MHB Solving Nilpotent Problem in Abstract Algebra

[cbarker1]

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

 

[Evgeny.Makarov]

"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.

 

[cbarker1]

$\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

 

[Euge]

Consider $\overline{ab}^{\,k}$.