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It is already asked in Nilpotent elements in $\mathbb{Z}_n$

I could not understand following. if a is nilpotent then a $n\nmid a$, $n \vert a^{k} $ let n = $p_1^{m_1} ...p_r^{m_r}$ then how can we say $p_i \vert a$ for all i = 1 to r ???

Magneto
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1 Answers1

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Since $p_i\mid n$ and $n\mid a^k$ for some $k$, that means $p_i\mid a^k$ (divisibility is transitive). One well-known (and in some contexts defining) property of prime numbers is that if $p$ is prime and $p\mid xy$ for some integers $x, y$, then either $p\mid x$ or $p\mid y$.

If we apply that to $p_i\mid a^k$, we see that either $p_i$ divides $a$, or it divides $a^{k-1}$ (because $a^k = a\cdot a^{k-1}$). If $p_i\mid a$, we're done. If not, then we have $p_i\mid a^{k-1}$. We do the same thing here, and keep going, and we are eventually forced to conclude that $p_i\mid a$

Arthur
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  • ok ... so we have this(pls correct if i am going wrong below) $p_1^{m_1} p_2^{m_2}... \vert a^k$ , $p_1^{m_1} $, $p_2^{m_2}$ are coprime to each other and so $p_1^{m_1} \vert a^k$ ... which means $p_1 \vert a$ ... All this mean $p_1 p_2 ....p_r \vert a$. But we have $n \nmid a $. If n is square free => $n = p_1 p_2 .. p_r$ => by above statement $n \vert a$ which is not so. So, n should have square of atleast 1 prime which divides n – Magneto Oct 12 '17 at 11:47
  • I suppose in fact Zn does not have non-zero nilpotent element. – Magneto Oct 12 '17 at 11:49
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    @anirudhb That looks correct. However, we don't need to point out that the $p_i^{m_i}$ are coprime to eachother. $ab\mid c \implies a\mid c$ regardless of coprimality between $a$ and $b$. It's when you have a product on the right side of the vertical line you have to worry about coprimality. That's why we can't immediately say $p_i\mid a^k \implies p_i\mid a$, but have to go through a proof that, among other things, hinges on the fact that $p_i$ is prime. – Arthur Oct 12 '17 at 12:01