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 $i_{1}<i_{2}<\\dots<i_{n}$ 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 $(A+B)^{k}=0$ .∎

单词	PropertiesOfNilAndNilpotentIdeals
释义	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 $i_{1}<i_{2}<\\dots<i_{n}$ 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 $(A+B)^{k}=0$ .∎
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