Relationship between units of a ring and primitive characters of the ring under addition

Question

Let $(R,+,\cdot)$ be a finite ring. Obviously, $(R,+)$ is an abelian group; however the unit (multiplicative) group need not be abelian.

My question is the following problem:

Given the dual group $\widehat{R}$ of $(R,+)$, under what conditions of $R$ is it true that

$$ \#\{\text{primitive* characters in ($\widehat{R},\cdot$)} =\#R^\times, $$ i.e. when does the number of primitive* characters of the abelian group $(R,+)$ equal the number of units in $R$?

*((Added)). A character $\chi\in\widehat{R}$ is said to be primitive with respect to a collection $\mathcal{C}$ of subgroups of $(R, +)$ if $\chi$ does not descend to a character on any quotient group $R/S$ for all $S\in\mathcal{C}$. To descend here is canonical in the sense that $\chi$ is invariant on the elements of any coset of $R/S$.

NOTES
The archetypal example is the ring $\mathbb{Z}/n\mathbb{Z}=(\{0,1,2,\dotsc,n-1\},+,\cdot)$ of integers modulo $n$ where the answer is positive when $\mathcal{C}$ contains all $\mathbb{Z}/d\mathbb{Z}$ for all $d$ divides $n$.

Answer 1

The rings $\mathbb{Z}/n$ are the only examples.

I assume that "primitive character" just means that it is faithful, or equivalently that it does not factor through a proper quotient; this is the meaning for Dirichlet characters, I think.

The fact that this is true for $R = \mathbb{Z}/n$ is sort of a coincidence. It comes from the fact that

• the primitive characters of $\mathbb{Z}/n$ can canonically be identified with the primitive $n^\text{th}$ roots of unity in $\mathbb{C}$, and there are $\varphi(n)$ of these, and
• the unit group $(\mathbb{Z}/n)^{\times}$ of $\mathbb{Z}/n$ also has size $\varphi(n)$.

These can't be canonically identified (the latter is a group but the former isn't). The unit group does act freely and transitively on the primitive $n^\text{th}$ roots of unity, because it is also the automorphism group $\operatorname{Aut}(\mathbb{Z}/n)$, and the Galois group $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, but to get an identification from here requires choosing a primitive $n^\text{th}$ root of unity, and to identify $\operatorname{Aut}(\mathbb{Z}/n)$ with the unit group involves the isomorphism $\operatorname{End}(\mathbb{Z}/n) \cong \mathbb{Z}/n$, where $\mathbb{Z}/n$ is an abelian group on the LHS and a ring on the RHS. So there are four interesting sets here all of which have cardinality $\varphi(n)$, as well as two different versions of $\mathbb{Z}/n$, the group and the ring.

In general the number of primitive characters of the additive group of $R$ is zero; among the finite abelian groups only the finite cyclic groups can have primitive characters (since these are the only finite subgroups of $\mathbb{C}^{\times}$). It is only nonzero if $(R, +) \cong \mathbb{Z}/n$, meaning it must be additively generated by a single $r \in R$, meaning $1$ must be an integer multiple of $r$, say $1 = kr$. This means $k$ and $r$ must be units $\bmod n$ so $r = k^{-1} \cdot 1$ and $R \cong \mathbb{Z}/n$ as a ring.