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Let $R$ be a commutative finite local ring of order $p^n$ ($p$ is a prime and $1\in R$). I'm struggling with the following two basic questions:

(a) Is it true that $x^n=0$ for every non-unit $x\in R$ ?

(b) Is there exists a nilpotent element $x\in R$ such that $x^{n-1}\neq0$ ?

My guess is that the answer to both questions is false, but unfortunately I can not find any appropriate counterexamples.

boaz
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  • Have you investigated the ring $R=\mathbb{Z}/p^n\mathbb{Z}$? What are the answers to your questions for this $R$? – kabenyuk Sep 09 '21 at 14:04
  • It seem to be true in this case: If $\bar{x}\in \mathbb{Z}/(p^n)$ is non-unit, then $p\mid x$, so $\bar{x}^n=\bar{0}$. conversely, $\bar{p}^{n-1}\neq \bar{0}$ in $ \mathbb{Z}/(p^n)$ – boaz Sep 09 '21 at 14:12
  • Very well. How does the general case differ from this special case? – kabenyuk Sep 09 '21 at 14:24
  • We can not say that $p\mid x$? does every non-unit of $R$ is of the form $py$ for some $y\in R$? it is not clear to me why $p^n=0$ in $R$ – boaz Sep 09 '21 at 14:29
  • Thanks @kabenyuk. I will be appreciate if you can explain it to me. I have only basic knowledge in ring theory. – boaz Sep 09 '21 at 14:32
  • Another good example of a local ring of $q=p^n$ elements is the field $F_q$. – kabenyuk Sep 09 '21 at 14:41
  • Since $R^{+}$ is abelian group of order $p^n$ we deduce that $p^n\bar{p}=0$, so $\bar{p}$ is non-unit. In addition, $\bar{p}R$ is an ideal in $R$. Is it maximal? If it does, then indeed every non-unit element is of the form $\bar{p}y$. Right? – boaz Sep 09 '21 at 14:44

1 Answers1

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I will try to answer your first question.

In any commutative local ring $R$ of order $p^n$ it is true that $x^n$=0 for every non-unit $x\in R$.

Let $R^* $ be the set of units of $R$ and $M$ be the maximal ideal of $R$. Then $|M|=p^k$ for some non-negative integer $k<n$ and $|R^*|=p^n-p^k=p^k(p^{n-k}-1)\geq p-1$.

Let $x\in M$. Let us prove that $x^n=0$.

Note first that if $x,\ldots,x^s$ are all nonzero, then $$ a_1x+\ldots+a_sx^s=0,\ a_i\in R^*\cup\{0\}\ \Rightarrow\ a_1=\ldots=a_s=0. $$ Indeed, if $a_1x+\ldots+a_sx^s=0$ and $a_1\neq0$, then $x(a_1+\ldots+a_sx^{s-1})=0\ \Rightarrow\ x=0$.

It follows that elements of the form $a_1x+\ldots+a_sx^s=0,\ a_i\in R^*\cup\{0\}$ are at least $p^s$.

So $s<n$, and then $x^n=0$.

kabenyuk
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  • thanks @kabenyuk. You claim that the set of elements $${a_1x+\ldots+a_sx^s:a_i\in R^{\ast}\cup{0}}$$ are distinct? why? – boaz Sep 09 '21 at 16:36
  • what about the second question? – boaz Sep 09 '21 at 17:56
  • It is sufficient that the set of elements $$ a_1x+\ldots+a_sx^s,\ a_i=t_i\cdot1,\ 0\leq t_i<p, $$ are distinct. The answer to the second question is negative. For example, let $R=F_q$ be a field of order $q=p^n$, $n>1$. It has no nilpotent elements with the property $x^{n-1}\neq0$. – kabenyuk Sep 10 '21 at 04:01
  • thanks @kabenyk. – boaz Sep 10 '21 at 05:26