Nilpotent Problem

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  • #1
cbarker1
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Dear Everyone,

I am stuck with an exercise problem. The problem states from Dummit and Foote Ed. 2 Abstract Algebra: "An element $x$ in $R$ (where $R$ is a ring with 1) is called nilpotent if $x^{m}=0$ for some $m \in \Bbb{Z}^{+}$. Show that if $n=a^{k}b$ for some $a,b \in \Bbb{Z}$, then $\overline{ab}$ is a element of $\Bbb{Z}/n\Bbb{Z}$."

My attempt:

Example: When $n=6=3\cdot 2$, then the only element will be $\overline{3*2}$ in $\Bbb{Z}/6\Bbb{Z}$.

Proof: Suppose $n=a^{k}b$ for some $a,b \in \Bbb{Z}$. (Do I need some cases? If so, three cases?)

Thanks,
Cbarker1
 
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Answers and Replies

  • #2
Evgeny.Makarov
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"An element $x$ in $R$ (where $R$ is a ring with 1) is called nilpotent if $x^{m}=0$ for some $m \in \Bbb{Z}^{+}$. Show that if $n=a^{k}b$ for some $a,b \in \Bbb{Z}$, then $\overline{ab}$ is a element of $\Bbb{Z}/n\Bbb{Z}$."
Could you define the notation $\overline{ab}$? Also, it is strange that the statement one is asked to prove does not mention the concept "nilpotent", which is defined right before that.
 
  • #3
cbarker1
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$\overline{ab}:=a * b $(mod n)

Sorry. I typed quickly and I forgot about the most important information. If $n=a^kb$ for some $a,b\in \Bbb{Z}$, then $\overline{ab}$ is a nilpotent element of $\Bbb{Z}/n\Bbb{Z}$.


Again, I am sorry.


Cbarker1
 
Last edited:
  • #4
Euge
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Consider $\overline{ab}^{\,k}$.
 

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