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This is an exercise from Beachy and Blair: Abstract algebra.

An element $[a]\in\mathbb{Z}_n$ is called nilpotent if $[a]^k=[0]$ for some $k$. Show that $\mathbb{Z}_n$ has no nonzero nilpotent elements if and only if $n$ has no factor that is a square (except $1$).

I was thinking like how can we use the fact that $n$ is squarefree (I assume this means that it is a product of different primes)? We maybe could solve some simultaneous congruences in the form $$ x\equiv a_i \ \bmod p_i \qquad n=\prod p_i, $$ and show that $a$ has to be congruent with $0$ or something like that, but I failed to do so.

Could you give me some HINTS or intuitions about this? Anything is appreciated. Thank you.

Vinyl_cape_jawa
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1 Answers1

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Hint: First of all, remember that $[x] = 0$ in $\mathbb{Z}_n$ if and only if $n \mid x$

So, now suppose that $n$ has a square factor, i.e. $n=k^2m$, what happens to $[km]^2$?

For the other direction, if $n$ is square free then $n=p_1\times \cdots \times p_k$. Now, if $[r]$ is some element in $\mathbb{Z}_n$ that is non-zero, a prime factor of $n$ must be missing in the prime decomposition of $r$. What does that tell you about $[r]^k$? Is it possible that $n \mid r^k$?

stressed out
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