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Could someone help me to figure out this problem?I am really appreciated it!

Let R be a ring. We say that x belongs to R is nilpotent if there exists n > 0 such that x^n = 0R. Let R be a commutative ring and let I denote the set of nilpotent elements of R. Prove that if x, y belong to I and r belongs to R, then x+y belongs to I and rx, xr belong to I. [Hint: In order to prove that x+y belongs to I, argue that if a^n = 0 and k is greater than or equal to n, then a^k = 0. You will also need to use the Binomial Theorem.]

rschwieb
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Jeff Hoffman
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  • Before posting, spend a few minutes searching to see if it's already out there. Before submitting, spend a few minutes adding context about steps you've taken already, and writing a descriptive title. Even if this question was not a duplicate, it would might get a rough welcome being unformatted and unmotivated. – rschwieb Mar 10 '17 at 21:16
  • Another thing to remember for ring questions is to be specific about when your ring is commutative, when possible. This problem is not true for noncommutative rings. – rschwieb Mar 10 '17 at 21:17

1 Answers1

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If $x^n=0$ and $y^m=0$ the binomial theorem (in a commutative ring) gives: $$(x+y)^{n+m}=\sum_{k=0}^{n+m}\binom{n+m}{k}x^k y^{n+m-k}$$

So: or $k\ge n$ and $x^k=0$, or $k<n$ and ${n+m-k}>m$ and $y^ {n+m-k}=0$. This implies that $(x+y)^{n+m}=0$ and is nilpotent.

For the second claim note that, if the ring is commutative than $(rx)^n=r^nx^n=0$

Emilio Novati
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