Nilradical and Jacobson's radical. Let A be a commutative ring with 1. 
1) Prove that a sum of a nilpotent element and an invertible element is invertible. 
2) Prove that if $f=a_0+a_1x+\dots+a_nx^n \in A[x]$
a) $\exists f^{-1}\in A[x] \Leftrightarrow a_0$ is invertible and the other coefficients are nilpotent.
b) f is nilpotent $\Leftrightarrow $ all its coefficients are nilpotent.
p.s. Those are the first two in a series of problems. The rest easily follow from each other. I'm only struggling with the first two.
 A: Here is an abstract argument, which I really like. All you have to know is, that the intersection of all prime ideals of $A$ (called "nilradical") consists exactly of the nilpotent elements of $A$.


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*Let $u,n \in A$, $u$ invertible, $n$ nilpotent. Let further $p$ be any prime ideal of $A$. Since $n \in p$ we have $u+n \equiv u \not\equiv 0\mod p$. So $u+n$ lies in no prime ideal of $A$ and must therefore be a unit.

*a) "$\Leftarrow$" follows from 1. For "$\Rightarrow$" let $p$ be any prime ideal of $A$. Since $A/p$ is an integral domain and the reduction $\overline{f} \in (A/p)[x]$ remains invertible, it follows $\deg(\overline{f}) = 0$, which means $a_0 \notin p$ and $a_i \in p$ for $i = 1, \dots, n$. Since this holds for any prime ideal $p$ of $A$ we conclude that $a_0$ is invertible while $a_i$ is nilpotent for $i = 1, \dots, n$.
b) Proceed similar to a) and use the fact that the only nilpotent element of an integral domain is $0$.

A: Hints: Supposing $a^n=0$, in order to show that $u+a$ is a unit for some unit $u$, it suffices to show that $1+au^{-1}$ is a unit. (Note that $au^{-1}=b$ has the same index of nilpotency that $a$ has.
To show that $1+b$ is a unit, use the identity $(1+b)(1-b+b^2-b^3+\dots)=1$
For the second part, to show that the polynomial is invertible, if $a_0$ is a unit, then $a_1x$ is clearly nilpotent, and obviously (by what you have been working on!) $a_0+a_1x$ must be a unit. Proceed by induction.
(Trying to think of a good hint for the converse...)
