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!



What are the units of $\mathbb{Z}/4\mathbb{Z}[x]$?
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)$.


