To show the following statement in ring R.

Question

For P, n $\in N$ with P, a prime , show that every zero-divisor in $Z_{p^n}$ is nilpotent. I know the definition of nilpotent element but don't know how to proceed.

Answer 1

I am not fluent in algebra, but let me try: Take $a,b\in\mathbb Z_{p^n}$ with and $ab\equiv0$, so $a$ and $b$ are zero divisors. Written with usuals integers (and used $a$ and $b$ interchangeably with some representatives from $\{0,..,p^n-1\}$) , this means

$$ ab=k\cdot p^n $$

for some $k\in\mathbb Z$. From this you can conclude

$$ a=k_1 \cdot p^{n_1},\quad b=k_2\cdot p^{n_2} $$

for some $k_1,k_2,n_1,n_2 \in\mathbb Z$ with $k_1k_2=k$ and $n_1+n_2=n$. Note that $n_1,n_2>0$ because $a,b

$$ a^n=(k_1\cdot p^{n_1})^n =k_1^n\cdot p^{n_1\cdot n}=k_1^n\cdot (p^n)^{n_1}\equiv k_1^n \cdot 0^{n_1}=0$$

and equivalently for $b$.

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The essential observation is, in my opinion, that for $n\geq 2$ there are elements in $\mathbb Z_{p^n}$ that contain a factor $p$ (again speaking about the representatives). This factor can be used to make powers of these elements zero.

Answer 2

Hint $ $ If $a$ is a z.d. then $\,p\mid a\,$ (else $a$ is coprime to $p^n$ so a unit). $ $ Thus $\,p\mid a\,\Rightarrow\,p^n\mid a^n$