Let $p$ be a prime integer and $R$ be a finite local ring. Assume that $p||R^\times|$. Then by Cauchy's Theorem, there always exists a primitive $p$ root of unity in $R^\times$. Here $R^\times$ is a unit group of $R$.

Is it possible to find a finite local ring $R$ which is not a finite field and a prime integer $p>2$ dividing $|R^\times|$ such that there exists a primitive $p$ root of unity $u \in R^\times $ for which $u - 1$ is still a unit in $R$?


Sure: just take a finite field and then add some nilpotents to get a local ring that is not a field. For instance, let $F$ be a finite field such that $p$ divides $|F^\times|$ and then let $R=F[x]/(x^2)$.

You can also get examples by taking $\mathbb{Z}/q^n$ where $q$ is a prime such that $p\mid q-1$. For instance, if $R=\mathbb{Z}/49$, then there are primitive cube roots of unity $u=18$ and $u=30$ such that $u-1$ is also a unit.

  • $\begingroup$ Let $R$ be a finite local ring with maximal ideal $M$ and residue $k = R/M$. If a prime $p| |k^\times|$, then we always have a primitive $p$ root $u$ such that $u - 1$ is also a unit in $R$. Is this right? and how do we prove it? Could you please give me some hint? Thank you $\endgroup$ – NongAm May 15 '17 at 4:38
  • $\begingroup$ I believe that is correct, though the proof I can see is fairly complicated. Here's an outline: first prove that a primitive $p$th root of unity must exist in $R$, then prove that any such primitive $p$th root cannot have image $1$ in $k$, then prove that $u-1$ is also a unit. $\endgroup$ – Eric Wofsey May 15 '17 at 4:56
  • $\begingroup$ How do we use $p||k^\times|$? Is it from $R^\times = k^\times \times (1 + M)$ as a group isomorphism? I do know how to get a $p$ root of unity in $R$. But I don't know how to show that the image is not $1$ in $k$. May be $p | |k^\times|$ is a key point. $\endgroup$ – NongAm May 15 '17 at 5:22
  • $\begingroup$ Yes, you need to use the fact that $p$ is a unit in $R$, since it is relatively prime to the characteristic. Briefly, if $u=1+n$ where $n\neq 0$ is nilpotent and $u^p=1$, then we have $u^p=1+pn+\text{higher powers of }n$ and so $0=pn+\text{higher powers of }n$. You then get a contradiction by multiplying by $n^{d-2}$ for the least $d$ such that $n^d=0$, since $p$ is a unit. $\endgroup$ – Eric Wofsey May 15 '17 at 5:40

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