1 + (x^2),x^2+(x^3+1)/(x^2) $ becomes equal to $1+(x^2+1), 1+x+ (x^3+1)/(x^2)$ becomes equal to $1+(x^2)$, so on and so forth.

I'm not sure about this though. Can someone check if this is correct?

Answer 1

Your computations make no sense to me so instead of commenting them, let us start from the beginning. Your ring is already a quotient ring, $A=$$\mathbb Z_2$$[x]/(x^3+1)$ and you want to re-quotient it by $I=(P)$ with $P=1,x,x^2$ or $x+1$ (more precisely, since $I$ is an ideal of $\mathbb Z_2[x]$, by its image in $A$, but this abuse of notation is usual).

$A/I=\mathbb Z_2[x]/(x^3+1,P)$ (intuitively, this means that in your new quotient, not only $x^3+1$ will be identified to $0$, but also $P$ will), so you have to compute the ideal $(x^3+1,P)$ of $\mathbb Z_2[x]$ generated by the two polynomials $x^3+1$ and $P$. For this, you may use the following tool (for any polynomials $Q$ and $R$):$$(PQ+R,P)=(R,P).$$

• $(x^3+1,1)=(0,1)$ (applying the tool to $Q=x^3+1$) and $(0,1)=(1).$ (Similarly, $(S,1)=(1)$ and $(0,S)=(S)$, for any polynomial $S$.)
• $(x^3+1,x)=(1,x)$ (applying the tool to $Q=x^2$), and $(1,x)=(1)$ (see previous bullet).
• $(x^3+1,x^2)=(1,x^2)$ (applying the tool to $Q=x$), and $(1,x^2)=(1).$
• $(x^3+1,x+1)=(0,x+1)$ (applying the tool to $Q=x^2-x+1$), and $(0,x+1)=(x+1).$

$\mathbb Z_2[x]/(1)$ is the trivial (null) ring.

$\mathbb Z_2[x]/(x+1)$ is isomorphic to $\mathbb Z_2.$