Possible Alternate Proof for Property of Polynomials with Nilpotent Coefficients? So I am currently working through Dummit & Foote exercise #7.3.33 on my own, and I ended up trying something that I couldn't find anywhere else online (there are a few posts on this problem, but I want to know if this particular method works), and wasn't sure how to make it work. I'm very curious to know if there's a way to get a solution out of this!
So the question states: (Let $R$ be a ring.)


*Assume $R$ is commutative. Let $p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_0$ be an element of the polynomial ring $R[x]$.


Prove that $p(x)$ is a unit in $R[x]$ if and only if $a_0$ is a unit and $a_1, a_2, ... , a_n$ are nilpotent in $R$.
I suppose I could have done something much simpler, but I ended up using the following:


*

*Part (b) of the same problem, $p(x)$ nilpotent in $R[x] \Leftrightarrow a_0, a_1, ... , a_n$ nilpotent in $R$. (I did this part first)

*Exercise 29, the collection of nilpotent elements $N(R)$ form an ideal in $R$.

*Exercise 17(b), if $\phi: R \rightarrow S$ is a ring homomorphism, and $\phi(1_R) = 1_S$, then $\phi$ maps units to units.

*Exercse 7.1.14, miscellaneous properties about nilpotents.


Here's my attempt:
Let $p(x) \in R[x]$, and $p(x) = \sum_{k=0}^{n} a_kx^k$. Let $N(R)$ denote the nilradical, or the ideal of $R$ consisting of all nilpotents. 
We consider the homomorphism $\phi: R[x] \rightarrow R/N(R)[x]$ defined by $\phi: \sum_{k=0}^{n} a_kx^k \mapsto \sum_{k=0}^{n} \overline{a_k}x^k$, where $\pi(a_k) = \overline{a_k} $, and $\pi$ is the natural projection from $R$ to $R/N(R)$.
It is clear that $\pi(1) = \overline{1}$.
$(\Leftarrow)$
Let $a_0$ be a unit of $R$, and $a_1,a_2,...,a_n \in N(R)$. By part (b), $\sum_{k=1}^{n} a_{k}x^k$ is nilpotent, and from exercise 7.1.4, the sum of a nilpotent element and a unit is a unit, so $p(x)$ is a unit.
$(\Rightarrow)$
Let $p(x)$ be a unit of $R[x]$. We assume the contrary, and that $a_0$ is not a unit, or at least one of $a_1, ... , a_n$ is not nilpotent.
If $a_0$ is not a unit then we are done, since for every $q(x) \in R[x]$, we have that the constant term of $p(x)q(x)$ is $a_{0}b_0$, where $b_0$ is the constant term of $q(x)$, and if $a_0$ is not a unit, then $a_{0}b_0 \neq 1$, which contradicts that $p(x)$ is a unit. So we may assume $a_0$ is a unit, and hence not nilpotent.
Let $a_i \notin N(R)$, $a_i \neq 0$, for some $1 \leq i \leq n$. Then, $\phi(p(x)) = \overline{a_0} + \overline{a_i}x^i + \overline{O(x)}$, where $\overline{O(x)}$ is the collection of all of the non-nilpotent terms. Since $p(x)$ is a unit, there exists $q(x) \in R[x]$ such that $p(x)q(x) = 1$, and so we must have $$\overline{1} = \phi(1) = \phi(p(x)q(x)) = \phi(p(x))\phi(q(x))$$
And here's where I get stuck; I was trying to force the $\overline{a_i}$ term to end up being nonzero to obtain a contradiction, but I can't seem to rule out the case that something from $\phi(q(x))$ multiplies the $\overline{a_i}$ term and ends up making it a nilpotent, or that $\overline{a_i}$ was a zero divisor all along and I accidentally multiplied its buddy.
Any ideas? Thanks in advance!
 A: Note that without loss of generality you can assume the constant term is $1$ since multiplying by a unit (to wit, $a_0^{-1}$) won't change whether something is a unit or non-unit.  Now consider a polynomial $h(x)$ in $R/N(R)[x]$ with constant term $1$, which you can write as $1-axf(x)$ in $R/N(R)[x]$, where $a$ is non-zero and $f$ is monic.  Suppose it is invertible with inverse $g(x)$ of degree $m-1$. 
Then, as $$(1-a^mf(x)^m)=(1-af(x))(1+af(x)+a^2f(x)^2+\cdot\cdot\cdot +a^{m-1}f(x)^{m-1})$$ , we can mutiply through by $g(x)$ to get  $$g(x)(1-a^mf(x)^m)=1+axf(x)+\cdot\cdot\cdot+a^{m-1}x^{m-1}f(x)^{m-1}$$ or
$$g(x)=1+axf(x)+\cdot\cdot\cdot+a^{m-1}x^{m-1}f(x)^{m-1}+a^mx^mf(x)^mg(x)$$.  Comparing terms of like degree, we see that all non-constant terms of $g$ have coefficient a multiple of $a$.  Comparing the highest degree term on the right, we see that it is zero, since it has degree at least $m$ and the left-hand side has smaller degree.  But the coefficient for that term is just $a^m$ times the leading coefficient of $g$, since $f$ is monic.  But if $g$ is non-constant, we have seen that its leading coefficient is a multiple of $a$, say $ab$.  Since $a^mab=0$ we have also $(ab)^{m+1}=0$, so the leading coefficient of $g$ is nilpotent.  
Since $R/N(R)$ is reduced, this is impossible.  Thus, $g$ must be constant, so $h$ is constant, so invertible polynomials in $R/N(R)[x]$ must be constants, so invertible elements of $R[x]$ must have non-constant coefficients nilpotent, since their images in $R/N(R)[x]$ must be constants. QED
