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This is a step from a proof I am reading that involves divisibility, not much in algebra.

Let $z \in r(m\mathbb{Z})$, the radical of $m\mathbb{Z}$ (here $m$ is an integer). Trying to show the radical here can be reduced to primes and eventually the general statement. In this step, we know for some $n$, the element $z^n = m\ell$ for some $\ell \in \mathbb{Z}$.

So $z^n$ is divisible by $m$. Suppose $m = p_1^{\alpha_1} \dots p_r^{\alpha_r}$, how does $m | z^n \implies p_1 \dots p_r | z$? How do we deal with $z^{n-1}$?

Lemon
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  • I'm not sure what you mean by the second question, but for the first part here's a hint: what happens if some $p_i$ doesnt divide $z$?Could $m$ still divide $z^n?$ – iYOA Jan 04 '18 at 23:41
  • @iYOA mean that $z^n = zz^{n-1} = m\ell = p_1^{\alpha_1} \dots p_r^{\alpha_r}\ell$ – Lemon Jan 04 '18 at 23:43
  • you mean the possibility of $p|z^{n-1}$ and not $z$? Yeah, Mark's answer deals with that, actually. – iYOA Jan 04 '18 at 23:51
  • @iYOA well originally I am seeing the above equation only implies $m | z^n$, was just wondering how to eliminate $z^{n-1}$ so I get $p_1 \dots p_r \ell' = z$ the divisibility condition. – Lemon Jan 05 '18 at 00:02

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Suppose $p|m$ and $m|z^n$ then $p|z^{n}$.

If $p$ is a prime then $p|z$ or $p|z^{n-1}$ in the first case you are already done. In the second case a simple induction shows that $p|z^{n-1}$ implies $p|z$ (you reduce the power by $1$ each time until $p|z^2$ implies $p|z$ or $p|z$).

Mark Bennet
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  • Just to make sure on something, your statement "If $p$ is a prime", in my case, aren't I taking a products of primes? I wouldn't be able to conclude the nice result that the product divides out the $z$ or $z^{n-1}$ right? Because that's where I am stuck. – Lemon Jan 05 '18 at 00:08
  • @Hawk For any prime you have $p|ab$ implies $p|a$ or $p|b$. Also if $a|n$ and $p|n$ then $n=ab$ and $p|a$ or $p|b$. If $a$ is the product of primes $p_1, p_2, \dots p_{r-1}$ and $p=p_r$ you therefore have $p_r|b$ and the product up to $p_r$ is a factor of $n$. You can therefore show that each prime in the product is a factor of $z$, hence the product of the primes is a factor of $z$. – Mark Bennet Jan 05 '18 at 00:26