Exhibit the ideals of $\mathbb{Z}[x]/(2,x^3+1)$

I start by trying to see the homomorphism between $\mathbb{Z}[x]$ and $\mathbb{Z_2}[x]$.I define the homomorphism by $\phi:(ax^i)=a(mod 2)x^i$.It is trivial to see that it a homomorphism. The kernel of the homomorphism is the ideal generated by $(2)$.The homomorphism is surjective also, so by the first isomorphism theorem we see that the two structures are isomorphic.

Hence we try to see the structure $\mathbb{Z_2}[x]/(x^3+1)$.The polynomials of this quotient ring is obtained by replacing $x^3 $ by $-1$, so we end up getting polynomials in 2nd degree. The possible polynomials are $x^2+x+1$,$x^2+1$,$x^2+x$,$x$,$x^2$,$x+1$,$1$.

$1$) So next we see that the ideal generated by $(x^2+x+1)$ is irreducible in $\mathbb{Z_2}$.By the third isomorphic theorem $\mathbb{Z_2}[x]/(x^3+1)/((x^2+x+1)/(x^3+1))$

is isomorphic to $\mathbb{Z_2}/(x^2+x+1)$ which is a field, as the polynomial is irreducible hence the only ideals are $0$ and itself -$\mathbb{Z_2}[x]/(x^2+x+1)$-So from this we can conclude that $(x^2+x+1)/(x^3+1)$ is an ideal in $\mathbb{Z_2}[x]/(x^3+1)$.

$2$)By the isomorphic theorem we see that $\mathbb{Z_2}[x]/(x^3+1)/(x) /(x^3+1)$
is isomorphic to $\mathbb{Z_2}[x]/(x)$ which is isomorphic to $ \mathbb{Z_2}$-a field.Then,$(x) /(x^3+1)$is also the only ideals.

3)Sinilarly I can conclude that $(x+1)/(x^3+1)$ is also an ideal.

$4$)Similarly,$(x^2+1)/(x^3+1)$ is also an ideal,But $(x^2+1)$ is not irreducible, then can I conclude something more than this?

Although there has been an answer to this question before but my attempt has been probably different from it and I have done it without consulting it, can someone just go through it and point out where I am wrong?

Let $I=(2,x^3+1)$ be the ideal in $\mathbb{Z}[x]$. Let $R$ be your ring $\mathbb{Z}[x]/I$. Since it is a quotient of $\mathbb{Z}/2\mathbb Z[x]$, all ideals of $R$ are principal. Let $(p(x)+I)$ be an ideal of $R$. If the degree of $p$ is at least $3$, we can reduce $p$ modulo $x^3+1$, so we can assume that the degree of $p$ is $0,1$ or $2$.We can also reduce all coefficients of $p$ modulo 2 and assume that all coefficients are $0$ or $1$. This gives the following possibilities for $p$: $0,1,x,x+1, x^2, x^2+1. x^2+x, x^2+x+1$. It is easy to check which of these ideals $(p+I)$ are different. For example since $x+I$ is invertible in $R$ ($xx^2\equiv 1 \mod I$) we have that $(x+I)=(x^2+I)=R=(1+I)$, $(x^2+1+I)= (x^3+x+I)=(x+1+I)$ and $(x+1+I)=(x^2+x+I)$. This leaves the following options for $p$: $0,1,x+1, x^2+x+1$. So there are at most 4 ideals in $R$. Note that $x^2+x+1+I$ is an idempotent in $R$ whose product by $ (x+1+R)$ is $0$. This implies that all four ideals are different.

Hint: The ideals of $\Bbb{Z}[x]/(2,x^3 +1)$ are of the form $I/(2,x^3+1)$ where $I$ is an ideal of $\Bbb{Z}[X]$ containing $2$ and $x^3 +1$. Hence, your job is to determine all such ideals.

Alternatively you can use the isomorphism $$\Bbb{Z}[x]/(2,x^3+1) \cong \Bbb{Z}_2[x]/(x^3+1)$$ and determine the ideals of the latter ring.