PlanetMath: properties of nil and nilpotent ideals

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properties of nil and nilpotent ideals

(Result)

Lemma 1 Let $A \subset B$ be ideals of a ring . If is nil and is nil, then is nil. If is nilpotent and is nilpotent, then is nilpotent.

Proof. Suppose that

and

are nil. Let $x \in B$ . Then $x^n \in A$ for some

, since

is nil. But

is nil, so there is an

such that $x^{nm} = (x^n)^m = 0$ . Thus

is nil.

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

Lemma 2 The sum of an arbitrary family of nil ideals is nil.

Proof. Let

be a ring, and let $\mathcal{F}$ be a family of nil ideals of

. Let $S = \sum_{I \in \mathcal{F}} I$ . We must show that there is an

with

for every $x \in S$ . Now, any such

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

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 and be nilpotent right ideals of a ring . Then there are natural numbers and such that and .

Let . Let $z_1, z_2, \dots, z_k$ be elements of . We may write for each , with $a_i \in A$ and $b_i \in B$ . If we expand the product $z_1z_2 \dotsm z_k$ we get a sum of terms of the form $x_1x_2 \dots x_k$ where each $x_i \in \{ a_i, b_i \}$ .

Consider one of these terms $x_1x_2 \dotsm x_k$ . Then by our choice of , it must contain at least of the 's or at least of the '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 . For $1 \le j \le n-1$ , define $y_j = x_{i_j} x_{i_j+1} \dotsm x_{i_{j+1}-1}$ , and define $y_n = x_{i_n}x_{i_n+1} \dotsm x_k$ . Since is a right ideal, $y_j \in A$ .