In my book, it is claimed without proof that if a ring (commutative with identity) $R$ has zero divisors, then $R[x]$ has invertible polynomials of degree greater than $0$. They give the following example in $\mathbb{Z}_4$: $$(1+2x)(1+2x)=1$$

This seems like a pretty lucky example: in this ring we have a zero divisor whose sum with itself is $0$. In general this doesn't seem like an easy question to tackle. First i thought that maybe could find some $a$ such that $a^2 = 0$ and use the factorization $(1+ax)(1-ax)$ but in general such $a$ does not exist: $\mathbb{Z}_6$ is a counterexample. No polynomial of form $(1+ax)$ has a degree $1$ inverse in $\mathbb{Z}_6[x]$ either. Trying out higher degrees seems like a nightmare.

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    I think having zero divisors is not enough, we actually need nilpotent elements to get non-constant polynomials that are units. So all the units in $\Bbb Z_6[X]$ have degree $0$ as $6$ is square-free. For a characterization of polynomials that are units, see [here](https://math.stackexchange.com/questions/19132/characterizing-units-in-polynomial-rings) – leoli1 Feb 26 '21 at 16:40

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This is false. A polynomial is a unit if it has the form $\sum_{i=0}^k a_i x^i,$ where $a_0$ is a unit and $a_i$ is nilpotent for $i>0.$ (see this problem set). In the ring $R=\mathbb{Z}_6$ there are no nilpotents, so there are no invertible polynomials of degree greater than $0.$

Igor Rivin
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