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What are the units of $\mathbb{Z}/4\mathbb{Z}[x]$? Anything of the form $\pm 1 + 2 x p(x)$ for $p(x) \in \mathbb{Z}_4[x]$ works, and is in fact its own inverse. It's easy to see that any unit must have constant coefficient $\pm 1$ and, if the unit is not constant, it must have highest non-zero coefficient $2$, which lends credence to the above form. I can show that if a unit is its own inverse it must be of this form, but I'm unable to complete the proof. Just in case, I wrote a script to assist me in finding counterexamples to this form and found none. I've also searched and found nothing useful.

If my form is correct, then the nontrivial units of $\mathbb{Z}_4[x]$ are precisely those elements with multiplicative order 2. Just since I'm curious, is it more than a coincidence that 2 is the only zero divisor of this ring? That is, is there a characterization of the units of $\mathbb{Z}_m[x]$ (or even a more general structure) which includes this observation as a special case?

The $\mathbb{Z}_4$ case is problem 3.10 from D.J.H. Garling's A Course in Galois Theory. I've otherwise completed the book and am now going back through to find answers to the questions I wasn't able to finish. Hints are certainly welcome in addition to solutions.

Thanks for any help!

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See…. Though the question is slightly different, the answer there is a complete answer to this question (essentially the same as Zev's answer below). – Anton Geraschenko Mar 24 '11 at 17:57
Awesome, thank you. It seems I should have widened my search to more general structures. – Josh Swanson Mar 25 '11 at 9:56
up vote 5 down vote accepted

Yes, these are the only units. Consider the obvious map ${\bf Z}_4[x] \to {\bf F}_2[x]$; if $f(x)\in ({\bf Z}_4[x])^*$ then $\psi(f(x))\in ({\bf F}_2[x])^*$ and so $\psi(f(x)) = \pm 1$. Thus $f(x) \pm 1 \in \ker(\psi)$ and so $f(x) = \pm 1 + 2g(x)$ for some $g(x)\in {\bf Z}_4[x]$.

More generally this works for polynomial rings over any local ring $L$ in which $rad(L)$ is nilpotent; then the units are of the form $ (\mbox{unit of} \ L) + (\mbox{something in} \ rad(L))g(x)$.

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Thank you for the elementary proof! – Josh Swanson Mar 25 '11 at 10:25
I noticed a very minor mistake: the $\psi(f(x)) = \pm 1$ should not have the $\pm$. I would ignore it if it weren't a convenient point for me to ask how to edit an answer. (The FAQ doesn't seem to be relevant and I don't see an "edit" button.) – Josh Swanson Mar 25 '11 at 10:53
I think you may need more 'reputation points' in order to edit answers but I'm not sure. – Seamus Mar 28 '11 at 10:03

Problem 1.2i in Atiyah-Macdonald states that for any ring $A$ and $f=a_0+\cdots+a_nx^n\in A[x]$, $f$ is a unit $\iff$ $a_0$ is a unit in $A$, and the $a_i$ are nilpotent.

Note that $A=\mathbb{Z}/n\mathbb{Z}$ has nontrivial nilpotents $\iff$ $n$ is not squarefree.

In your specific case, $\mathbb{Z}/4\mathbb{Z}$ has 0 and 2 as nilpotents, and 1 and 3 as units, hence the units of $(\mathbb{Z}/4\mathbb{Z})[x]$ are, as you predicted, exactly the polynomials of the form $\pm1+2xp$ for $p\in(\mathbb{Z}/4\mathbb{Z})[x]$.

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Ah, thanks for the correction Martin. – Zev Chonoles Mar 24 '11 at 10:26
The statement from Atiyah-Macdonald even holds for every graded ring $A = \bigoplus_{n \ge 0}A_n$ with unit: $a=(a_n)_{n \ge 0} \in A$ is a unit iff $a_0$ is a unit and $a_n$ is nilpotent for $n > 0$. – Ralph Mar 24 '11 at 21:42
Thank you! The result you quoted jogged my memory. It's also problem 7.3.33(a) of Dummit and Foote (3rd Ed.). Part (b) gives the related result that $p(x) \in R[x]$ is nilpotent if and only if each coefficient is nilpotent (which was much easier to show). – Josh Swanson Mar 25 '11 at 10:18

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