# 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.

#### Solutions Collecting From Web of "Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$?"

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.

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 $(1+a)b_n = 1$.

$N(R)$ 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.

Hint $\$ 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 a hint of this in 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 $1,\,$ so $\rm\,n = 0,\,$ a 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.

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 $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)$.