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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.

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For better readability, I turned your formulars into latex. I hope you don't mind. – Ralph Nov 24 '11 at 13:32
It seems there are trivial counterexamples : $A=(\mathbf{Z}/2\mathbf{Z})^n$ (for which $B=\{1\}$). Besides, you probably mean "if $|B|$ divides $|A|-1$" ? – François Brunault Nov 24 '11 at 13:46
This is a side remark, but the word "conjecture" in mathematics means something that you have a lot of circumstantial evidence for, that is true in all particular cases you can think of, and that you couldn't refute despite a few months of decent work on counterexamples. It doesn't mean "something that just came into my head and looks cute" or "something that would fill that damn gap in my failing proof". – fedja Nov 24 '11 at 14:32
François' example gives in case $n > 1$ also a counterexample for "$|B|$ divides $|A|-1$". – Ralph Nov 24 '11 at 14:46
I can see where the point is. Yes, I mean |B| divides |A|-1. Thanks for your answer. How about all the couterexamples? – Zhipeng Lu Nov 24 '11 at 15:22
up vote 3 down vote accepted

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 $m$.

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Nice argument! Some other examples for $m=3$ : $\mathbf{F}_2 \times \mathbf{F}_2 \times \mathbf{F}_4$, $\mathbf{F}_2 \times \mathbf{F}_4 \times \mathbf{F}_8$. And for $m=4$ : $\mathbf{F}_2 \times \mathbf{F}_2 \times \mathbf{F}_2 \times \mathbf{F}_8$, $\mathbf{F}_2 \times \mathbf{F}_2 \times \mathbf{F}_4 \times \mathbf{F}_4$. – François Brunault Nov 24 '11 at 20:41
In fact one can show that given any finite field $K=\mathbf{F}_{2^n}$ (or any finite product of such fields), there will be an arithmetic progression of integers $m$ such that the finite ring $A=(\mathbf{F}_2)^m \times K$ works. – François Brunault Nov 24 '11 at 21:01
Nice examples. I guess the OP's conjecture (or whatever the correct wording is) has at most a chance to be true when $|A|$ is odd and if the counterexample for $m=2$ is an exception. – Ralph Nov 24 '11 at 21:10
Just saw your counterexamples in the linked question. I would say, you've clearly shown that the conjecture is false in general. – Ralph Nov 24 '11 at 21:22

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