Characterization for quotients of the form $\mathbb{Z}[x]/p(x)$ to be isomorphic Question: If $\mathbb{Z}[x]/(p(x)) \cong \mathbb{Z}[x]/(q(x))$ as rings(assuming $1$ goes to $1$) then is $p(x)= q(\pm x + a)$ for some $a \in \mathbb{Z}$?
Certainly the converse is true.
For quadratics the above seems to hold for the examples I tried meaning when I take $p(x)$ and $q(x)$ which are not affine translates of each other then the quotients are not isomorphic.
Note that for quadratics the condition is same as having equal discriminants. Also for any isomorphism descending from a surjective map from $\mathbb{Z}[x] \to Z[x]$ answer is yes for such maps  $x \to \pm x +a$ (else we won't have surjection).
I came to this conjecture from first noticing that $\mathbb{Z}[x]/(p(x)) \cong \mathbb{Z}[x]/(q(x))$ implies that the degrees of $p(x)$ and $q(x)$ have to be the same by looking at the induced $\mathbb{Z}$ module isomorphism. Also if one is monic then so is the other (non monics give a non-finitely generated $\mathbb{Z}$ module).
If the answer is no, then are there some other weaker sufficient conditions for the quotients to be isomorphic?
 A: For a monic counter-example, let $\beta$ be a root of $f=x^4+x^2+x+1$ which is irreducible. Clearly $$\Bbb{Z}[\beta] = \Bbb{Z}[\beta^2]$$
The minimal polynomial of $\beta^2$ is $$g = y^4 + 2y^3 + 3y^2 + y + 1$$ so we have $$\Bbb{Z}[x]/(f)\cong \Bbb{Z}[y]/(g)$$
One can check that $g$ has only one root in $\Bbb{Q}(\beta)$, so $f= g(\pm x+a)$ would give $\beta^2=\pm \beta + a$ contradicting that $[\Bbb{Q}(\beta):\Bbb{Q}]=4$.
A: Reuns has given a non-monic counterexample in the comments. I will focus on the monic case here, but I won't give a conclusive answer; I believe someone with a strong background in algebraic number theory should be able to either find a counterexample or at least shed light on the problem with the framing I'm giving.
It will be impressive if this is true for $p, q$ monic, since it's blatantly untrue when you replace $\mathbb Z$ with $\mathbb Q$:
The analogous question over $\mathbb Q$ can be phrased as follows: if $p, q \in \mathbb Q[x]$ satisfy $\mathbb Q[x]/(p(x)) \cong \mathbb Q[x]/(q(x))$, then $p(x)=aq(bx+c)$ for some $a, b, c \in \mathbb Q$ ($a$ and $b$ necessarily being invertible except in the degenerate case of constant polynomials).
Assuming $p(x)$ is irreducible and we're searching for such a $q$, then $K := \mathbb Q[x]/(p(x))$ is a field extension of $\mathbb Q$; for simplicity let's assume it's Galois. Let $\alpha_1, \ldots, \alpha_n$ be the Galois conjugates of $x$ (so $\deg p = n$). The question becomes: is there a generator of $K$ which is not of the form $a \alpha_i + b$ for some $i=1, 2, \ldots, n$?
Let $F_1, \ldots, F_k$ be all the proper subfields of $K$ (there are finitely many due to Galois correspondence).
Then we are simply seeking an element of $K$ which is not in the finite union of subspaces
$$\bigcup_{i=1}^k F_k \cup \bigcup_{i=1}^n (\mathbb Q \alpha_i + \mathbb Q).$$
It's a fact of linear algebra that a vector space over an infinite field is not a union of proper vector-subspaces, so many such elements of $K$ exist as long as $n>2$ (otherwise the RHS subspaces are not proper). So in the nicest possible setting over $\mathbb Q$, with the exception of quadratic extensions, there will be many counterexamples.
Of course, such counterexamples over $\mathbb Q$ can't necessarily be lifted to $\mathbb Z$; suppose that we take $p, q$ to be the irreducible polynomials of $\alpha, \beta \in K$ as above, and suppose we take $\alpha, \beta$ to be algebraic integers so that $p, q$ have integer coefficients (this can be done by scaling $\alpha, \beta$ appropriately). Then we have embeddings
$$\mathbb Z[x]/(p(x)) \cong \mathbb Z[\alpha], \mathbb Z[x]/(q(x)) \cong \mathbb Z[\beta] \hookrightarrow K=\mathbb Q[x]/(p(x)) \cong \mathbb Q[x]/(q(x)),$$
but there's no reason to suppose that these two lattices inside $K$ are equal, and unequal lattices won't necessarily give isomorphic extensions of $\mathbb Z$ (e.g. $\mathbb Z[\sqrt{-1}] \not\cong \mathbb Z[2\sqrt{-1}]$). But if we can get the lattices to be equal, then we find a counterexample over $\mathbb Z$.
The scope of the search having been defined, here's a couple attempts by me that don't pan out:
First consider $K = \mathbb Q(\sqrt 2, \sqrt 3)$ and $\alpha = \sqrt 2 + \sqrt 3$. Then $\mathbb Z[\alpha]$ has $\mathbb Z$-basis given by $1, \sqrt 2, \sqrt 3, 2\sqrt 6$. Consider $\beta = \sqrt 2 + \sqrt 3 + 2 \sqrt 6$. Then it's easy to see, by taking powers and subtracting integer parts, that $\mathbb Z[\beta]$ contains $12 \sqrt 2 + 8 \sqrt 3 + 2 \sqrt 6$ and $83 \sqrt 2 + 81 \sqrt 3 + 78 \sqrt 6$. Taking fourth powers or higher won't gain anything for degree four algebraic integers, so it's just a matter of what the span of these three vectors is. Unfortunately, they span a proper subspace of that spanned by $\sqrt 2, \sqrt 3, 2 \sqrt 6$ (easy to check by checking the determinant of the appropriate $3 \times 3$ matrix).
Another obvious thing to try with the same $K$ is $\alpha = \sqrt 2 + \sqrt 3 + \sqrt 6$ and see whether an appropriate $\beta$ exists. It turns out that $\mathbb Z[\alpha]$ has basis $\sqrt 2 + \sqrt 3 + \sqrt 6, 2 \sqrt 3, 4 \sqrt 2$. An alternate generator must include the first basis element, and trying to throw in other basis elements without getting the same irreducible polynomial is fruitless.
There are three options that I see to continue the search for counterexamples:

*

*Stick to the scope of considering $\mathbb Q$-generators of Galois extensions $K/\mathbb Q$,

*Expand to the scope to considering $\mathbb Q$-generators of non-Galois field extensions $K/\mathbb Q$,

*Expand the scope to considering $p, q$ not irreducible.

All three are potentially fruitful for different reasons; the first approach is tempting because it only requires the easiest setting of Galois theory and seeking out a scenario where a sublattice is full. It's that last part that gives trouble unfortunately. The second approach might honestly be even easier, because there will be "fewer" extensions $F_k$ to consider in the union above; to do Galois theory properly does require taking a Galois closure, but lattice of subfields is all you really need here. And the third approach is something else entirely (probably equivalent to the first two if you take $p, q$ to be square free).
