# properties of nil and nilpotent ideals

###### Lemma 1.

Let $A\subset B$ be ideals of a ring $R$. If $A$ is nil and $B/A$ is nil, then $B$ is nil. If $A$ is nilpotent and $B/A$ 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 $z_{1},z_{2},\dots,z_{k}$ 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 $z_{1}z_{2}\cdots z_{k}$ we get a sum of terms of the form $x_{1}x_{2}\dots x_{k}$ where each $x_{i}\in\{a_{i},b_{i}\}$.

Consider one of these terms $x_{1}x_{2}\cdots x_{k}$. 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 $i_{1} with $x_{i_{j}}\in A$ for each $j$. For $1\leq j\leq n-1$, define $y_{j}=x_{i_{j}}x_{i_{j}+1}\cdots x_{i_{j+1}-1}$, and define $y_{n}=x_{i_{n}}x_{i_{n}+1}\cdots x_{k}$. Since $A$ is a right ideal, $y_{j}\in A$.

Then $x_{1}x_{2}\cdots x_{k}=x_{1}x_{2}\cdots x_{i_{1}-1}y_{1}y_{2}\cdots y_{n}\in x% _{1}x_{2}\cdots x_{i_{1}-1}A^{n}=0$.

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

Title properties of nil and nilpotent ideals PropertiesOfNilAndNilpotentIdeals 2013-03-22 14:12:54 2013-03-22 14:12:54 mclase (549) mclase (549) 6 mclase (549) Result msc 16N40 KoetheConjecture NilIsARadicalProperty