Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$?

Question

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.

Answer 1

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.

Answer 2

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 $(1+a)b_n = 1$.

Answer 3

$N(R)$ 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.

Answer 4

HINT $\ $ Note that a nilpotent element $\rm\:n\:$ lies in every prime ideal $\rm\:P\:$, since $\rm\: n^k = 0\in P\ \Rightarrow\ n\in P\:.\:$ In particular, $\rm\:n\:$ lies in every maximal ideal. So $\rm\:n+1\:$ is a unit, since it lies in no maximal ideal $\rm\:M\:$ (else $\rm\:n,n+1\in M\ \Rightarrow (n+1)-n = 1\in M\:$); 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 $\rm\:1 + n\:$ is coprime to all primes, so it must be the unit $1\:,\:$ so $\rm\:n = 0\:,\:$ a contradiction.

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 $\rm\:J\:$ such that $\rm\:1+J \subset U(R)\:,\:$ i.e are all units, are precisely those ideals contained in $\rm\:Jac(R)\:.\:$ Indeed, we have the following theorem, 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.

Answer 5

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