Commutative ring $R$ such that $R[x]$ has nonconstant units. I have an exercise asking to give an example of a commutative ring $R$ such that $R[x]$ has nonconstant units. At first glance, surely $\mathbb{Z}[x]$ is a commutative ring and would give nonconstant units, since the multiplicative inverse of a polynomial is nonconstant. Perhaps I am misunderstanding the question? I don't want the answer flat out, unless I am correct, I am just looking for guidance.
 A: Hint: They mean polynomials of the form $a_nx^n+\cdots+a_0$ with $a_n\ne 0$ and $n>1$. What think about a suitable $\mathbb{Z}_n$ for which $2x+1$ is its own inverse!
A: The  ring $R[x]$ contains  a non-constant invertible polynomial $f(x)\in R[x]\setminus R$ if and only if the ring $R$ is non reduced, that is if there exists  some non-zero $0\neq r\in R$ and some $n\gt 0$ with $r^n=0$.
This a not  so easy  theorem , given as exercise 2 in Chapter 1 of Atiyah-Mcdonald's Commutative Algebra.
Taking this theorem as a guide, it is very easy ( i.e. without solving the exercise!) to give an example of such an invertible $f(x)$ in a non-reduced ring of your choice: Alex gives you an indication for a possible answer.
A: You're misunderstanding the question; there is no multiplicative inverse of a non-constant polynomial in $\mathbb{Z}[x]$. That is, $\frac{1}{x}$ (for example) is not an element of $\mathbb{Z}[x]$, so $x$ is not a unit in $\mathbb{Z}[x]$.
Just to be clear: for a ring $R$, the polynomial ring $R[x]$ is
$$R[x]=\{a_0+a_1x+\cdots+a_nx^n\mid a_i\in R,\, n\geq 0\}$$
and for a ring $S$, an element $a\in S$ is a unit when there is some $b\in S$ such that $ab=1$.
A: Here's a simple example of a nonconstant unit polynomial: $\rm\:   1\, =\, (1-nx)(1+nx)\ $ over $\rm\:\Bbb Z/n^2.$
Generally, $\rm\ r_0 \!+ r_1 x\! +\:\cdots + r_n x^n\ $ is a unit in $\rm\:R[x] \iff r_0\:$ is a unit and $\rm\: r_k $ is nilpotent for $\rm\ k\ge 1.$
Proof $\ $ (sketch) $ $ If $\rm\:R\:$ is a domain then easily $\rm\:f(x)\:$ a unit $\rm\Rightarrow r_k = 0\:$ for $\rm\:k>0\:.\: $ Now $\rm\ R\to R/P,\ $ for $\rm\,P\,$ prime, reduces to the domain case, yielding that the $\rm\:r_k,\ k>0\:$ are in every prime ideal. But the intersection of all prime ideals is the nilradical, the set of all nilpotent elements -  see here.
See also this post on the general method of reduction to domains by factoring out prime ideals.
