Let $r\in R$ and if $x:=r+\frakN(R) \in R/\frakN(R)$ is a nilpotent element of $R/\frakN(R)$, then there exists an integer $n$ such that
Thus we have
This means that $r^n$ is an nilpotent element of $R$, and hence there exists an integer $m$ such that
Therefore $r$ is an nilpotent element of $R$, that is $r\in \frakN(R)$ and we have $x=\frakN(R)$, which is the zero element in $R/\frakN(R)$.
Is the Set of Nilpotent Element an Ideal?
Is it true that a set of nilpotent elements in a ring $R$ is an ideal of $R$?
If so, prove it. Otherwise give a counterexample.
We give a counterexample.
Let $R$ be the noncommutative ring of $2\times 2$ matrices with real […]
Boolean Rings Do Not Have Nonzero Nilpotent Elements
Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a Boolean ring.)
(a) Prove that $x^n=x$ for any positive integer $n$.
(b) Prove that $R$ does not have a nonzero nilpotent […]
Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Since $a$ […]
Every Prime Ideal of a Finite Commutative Ring is Maximal
Let $R$ be a finite commutative ring with identity $1$. Prove that every prime ideal of $R$ is a maximal ideal of $R$.
We give two proofs. The first proof uses a result of a previous problem. The second proof is self-contained.
Let $I$ be a prime ideal […]