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Let $p$ be a prime numer and let $k$ be a natural numer such that $k\geq 2$. I wish to descripe all zero's divisors in $\mathbb Z_{p^k}$. Obviously elements of the form $np$, where $n=0,...,p^{k-1}-1$, are zero divisors, because $p^k|np^k$. Are there others?

Alex
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1 Answers1

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Suppose $ab=0$, and $p\nmid a$. Well $p^k|ab$, so $p^k|b$, so $b=0$. Hence you found them all.

vadim123
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  • Thanks. Maybe still one trivial question to the answer: why if $p\nmid a$ and $p^k|ab$ then $p^k|b$ (I know than then $p|b$ but why $p^k|b$? – Alex Oct 05 '13 at 14:51
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    @Alex I is true from basic theorems on prime factorisation. But you can also see it like this. Suppose $k\gt 1$, then $b=pc$ and $p^{k-1}|ac$, whence $p|c$ - and following this through all the factors of $p$ must be contained in $b$. – Mark Bennet Oct 05 '13 at 15:19