Show that $(A[X])^\times=A^\times+nil(A[X])$ I am currently reading through the book Basic Commutative Algebra, by Balwant Singh, wherein the exercise I.XVI reads like:  

Show that $(A[X])^\times=A^\times+nil(A[X])$.   

Here, for a ring $A$, $A^\times$ means the set of all units, and $nil(A)$ is the intersection of all prime ideals, or the nilpotent radical, the radical of the zero ideal $(0)$.  
P.S. I have shown that $A^\times=A^\times+nil A$, but cannot conquer this exercise.
Thanks for any help.
 A: As you say, the sum of a unit and a nilpotent element is a unit. This immediately gives one inclusion. 
Here's a sketch for a proof of the reverse inclusion. Let $f(X) = a_nX^n + \cdots + a_0$ be invertible. Let $g(X) = b_mX^m + \cdots + b_0$ be the inverse of $f$. Clearly $b_0 \in A^*$. Let's show that $a_n$ is nilpotent. To begin with, $a_nb_m = 0$. Looking at the coefficient of $X^{n + m - 1}$ in the $fg$, we see that $a_nb_{m - 1} + a_{n - 1}b_m = 0$. Multiplying this by $a_n$, we get
\[
a_n^2b_{m - 1} + a_{n - 1}a_nb_m = a_n^2b_{m - 1} = 0.
\]
The idea is to continue doing this until we reach $a_n^{m + 1}b_0 = 0$.
A: Let me show without calculation the difficult part:   
If $f(X)=a_0+a_1X+...+a_n X^n\in (A[X])^\times$, then for all   $i\gt0$, the coefficient  $a_i$ is   nilpotent.  
Proof
For an arbitrary prime ideal $\mathfrak p\subset A$, the class $\bar f(X)\in A/\mathfrak p [X]$ is invertible in $A/\mathfrak p [X]$.
Since $A/\mathfrak p [X]$ is a domain, this implies $\bar a_i=0$  (because degree considerations show that an invertible polynomial with coefficients in a domain is a constant).
But this says that $a_i\in \mathfrak p$ for all prime ideals in $\mathfrak p \subset A$. Hence $a_i$ is nilpotent.
