TITLE: Is $\mathbb{Z}[x] / \langle (x^2 + 1)^2 \rangle$ isomorphic to a familiar ring?
QUESTION [9 upvotes]: The quotient ring $\mathbb{Z}[x] / \langle (x^2 + 1)^2 \rangle$ was brought up in class today to contrast it with $\mathbb{Z}[x] / \langle x^2 + 1 \rangle$ after a discussion about adjoining elements to rings.  That is, the second quotient ring given here is of course isomorphic to $\mathbb{Z}[i]$.  We are adjoining $i$ to $\mathbb{Z}$ because $i$ satisfies $x^2 + 1 = 0$.  However, $i$ also satisfies $(x^2 + 1)^2 = 0$.  But surely $\mathbb{Z}[x] / \langle (x^2 + 1)^2 \rangle \not\cong \mathbb{Z}[x] / \langle x^2 + 1 \rangle$.  So is $\mathbb{Z}[x] / \langle (x^2 + 1)^2 \rangle$ isomorphic to some familiar ring?
I've tried to think of some homomorphisms from $\mathbb{Z}[x]$ to other rings (e.g. $\mathbb{Z}[i] \times \mathbb{Z}[i]$) trying to get a kernel of $\langle (x^2 + 1)^2 \rangle$, but no luck.  Intuitively, it seems like we are "adjoining $i$ twice", to get a sort of $4$ dimensional structure, as compared to $\mathbb{Z}[i]$ which is like a $2$ dimensional structure.
EDIT:  A thought: perhaps the quotient is isomorphic to the Lipschitz quaternions?  I can't seem to prove this claim.

REPLY [9 votes]: The ring
$$ \mathbb{Q}[x] / (x^2 + 1)^2 $$
is much easier to understand: with some work (e.g. Newton's method), we can discover that it contains a square root of unity:
$$ i := \frac{1}{2} x (x^2 + 3) $$
We can rewrite the ring as
$$ \mathbb{Q}(i)[x] / (x - i)^2 $$
This ring has an evident square root of zero (see the "dual numbers"), which I will call $e$. So this ring is simply
$$\mathbb{Q}[i, e]$$
The image of $x$ in this ring is $i+e$. We can verify that we get a homomorphism by expanding to check
$$ ((i+e)^2 + 1)^2 = 0$$
This ring can't be written as a product of rings: if we try to solve for idempotents
$$ (a + be)^2 = (a + be) $$
where $a,b \in \mathbb{Q}(i)$, the only solutions are $b=0$ and $a=0,1$.

Going back to the original ring, we can write
$$ \mathbb{Z}[x] / (x^2 + 1)^2 \cong \mathbb{Z}[i + e] $$
Of course, maybe this doesn't qualify as a 'familiar' ring....