Show that if $m\in M_n$ and $k \in \Bbb Z_n$ then $mk\in M_n$.

Question

Let $M_n = \{a\in \Bbb Z_n\mid \text{ there exists a non-zero integer $k$ with the property that }a^k\equiv 0 \pmod n \}.$

Show that if $m \in M_n$ and $k \in \Bbb Z_n$, then $mk\in M_n$. For which integers $n$ is $M_n=0$?

This is for an elementary abstract algebra class. Having some difficulty showing this. Some scratch work:

$M_n$ is a ring with no multiplicative identity since $1\not\in M_n \forall n$

A hunch is that $M_n= 0$ if $n$ has any integer roots.

$m\in M_n \iff \exists r\not =0 \in \Bbb Z$ such that $n$ divides $m^r$ $\iff \exists q \in \Bbb Z$ such that $m^r = nq$

Suppose $mk \in M_n$

then $\exists s \not =0 \in \Bbb Z$ such that $n$ divides $(mk)^t \iff \exists p \in \Bbb Z$ such that $(mk)^s = np \iff m^tk^t=np$

Let $s=r$

$m^rk^r=np$

$m^r=nq\Rightarrow qk^r=p$

q is a fixed integer, r is a fixed non-zero integer, p is any integer. So, $qk^r$ is an integer for all k. Therefore $mk\in M_n$

I'm fairly certain this is all wrong. Any help is greatly appreciated, thank you.

Answer 1

$\begin{eqnarray}{\bf Hint} &&p_1^{e_1}\cdots p_j^{e_j}\!&\mid a^k\,\ \rm for\ some\ k\\ \iff &&\ \forall\, i\!:\ \ \ p_i^{e_i} &\mid a^{k_i}\ \rm for\ some\ k_i \\ \iff &&\ \forall\, i\!:\ \ \ p_i&\mid a \end{eqnarray}$

Answer 2

If $m\in M_n$ (so $m^r\equiv 0\pmod n$ for some $r$) and $k\in \mathbb Z_n$ then $(mk)^r\equiv m^rk^r\equiv 0\pmod n$.

Answer 3

The set $M_n$ is the set of nilpotent elements of $\mathbb Z_n$. The key observation is that $m\in M_n$ iff all prime factors of $n$ divide $m$. This comes from $p \mid n \mid m^r$.