A unit in $\mathbb{Z}_{9}[x]$ Show that $1+3x$ is a unit in $\mathbb Z_9[x]$. Hence corollary 4.5 may be false if $R$ is not an integral domain. 
The corollary 4.5 is:
Let $R$ be an integral domain and $f(x) \in R[x]$. Then $f(x)$ is a unit in $R[x]$ if and only if $f(x)$ is a constant polynomial that is a unit in $R$. In particular, if $F$ is a field, the units in $F[x]$ are the nonzero constants in $F$.
The biggest problem I am having is finding how what it means to be a unit in $\mathbb{Z}_{9}[x]$. If I knew that I think I could figure it out from there. 
 A: A unit of a (commutative) ring $R$ is an element $u \in R$ such that there exists $v \in R$ such that $$uv = vu = 1$$
Informally, the units of $R$ are the invertible elements of $R$ under the multiplication operation. With this in mind, a hint is: consider $1+6X \in (\mathbb{Z}/9\mathbb{Z})[X]$.
A: First note that the corollary you cite requires $R$ to be an integral domain, which is not the case here. $\mathbb Z_9$ has a zero-divisor ($3 \cdot 3=0$).
Being a unit is the same as having an inverse.
So there should exist some polynomial $g(x) \in \mathbb Z_9[x]$ such that $(1+3x)g(x)=1$.
Try to write out what this means and find this polynomial.
A: You want to find a polynomial which when multiplied by $1+3x$ gives $1$ modulo 9.
You can try to guess with something simple: $a+bx$. So you want $(1+3x)(a+bx)\equiv1$ modulo 9.
Multiply this out and you'll get some equations for $a$ and $b$ modulo 9. You may find at this point that you can guess $a$ and $b$.
A: Here's how I would come up with a solution.
Formally
$$\frac{1}{1+3x}=\frac{1}{1-(-3x)}=\sum_{n=0}^{\infty}(-3x)^n=1+(-3x)+9x^2-27x^3\dots=1-3x$$ when $9$ is set to zero. It is relatively easy to show such logic will always work in cases like this (when the first coefficient is a unit and the rest are nilpotent) and showing the more general characterization of units of polynomial rings (see the first couple exercises in Atiyah-Macdonald) can be done using a similar argument.
A: $(1+3x)(1-3x) = 1-9x^2$. Using congruence mod 9 and it's done
