Jacobson radical of $A[x]$.

Solution 1:

Careful! For instance, if $A=\mathbb{Z}/4\mathbb{Z}$, then $2x+1$ has positive degree, but it is a unit in $A[x]$. In fact, we have $(2x+1)^2=4x^2+4x+1=1$.

The fact that $\deg(f)>0$ implies that $f$ is not a unit, is not true in general. (It is true, however, if $A$ is a field.) One way to generalize that statement is the following (which is Exercise 1 from Chapter 1 of Atiyah-Macdonald):

Let $A$ be a ring, and $A[x]$ the ring of polynomials. Then $f=a_0+a_1x+\dots+a_nx^n$ is a unit if and only if $a_0$ is a unit in $A$, and $a_1, \dots, a_n$ are nilpotent.

Solution 2:

Thing is, if $A$ is not an integral domain then polynomials of positive degree can still be invertible. In general, a polynomial $\sum_{i=0}^n a_ix^i$ in $A[x]$ is invertible if and only if $a_0\in A^{\times}$ and $a_1,...,a_n$ are nilpotent.

For example, take $A=\mathbb{Z}/4\mathbb{Z}$. Then the polynomial $(2+4\mathbb{Z})x$ belongs to the Jacobson radical of $A[x]$. Indeed, take any $g\in A[x]$. Then the free term of $(1+4\mathbb{Z})-(2+4\mathbb{Z})xg(x)$ is a unit, while the other coefficients are nilpotent. (as they are divisible by the nilpotent elements $2+4\mathbb{Z}$)