Is it true that $q\nmid k$ where $q$ is the characteristic of a ring and $k$ is the order of any unit?

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.

• 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 $u=1$. Aug 3 '19 at 13:38
• @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? Aug 4 '19 at 3:03
• @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.
– Ben
Aug 6 '19 at 4:52
• 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 Aug 6 '19 at 5:25

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 $$C_3\times C_2$$.

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 $$1+I \subset R^\times\times I$$.

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.

• Is it possible to characterize those rings in which the orders of unit groups divide the characteristic ? Aug 3 '19 at 6:44
• @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.
– Ben
Aug 3 '19 at 13:47
• 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" ? Aug 4 '19 at 2:55
• @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.
– Ben
Aug 5 '19 at 3:29
• 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 $R \cong Re \times R(1-e)$.
– Ben
Aug 6 '19 at 9:45