Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$? Let $R$ be a commutative ring with 1, we define
$$N(R):=\{ a\in R \mid  \exists k\in \mathbb{N}:a^k=0\}$$
and
$$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$
Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?
I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that.
 A: This really is a problem in disguise: How did you derive the formula for the sum of the geometric series in year (something) at school??
$(a + 1)(a^{k-1} - a^{k-2} + \ldots 1) = 1-(-a)^n$
but as $a^n = 0$, we have (it does no matter whether $n$ is even or odd) that $(a+1)$ is invertible. Prove the following analogous problem, it may strengthen your understanding:
Let $A$ be a square matrix. If $A^2 = 0$, show that $I - A$ is invertible. 
If $A^3 = 0$, show that $I - A$ is invertible.
Hence in general show that if $A^n = 0$ for some positive integer $n$, then $I -A$ is invertible.
A: Note the following:
$(1+a)(1-a) = 1-a^2$.
$(1-a^2)(1+a^2) = 1-a^4$
$(1-a^4)(1+a^4) = 1-a^8$
...
Thus, continuing in this way, we may find some $b_n$ such that
$(1+a)b_n = 1-a^{2^n}$ 
For large enough $n$, this will give us $(1+a)b_n = 1$.
A: $N(R)$ is at least in some texts referred to as the nilradical of $R$. It is contained in all prime ideals (in fact, it is the intersection of all prime ideals, Atiyah, MacDonald prop. 1.8), so taking a nilpotent element $a$, since it is contained in all maximal ideals, $a+1$ is not in any maximal ideal. Then the ideal generated by $a+1$ must neccesarily be the whole ring, which means that it specifically generates 1 at some point.
A: Let $a\in N(R)$ and $k$ such that $a^k=0$. We have 
\begin{align*}(1+a)\left(\sum_{j=0}^{k-1}(-a)^j \right)&=\sum_{j=0}^{k-1}(-a)^j+\sum_{j=0}^{k-1}-(-a)^{j+1}\\
& =\sum_{j=0}^{k-1}(-a)^j-\sum_{j=1}^k(-a)^j\\
&=1-(-a)^k=1+(-1)^ka^k=1,
\end{align*}
and we have $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)=1$ by the same computation (it's true even if the ring is not commutative), hence $1+a$ is invertible, and it's inverse is $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)$.
A: Here's a more conceptual structural view (vs.  common proof by simpler multiples & geometric series).
Hint $\ $ Notice that a nilpotent  $\rm\,n\,$ lies in every prime ideal $\rm\,P,\,$ because $\rm\, n^k = 0\in P\ \Rightarrow\ n\in P.\,$ In particular, $\rm\,n\,$ lies in every maximal ideal. Hence $\rm\,n\!+\!1\,$ is a unit, since it lies in no maximal ideal $\rm\,M\,$ (else $\rm\,n\!+\!1,\,n\in M\, \Rightarrow\, (n\!+\!1)-n = 1\in M),\,$ i.e. elements coprime to every prime are units. 
You may recognize this in modern proofs of Euclid's theorem that that are infinitely many primes. Namely, if there are only finitely many primes then their product $\rm\,n\,$ is divisible by every prime, so $\rm\,n\!+\!1\,$ is coprime to all primes, so it must be the unit $1,\,$  contradiction. 
Remark $ $ You'll meet related results later when you study the structure theory of rings. There the intersection of all maximal ideals of a ring $\rm\,R\,$ is known as the Jacobson radical $\rm\,Jac(R).\,$ The ideals $\rm\,J\,$ with $\rm\,1+J \subset U(R)= $ units of $\rm R,\,$ are precisely those ideals contained in $\rm\,Jac(R).\,$ Indeed, we have the following theorem, excerpted from my post on the fewunit ring theoretic generalization of Euclid's proof of infinitely many primes.
THEOREM $\ $ TFAE in ring $\rm\,R\,$ with units $\rm\,U,\,$ ideal $\rm\,J,\,$ and Jacobson radical $\rm\,Jac(R).$
$\rm(1)\quad  J \subseteq Jac(R),\quad $ i.e.  $\rm\,J\,$  lies in every max ideal $\rm\,M\,$ of $\rm\,R.$
$\rm(2)\quad  1+J \subseteq U,\quad\ \ $  i.e. $\rm\, 1 + j\,$  is a unit for every $\rm\, j \in J.$
$\rm(3)\quad  I\neq 1\ \Rightarrow\  I+J \neq 1,\qquad\ $  i.e.  proper  ideals survive in $\rm\,R/J.$
$\rm(4)\quad M\,$ max $\rm\,\Rightarrow  M+J \ne 1,\quad $  i.e. max ideals survive in $\rm\,R/J.$
Proof $\, $ (sketch) $\ $  With  $\rm\,i \in I,\ j \in J,\,$ and max ideal $\rm\,M,$
$\rm(1\Rightarrow 2)\quad  j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\,$ unit.
$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\,$ unit $\rm\,\Rightarrow  I = 1.$
$\rm(3\Rightarrow 4)\ \,$  Let $\rm\,I = M\,$ max.
$\rm(4\Rightarrow 1)\quad  M+J \ne 1 \Rightarrow\ J \subseteq M\,$  by  $\rm\,M\,$ max.
