Rings such that "$ x^d - 1 \mid x^n - 1 \iff d \mid n $" holds

Question

There is a classical exercise in elementary number theory or abstract algebra (cf. this question or this question):

Let $ d, n \in \mathbb{Z}_+ $.

• For integers $a \ge 2$, we have: $a^d - 1 \mid a^n - 1 \iff d \mid n$.
• In a polynomial ring $R[x]$ where $R$ is an integral domain, we have: $ x^d - 1 \mid x^n - 1 \iff d \mid n $.

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$\impliedby$ are tivial. For $\implies$, consider $ n = dq + r $, then we have $$ x^d - 1 \mid x^r - 1 $$ or $ a^d - 1 \mid a^r - 1 $. Then we may use $0 \le r < d$ to get the conclusion.

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The proof on $\mathbb{Z}$ and the one on $R[x]$ seems similar but not completely the same. I want to generalize this conclusion to some abstract ring $A$ and give a general proof.

Some of my observation: For $\mathbb{Z}$, we use the fact that $a \mid b \implies |a| \le |b|$; for $R[x]$, we use the fact that $f(x) \mid g(x) \implies \deg(f) \le \deg(g)$. So it seems that the essence of the proof is a map $$ v : A \longrightarrow \mathbb{Z}_{\ge 0}, $$ such that $\forall \, x, y \in A$, we have $x \mid y \implies v(x) \le v(y)$. Is there a name of such a property? Or is there a general kind of rings such that such a property holds?

Answer 1

Partial answer.

Let $R$ be any commutative ring. Define the gcd of two elements as usual as an element with the property $d \mid \mathrm{gcd}(x,y) \iff d \mid x \land d \mid y$. It is unique up to two-sided divisibility $\sim$ defined by $a \sim b \iff a \mid b \mid a$. This proof proof shows that, as soon as $a^n - 1$ and $a^m - 1$ have a gcd, then it divides $a^{\mathrm{gcd}(n,m)} - 1$ and vice versa. Hence, $a^{\mathrm{gcd}(n,m)}-1$ is a gcd of these.

Now assume that $a^m - 1$ divides $a^n - 1$. Then a gcd exists, namely $a^m - 1$. On the other hand, $a^{\mathrm{gcd}(n,m)} - 1$ is a gcd. Hence, $a^m - 1 \sim a^{\mathrm{gcd}(n,m)} - 1$. The converse is also true: If $a^m - 1 \sim a^{\mathrm{gcd}(n,m)} - 1$, then $a^m - 1$ divides $a^{\mathrm{gcd}(n,m)} - 1$, which divides $a^n - 1$.

Hence, the divisibility condition is equivalent to $a^m - 1 \sim a^{\mathrm{gcd}(n,m)} - 1$. The question is when this implies $m \mid n$, which is just $\mathrm{gcd}(n,m) = m$. Notice that $a^m = (a^{\mathrm{gcd}(n,m)})^k$ for some $k \geq 1$.

This reduces the problem to:

Which commutative rings $R$ and elements $a$ have the property that for all $n,k \geq 1$ we have $a^n - 1 \sim a^{nk} -1 \implies k=1~$?

By substituting $a^n$, it is sufficient to answer:

Which commutative rings $R$ and elements $a$ have the property that for all $k > 1$ we have $a - 1 \not\sim a^{k} -1 ~$?

This is clearly satisfied for the two examples that you mentioned and unifies their proofs also to some degree.

Answer 2

I prefer to think about this type of result not in terms of divisibility, but in terms of greatest common divisors and ideals: (i) when $a \geq 2$ in $\mathbf Z$ we have $\gcd(a^m-1,a^n-1) = a^{(m,n)}-1$ and (ii) in $R[x]$ where $R$ is an integral domain, the ideal $(x^m-1,x^n-1)$ equals the ideal $(x^{(m,n)}-1)$.

Here is a generalization to all commutative rings.

Theorem. If $A$ is a commutative ring, $a \in A$, and $m$ and $n$ are positive integers, then the ideals $(a^m-1,a^n-1)$ and $(a^{(m,n)}-1)$ are equal.

Proof. Let $I = (a^m-1,a^n-1)$ and $J = (a^{(m,n)}-1)$, so we want to prove $I = J$. Obviously $a^m-1$ and $a^n-1$ are multiples of $a^{(m,n)}-1$ in $A$, so $I \subset J$. To prove $J \subset I$, we will show $a^{(m,n)} \equiv 1 \bmod I$.

In the quotient ring $A/I$, we have $\overline{a}^m = \overline{1}$ and $\overline{a}^n = \overline{1}$, so $\overline{a}$ is a unit in $A/I$ with order dividing $m$ and $n$. Thus its multiplicative order divides $(m,n)$, so $\overline{a}^{(m,n)} = \overline{1}$, which says $a^{(m,n)} \equiv 1 \bmod I$ and we're done. QED

Example. Let $A = \mathbf Z$ and $a \geq 2$. Then the ideal $(a^m-1,a^n-1)$ in $\mathbf Z$ has a unique positive generator, which general theory says is the gcd of $a^m-1$ and $a^n-1$. At the same time, the theorem above says this ideal is $(a^{(m,n)}-1)$ and $a^{(m,n)} - 1$ is positive. Thus $\gcd(a^m-1,a^n-1) = a^{(m,n)}-1$.

Example. Let $A = R[x]$ where $R$ is any commutative ring, not necessarily an integral domain. The theorem says for all $f(x) \in A$ that $(f(x)^m-1,f(x)^n-1) = (f(x)^{(m,n)}-1)$ as ideals in $R[x]$. In particular, we have $(x^m-1,x^n-1) = (x^{(m,n)}-1)$ as ideals in $R[x]$.