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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.

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  • $\begingroup$ You are adjoining a nonzero square root of zero, which we could call $\tilde{0}$. Also, you are adjoining a number we could call $\tilde{\imath}$, which is a square root of $-1+\tilde{0}$. $\endgroup$ – alex.jordan Feb 1 '14 at 1:04
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    $\begingroup$ Regarding your edit, I don't think you can find any square roots of $-1$ in your ring, or square roots of $0$ in the Lipschitz quaternions. Indeed the latter lacks zero divisors...and isn't even commutative! $\endgroup$ – Kevin Carlson Feb 1 '14 at 1:10
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    $\begingroup$ @Kevin: How interesting! $(1/2) x (x^2 + 3)$ is a square root of $-1$, but alas that lies in the ring of fractions, not in the ring itself! $\endgroup$ – Hurkyl Feb 1 '14 at 1:28
  • $\begingroup$ ... so we can get an alternate characterization of this ring as $$\mathbb{Z}[2i][x] / \langle x^2 - 2ix - 1 \rangle $$ $\endgroup$ – Hurkyl Feb 1 '14 at 1:36
  • $\begingroup$ Very related: math.stackexchange.com/questions/414831 $\endgroup$ – Martin Brandenburg Feb 1 '14 at 10:04
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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....

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  • $\begingroup$ (+1) Very cool. That ring isn't familiar to me (I'm sure it is to someone), but it is still satisfying that there is a tangible ring to point to. Thanks! $\endgroup$ – tylerc0816 Feb 1 '14 at 3:04
  • $\begingroup$ Ah, that's great. $\endgroup$ – Kevin Carlson Feb 1 '14 at 9:49
  • $\begingroup$ Since $\mathbb{Q}(i,e)$ is not a field, I would choose another notation. $\endgroup$ – Martin Brandenburg Feb 1 '14 at 10:09
  • $\begingroup$ If $k$ is a commutative ring with $2 \in k^*$, then there is an isomorphism of $k$-algebras $k[x]/(x^2+1)^2 \cong k[i,e]/(i^2+1,e^2)$. See math.stackexchange.com/questions/414831 for a proof. $\endgroup$ – Martin Brandenburg Feb 1 '14 at 10:11
  • $\begingroup$ @Martin: Thanks. I had typed $\mathbb{Q}[i,e]$, but accidentally went through the habit of changing the $[]$ to $()$. :( $\endgroup$ – Hurkyl Feb 1 '14 at 19:14

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