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?

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.
