Find all ideals in the following ring containing an ideal Can someone help me find the complete list of proper ideals in $\mathbb{Z}[x]$ containing $(x^3+1,6)$? I have been struggling on this for hours. I want to use this to find all ideals of $\frac{\mathbb{Z}[x]}{(x^3+1,6)}$. Should there be 15?
This is what I have:
$$(x^3+1,6)$$
$$(x+1,6)$$
$$(x^2-x+1,6)$$
$$(x^3+1,3)$$
$$(x^2-x+1,3)$$
$$(x+1,3)$$
$$(x^3+1,2)$$
$$(x^2-x+1,2)$$
$$(x+1,2)$$
$$(x+1,x^2-x+1,2)$$
$$(x+1,x^2-x+1,3)$$
$$(x+1,x^2-x+1,6)$$
Is this right so far? What are the others? This would really help me out. I know that $\frac{\mathbb{Z}[x]}{(x^3+1,6)} \cong \frac{\mathbb{Z}_2[x]}{(x^3+1)} \oplus \frac{\mathbb{Z}_3[x]}{(x^3+1)}$ but I do not know how this can help me. Is it possible for a principle ideal to be on this list? Can someone explain to me how to use $\frac{\mathbb{Z}[x]}{(x^3+1,6)} \cong \frac{\mathbb{Z}_2[x]}{(x^3+1)} \oplus \frac{\mathbb{Z}_3[x]}{(x^3+1)}$ to do this? I know the factors of $x^3+1$ are $x+1$ and $x^2-x+1$ so there are $16$ ideals determined by $(1),(x^3+1),(x^2-x+1),(x+1)$ but I cannot figure out how to use this with the primes $2,3$. I know the first $9$ are correct, but I am having trouble determining the others. I tried ideals of the form $(f(x),2,3)$ but I know this generates the whole ring.
 A: As you pointed out:
$$\frac{\mathbb{Z}[x]}{(x^3+1,6)} \cong \frac{\mathbb{Z}_2[x]}{(x^3+1)} \oplus \frac{\mathbb{Z}_3[x]}{(x^3+1)}$$
Now $$\frac{\mathbb{Z}_2[x]}{(x^3+1)} \cong \frac{\mathbb{Z}_2[x]}{(x+1)}\oplus \frac{\mathbb{Z}_2[x]}{(x^2+x+1)}\cong \mathbb{F}_2\oplus\mathbb{F}_4,$$
and   $$\frac{\mathbb{Z}_3[x]}{(x^3+1)}\cong \frac{\mathbb{Z}_3[x]}{(x+1)^3}\cong \frac{\mathbb{Z}_3[y]}{(y^3)}.$$
Each field has $2$ ideals, including the whole field, and $\frac{\mathbb{Z}_3[y]}{(y^3)}$ has $4$ ideals (including itself): $$(1),(y),(y^2),(0).$$
Thus in total you get $2\times 2\times 4=16$ ideals.  However one of these is the ring itself, so you have $15$ proper ideals in total.

To actually find these ideals, note that $$3+4=1,\qquad 3^2=3, \qquad4^2=4,\qquad {\rm in}\quad \frac{\mathbb{Z}[x]}{(x^3+1,6)}.$$  Thus $3\in\mathbb{Z}[x]$ represents the identity of $\frac{\mathbb{Z}_2[x]}{(x^3+1)}$ and $4\in\mathbb{Z}[x]$ represents the identity of $\frac{\mathbb{Z}_3[x]}{(x^3+1)}$.
Thus $$4,\qquad 4(1+x),\qquad 4(1+x)^2$$
represent the generators of the ideals in $\mathbb{Z}_3[y]/{y^3}$.
We have $$(x^2+x+1)-x(x+1)=1.$$
Thus $(x^2+x+1)\in \frac{\mathbb{Z}_2[x]}{(x^3+1)}$ represents the identity in $\mathbb{F}_2$ and $-x(x+1)\in \frac{\mathbb{Z}_2[x]}{(x^3+1)}$ represents the identity in $\mathbb{F}_4$.
Putting it all together, we obtain all $15$ ideals by:
$(1)$ Starting with $x^3+1,6$.
$(2)$ Adding one of $4, 4(1+x),4(1+x)^2$ or none of them.
$(3)$ Adding $3(x^2+x+1)$ or not.
$(4)$ Adding $-3x(x+1)$ or not.
$(5)$ Throw away the case where you added $4, 3(x^2+x+1), -3x(x+1)$ (as this is not a proper ideal).
A: By the ideal correspondence theorem, the ideals of $\mathbb{Z}[x]$ which contain $(x^3 + 1, 6)$ correspond to the ideals of $\mathbb{Z}[x]$ in a bijective manner.
So we need to find all the ideals of $\mathbb{Z}[x] / (x^3 + 1, 6)$. To do this, we will find the ideals of $\mathbb{Z}_2[x] / (x^3 + 1) \times \mathbb{Z}_3[x] / (x^3 + 1)$.
Note that over $\mathbb{Z}_3[x]$, we have $x^3 + 1 = (x + 1)^3$. So we are analysing the ideals of $\mathbb{Z}_2[x] / ((x + 1) (x^2 + x + 1)) \times \mathbb{Z}_3[x] / ((x + 1)^3)$.
The associate classes of $\mathbb{Z}_3[x] / ((x + 1)^3)$ are therefore $1 \mid (x + 1) \mid (x + 1)^2 \mid(x + 1)^3 = 0$.
And the associate classes of $\mathbb{Z}_2[x] / ((x + 1)(x^2 + x + 1))$ are $1$, $x + 1$, $x^2 + x + 1$, and $(x + 1)(x^2 + x + 1) = 0$. And let us note that $(x + 1, x^2 + x + 1) = (1)$.
In general, the ideals of $R_1 \times R_2$ are of the form $I_1 \times I_2$ where $I_i$ is an ideal in $R_i$.
This gives us 16 distinct ideals of the product ring - an ideal $(a, b)$ for each associate class pairing.
So there are indeed a total of 15 proper ideals.
To translate an ideal back to an ideal of $\mathbb{Z}[x]$, we must consider the map $f : \mathbb{Z}[x] \to (\mathbb{Z}_2[x] / (x^3 + 1), \mathbb{Z}_3[x]/(x^3 + 1))$. We take an ideal $I$ of $\mathbb{Z}_2[x] / (x^3 + 1), \mathbb{Z}_3[x]/(x^3 + 1)$ and take its inverse image $f^{-1}(I)$.
To compute all the ideals, we must first find one solution $f(P) = (x + 1, 1)$ and another for $f(Q) = (x^2 + x + 1, 1)$. We must then find one solution of $f(R) = (1, x + 1)$.
From here, for each $i \in \{0, 1, 2, 3\}$, each $j \in \{0, 1\}$, and each h $k \in \{0, 1\}$, we have the ideal $(P^j Q^k R^i, x^3 + 1, 6) = f^{-1}(((x + 1)^j (x^2 + x + 1)^k, (x + 1)^i))$.
This method gives us all 16 ideals of $\mathbb{Z}[x]$ which contain $6$ and $x^3 + 1$. We then throw out $(1)$ to get the 15 necessary ideals.
