A question on finite commutative rings

Question

I have considered about Lehmer's conjecture that φ(n)|(n-1) if and only if n is a prime. I generalized the conjecture onto genernal finite commutative ring. The idea is below.

Set $A$ is a finite commutative ring, and $B$ is its unit group. Then there is a partition of $A$ by $B$ in multipication. Choose the representatives $\lbrace a_1,a_2,...a_k\rbrace$, then define a subgroup $B_i$ of $B$ as $B_i=\lbrace b \in B \mid b\cdot a_i=a_i\rbrace$, for $1 \le i \le k$. Clearly, $\sum_{i=1}^k [B:B_i]=|A|-1$, for $\lbrace a_i \rbrace$ is the collection of representatives. With a little more consideration, this converts to the well-known result in basic number theory that $\sum_{d|n}\phi(d)=n$ (using Chinese residue theorem, or fact that every artin ring is a direct sum of local rings). Now, what I conjectured is, if $|B|$ can divide $|A|$, then $A$ must be a field.

I don't know much about finite commutative rings, but I can see its significance in number theory here. Moreover, I got to know some analytic methods used in the attacking of Lehmer's conjecture, such as Kevin Ford from UIUC. When faced the problem about rings, some representation theory may work. I don't know.

Answer 1

Suppose that $|A^\times|$ divides $|A| -1$, where $A^\times = B$ is the group of units.

Since a finite ring has a unique factorization into local rings, we can write $A = \prod_{i=1}^m A_i$ with local rings $(A_i,\mathfrak{m}_i)$ and find for the unit group

$$A^\times = \prod_{i=1}^m \hspace{1pt} A_i^\times.$$

[This editor makes me crazy: Without puting this formular into a single line it produces rubbish, having it in a single line, either with single-dollar or double-dollar, it works !?]

Since $(\mathfrak{m}_i,+)$ is a subgroup of $(A_i,+)$, we know that $|\mathfrak{m}_i|$ divides $|A_i|$, say $|A_i| = k_i |\mathfrak{m}_i|$. Now $A_i^\times = A_i \setminus \mathfrak{m}_i$ implies $|A_i^\times| = (k_i-1)|\mathfrak{m}_i|$. Since $|A^\times|$ divides $|A| -1$ we conclude that $|\mathfrak{m}_i|$ divides $-1$, what is only possible for $\mathfrak{m}_i = 0$. Thus $A_i$ is a field and we have shown:

$A$ is a direct product of fields

Let $|A_i|=q_i$. Then the assumption above is equivalent to $$ \prod_{i=1}^m (q_i -1) \quad \text{divides} \quad \prod_{i=1}^m q_i -1. \quad\quad\quad (\ast)$$

As I learned from A Haynes in the following link - and was correctly suggested by the OP - $(\ast)$ is a generalization of the Lehmer totient problem and still unsolved.

A question on divisibility of a product of primes

In case $m=2$ it's easy to see that the only possibilities for $A$ are
$$\mathbb{F}_2 \times \mathbb{F}_2, \quad\quad \mathbb{F}_3 \times \mathbb{F}_3$$ Moreover, as pointed out by François in his comment, $A= \mathbb{F}_2^m$ satisfies the assumption for every $m$.