# properties of nil and nilpotent ideals

###### Lemma 1.

Let $A\mathrm{\subset}B$ be ideals of a ring $R$. If $A$ is nil and $B\mathrm{/}A$ is nil, then $B$ is nil. If $A$ is nilpotent^{} and $B\mathrm{/}A$ is nilpotent, then $B$ is nilpotent.

###### Proof.

Suppose that $A$ and $B/A$ are nil. Let $x\in B$. Then ${x}^{n}\in A$ for some $n$, since $B/A$ is nil. But $A$ is nil, so there is an $m$ such that ${x}^{nm}={({x}^{n})}^{m}=0$. Thus $B$ is nil.

Suppose that $A$ and $B/A$ are nilpotent. Then there are natural numbers^{} $n$ and $m$ such that ${A}^{m}=0$ and ${B}^{n}\subseteq A$. Therefore, ${B}^{nm}=0$.
∎

###### Lemma 2.

The sum of an arbitrary family of nil ideals is nil.

###### Proof.

Let $R$ be a ring, and let $\mathcal{F}$ be a family of nil ideals of $R$. Let $S={\sum}_{I\in \mathcal{F}}I$. We must show that there is an $n$ with ${x}^{n}=0$ for every $x\in S$. Now, any such $x$ is actually in a sum of only finitely many of the ideals in $\mathcal{F}$. So it suffices to prove the lemma in the case that $\mathcal{F}$ is finite. By induction^{}, it is enough to show that the sum of two nil ideals is nil.

Let $A$ and $B$ be nil ideals of a ring $R$. Then $A\subset A+B$, and $A+B/A\cong B/(A\cap B)$, which is nil. So by the first lemma, $A+B$ is nil. ∎

###### Lemma 3.

The sum of a finite family of nilpotent left or right ideals^{} is nilpotent.

###### Proof.

We prove this for right ideals. Again, by induction, it suffices to prove it for the case of two right ideals.

Let $A$ and $B$ be nilpotent right ideals of a ring $R$. Then there are natural numbers $n$ and $m$ such that ${A}^{n}=0$ and ${b}^{m}=0$.

Let $k=n+m-1$. Let ${z}_{1},{z}_{2},\mathrm{\dots},{z}_{k}$ be elements of $A+B$. We may write ${z}_{i}={a}_{i}+{b}_{i}$ for each $i$, with ${a}_{i}\in A$ and ${b}_{i}\in B$. If we expand the product^{} ${z}_{1}{z}_{2}\mathrm{\cdots}{z}_{k}$ we get a sum of terms of the form
${x}_{1}{x}_{2}\mathrm{\dots}{x}_{k}$ where each ${x}_{i}\in \{{a}_{i},{b}_{i}\}$.

Consider one of these terms ${x}_{1}{x}_{2}\mathrm{\cdots}{x}_{k}$. Then by our choice of $k$, it must contain at least $n$ of the ${a}_{i}$’s or at least $m$ of the ${b}_{i}$’s. Without loss of generality, assume the former. So there are indices $$ with ${x}_{{i}_{j}}\in A$ for each $j$. For $1\le j\le n-1$, define ${y}_{j}={x}_{{i}_{j}}{x}_{{i}_{j}+1}\mathrm{\cdots}{x}_{{i}_{j+1}-1}$, and define ${y}_{n}={x}_{{i}_{n}}{x}_{{i}_{n}+1}\mathrm{\cdots}{x}_{k}$. Since $A$ is a right ideal, ${y}_{j}\in A$.

Then ${x}_{1}{x}_{2}\mathrm{\cdots}{x}_{k}={x}_{1}{x}_{2}\mathrm{\cdots}{x}_{{i}_{1}-1}{y}_{1}{y}_{2}\mathrm{\cdots}{y}_{n}\in {x}_{1}{x}_{2}\mathrm{\cdots}{x}_{{i}_{1}-1}{A}^{n}=0$.

This is true for all choices of the ${x}_{i}$, and so ${z}_{1}{z}_{2}\mathrm{\cdots}{z}_{k}=0$. But this says that ${(A+B)}^{k}=0$. ∎

Title | properties of nil and nilpotent ideals |
---|---|

Canonical name | PropertiesOfNilAndNilpotentIdeals |

Date of creation | 2013-03-22 14:12:54 |

Last modified on | 2013-03-22 14:12:54 |

Owner | mclase (549) |

Last modified by | mclase (549) |

Numerical id | 6 |

Author | mclase (549) |

Entry type | Result |

Classification | msc 16N40 |

Related topic | KoetheConjecture |

Related topic | NilIsARadicalProperty |