- #1

mathmari

Gold Member

MHB

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I am looking at the following exercise:

An element $r\in R$ is called nilpotent if $r^n=0$ for some integer $n=1,2, \dots $.

- Show that if $r$ is nilpotent then $1-r$ is invertible in $R$.
- Show that if $R$ is commutative then the set $N(R)$ of nilpotent elements is an ideal of $R$. Give an example where that is not true when the ring is not commutative.
- Show that $N(\mathbb{Z}_m)=0$ if and only if $m$ is not divided by a square of any prime.

- Since $r\in R$ is nilpotent we have that $r^n=0$ for some $n=1, 2, \dots $.

Then $$1=1-r^n=(1-r)(r^{n-1}+\dots +1)$$

So, $1-r$ is invertible in $R$.

$$$$

- To show that $N(R)$ is an ideal of $R$ we have to show that $N(R)$ is a left and a right ideal, right? (Wondering)

So, we have to show that $ra\in N(R)$ and $ar\in N(R)$, for $r\in R$ and $a\in N(R)$.

Since $R$ is commutative, we have that $(ra)^n=r^na^n$. Since $a\in N(R)$ we have that $a^n=0$. Therefore, $(ra)^n=0$.

Since $R$ is commutative, we have that $(ar)^n=a^nr^n$. Since $a\in N(R)$ we have that $a^n=0$. Therefore, $(ar)^n=0$.

So, $N(R)$ is an ideal of $R$.

Is this correct? (Wondering)

How could we find an example where that is not true when the ring is not commutative? (Wondering)

$$$$

- Suppose that $N(\mathbb{Z}_m)=0$, then $r^n\neq 0$, for all $n\in \mathbb{N}$, where $r\in \mathbb{Z}_m$, right? (Wondering)

How could we continue? (Wondering)