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An element $x$ in a ring $R$ is called nilpotent if $x^n=0$ for some $n\in \mathbb N$. Let $R$ be a commutative ring and $N=\{x\in R\mid \text{x is nilpotent}\}$.

(a) Show that $N$ is an ideal in $R$.

(b) Show that the quotient ring $R/N$ has no non-zero nilptoent elements.

What's the steps to prove (a) and (b)?

user26857
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Sara
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    Have you tried it yourself? Where did you get stuck? – Prasun Biswas Apr 08 '15 at 12:46
  • This is answered several times already on the site, in particular http://math.stackexchange.com/q/132349/29335 and http://math.stackexchange.com/q/132369/29335. Please, please do some searches before you post your questions. And include what you have tried so far when you do wind up posting questions. Regards – rschwieb Apr 08 '15 at 12:49

2 Answers2

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You should explain what you tried for answering both these questions. Namely, the first one (a) is quite easy once you go back to the definition of an ideal. The second one should not take much longer: assume that $x \in R/N$ is nilpotent, let $\widehat{x} \in R$ be an antecedent of $x$, and look at what the assertion “$x$ is nilpotent” means for $\widehat{x}$.

Circonflexe
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The steps to prove (a) are simple:

  • Take two elements $x,y\in N$ and show that $x+y\in N$
  • Take an element $n\in N$ and an element $r\in R$ and show that $nr\in N$.

In order to prove (b), personally, I would take a nilpotent element in $R/N$ and show that it is the zero element.

5xum
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