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Let $(R,+,\times)$ be a finite ring with an identity. (Dis)Prove that $q\nmid k$ where $q=\text{Char}(R)$ is the characteristic of a ring and $k=\text{Ord}(r)$ is the order of a unit $r$.

If $R=\mathbb{Z}/n\mathbb{Z}$, then $q=n$ and $k\mid \varphi(n)$ where $\varphi$ is the Euler function. It holds.

If $R=\mathbb{F}_{p^s}$, then $q=p$ and $k\mid p^s-1$ where $p$ is prime. It holds.

If $R=GR(p^s,r)$, then $q=p^s$ and $k\mid (p^r-1)p^{(s-1)}$, where $p$ is prime. It holds.

I am wondering if it holds for all rings.

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    $\begingroup$ It holds for integral domains or any ring of prime characteristic without nilpotent element : if $u^p = 1$ then $(u-1)^p=0$ which implies $u=1$. $\endgroup$
    – reuns
    Aug 3 '19 at 13:38
  • $\begingroup$ @reuns it seems that these conditions are sufficient but not necessary. Can you verify whether the two examples given by Ben meet these conditions? I think $\mathbb{F}_p[x]/(x^p)$ does. How about the other example? $\endgroup$ Aug 4 '19 at 3:03
  • $\begingroup$ @reuns Does $p$-nilpotent just mean $x^p=0$ (not also $x^q\neq 0$ for $q<p$)? Else in $F_3[x]/(x^2)$, $1+x$ is a unit of order 3. $\endgroup$
    – Ben
    Aug 6 '19 at 4:52
  • $\begingroup$ Right it doesn't work because there are things like $R=\Bbb{F}_p[x]/(x^2-1) \times \Bbb{F}_2$ where $(x ,1)$ is of order $2$ in $R^\times$ but it is of order $1$ in $R/(2)^\times$ @Ben $\endgroup$
    – reuns
    Aug 6 '19 at 5:25
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No, you can take products of two rings whose unit groups orders divide each others characteristic.

  • Take $F_4$ has units $C_3$ and $F_3$ has units $C_2$, so $F_4\times F_3$ which has characteristic 6 has units $C_3\times C_2$.

ADDED:

Another thing you can do is take nilpotent extensions to add more units to a given ring, of orders divisible by the characteristic. For example

  • $F_{p}[x]/(x^p)$, which has a unit $1+x$ of order $p$.

  • Hochschild square-zero extension of $R$ by a module $I$ with some element of order divisible by the characteristic of $R$. These extensions have isomorphic copy of $I$ in their unit group as $1+I \subset R^\times\times I$.

ADDED LATER:

For finite commutative rings, being reduced is equivalent to semisimple (cf. Jacobson radical; Semisimple), in other words a product of finite fields (by Artin-Wedderburn), like in our first example.

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  • $\begingroup$ Is it possible to characterize those rings in which the orders of unit groups divide the characteristic ? $\endgroup$ Aug 3 '19 at 6:44
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    $\begingroup$ @zongxiangyi I doubt anyone has tried to do this. I added an example of indecomposable rings just to show they exist. There is no classification of finite rings to build off of, so it sounds hard. $\endgroup$
    – Ben
    Aug 3 '19 at 13:47
  • $\begingroup$ I do not think that $F_{p}[x]/(x^p)$ is not indecomposable ring and it has $$\mathbb{F}_{p}[x]/(x^p)\cong \mathbb{F}_p \times \mathbb{F}_p \times \cdots \times \mathbb{F}_p.$$. How do you define "a ring is indecomposable" ? $\endgroup$ Aug 4 '19 at 2:55
  • $\begingroup$ @zongxiangyi That is an isomorphism of vector spaces but not of rings. By indecomposable I mean its not isomorphic (in the category of rings) to a nontrivial product of two rings. This is equivalent to existence of nontrivial idempotents. For instance if $p=2$ none of $0,1,x,1+x$ are idempotent so its not a product of rings. $\endgroup$
    – Ben
    Aug 5 '19 at 3:29
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    $\begingroup$ There is something similar for groups (planetmath.org/indecomposablegroup), and you can look up the notion of indecomposable module (en.wikipedia.org/wiki/Indecomposable_module). A ring $R$ is "indecomposable" iff its (right) regular module is an indecomposable $R$-module. This is because $R = End(R_R)$, and if $R$ has a nontrivial idempotent $e \in R$ then $R \cong Re \times R(1-e)$. $\endgroup$
    – Ben
    Aug 6 '19 at 9:45

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