The unit group of a polynomial ring and the unit group of the coefficient ring Show that if $R$ is an integral domain then $R[X]^\ast=R^\ast$. Show that this is false if $R$ is not an integral domain.

Note1: this is not homework.Perhaps I must have mentioned that my biggest problem with attempting this question was the interpretation of $R[X]^\ast$ and $R^\ast$. It's for this reason that I noted this below yesterday. 

Note2: this question actually comes from the notes of Iain Gordon(http://www.maths.ed.ac.uk/~igordon/). 

$R^*$ here means the group of units of the ring $R$.
Suppose $f(x)\in R[x]$ is invertible; write it as $f(x)=a+xg(x)$ and suppose $b+xh(x)$ is the inverse. Then
$$
ab+x(ah(x)+bg(x)+xg(x)h(x))=1
$$
and, comparing alike terms, we get $ab=1$. Therefore $a$ is invertible.

Until now we have not used the fact that $R$ is a domain. Suppose it is and that $f(x)$ has positive degree. Then $f(x)k(x)$ has positive degree: indeed the leading coefficient of the product is the product of the leading coefficients, which cannot be zero. Thus an invertible polynomial must have degree $0$ and, by what we saw before, it is an invertible constant.

Note that we are identifying constant polynomials in $R[x]$ with $R$.

In order to find a counterexample, we look at the simplest nondomain, that is, $R=\mathbb{Z}/4\mathbb{Z}$. The first case we can try is $1+2x$: the constant term must be invertible, and the leading coefficient must be a zero divisor, by the arguments above.

Now
$$
(1+2x)^2=…
$$



The general result that, if $R$ is not a domain, then the set of invertible polynomials contains non constant polynomials is false. See https://math.stackexchange.com/a/30390/62967, where it is proved that a polynomial $a_0+a_1x+\dots+a_nx^n$ is invertible if and only if $a_0$ is invertible and $a_i$ is nilpotent for $i&gt;0$.

So if $R$ is not a domain but it has no (nontrivial) nilpotent element, then the invertible polynomials are just the invertible constants. A ring with such a property is the product of two fields, for instance.
egreg's argument above shows that any unit in $R[x]$ is also a unit in $R$.  If $R$ has no multiplicative identity, then neither does $R[x]$.

Since the degree-polynomials form a subring of $R[x]$ that is isomorphic to $R$, then $R[x]$ is commutative if and only if $R$ is commutative. Similarly, if $a, b \in R$ are non-trivial divisors of 0, then $a, b \in R[x]$ are likewise non-trivial divisors of 0.

Summarizing, if $R$ lacks any of the properties that distinguish a ring from an integral divisor, then $R[x]$ also lacks that property.