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Let R be a commutative ring with identity, and J an ideal generated by the members $a^n-1$ and $a^m-1$ for some $a \in R$ and $n, m$ positive integers.

I want to establish that the principal ideal generated by $a^d - 1$ where $d = gcd(m, n)$ (lets call it $I$) is infact equal to $J$.

Since I am not necessarily dealing with an integral domain (e.g. I could have zero divisors) and in particular its not necessarily a GCD domain, I do not get gcd-like elements "for free".

However, assuming $n > m$, I can express $a^n -1$ in terms of a linear combination (in $R$) of $a^{n-m}-1$ and $a^m-1$ and similary can express $a^{n-m}-1$ in terms of $a^n-1$ and $a^m-1$. Therefore the subring (ideal) generated by $(a^n-1, a^m-1)$ is the same as the one generated by $(a^{n-m}-1, a^m-1)$. I can now continue this downward iteration in similar fashion to the euclidean algorithm for finding GCD, and end up at a pair that looks like $(a^{gcd(m, n)}-1, a^{gcd(m, n)}-1)$ whence I deduce that the ideal generated by $a^d-1$ is the same as the one generated by $a^{n-m}-1$ and $a^m-1$.

Assuming we take the validity of the euclidean algorithm for finding an integral gcd as a given, does this reasoning apply? I am skeptical of my own way since I worry that there is a detail to do with possibility of zero divisors, and also the original question posed to me suggests a more roundabout route of establishing that $a$ is a unit in the quotient ring R/J and has order that is multiple of $d$ with respect to the multiplicative group of units in that quotient ring.

giorgio
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  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque May 11 '24 at 03:05
  • Most of the proofs in the dupe also work here. Alternatively specialize (evaluate) the (universal) Bezout equation in $\Bbb Z[x]$ to deduce that the gcd persists for elements of any ring, analogous to here. Similarly we can specialize any ideal equality in $\Bbb Z[x],,$ i.e. they are universal ideal equalities. – Bill Dubuque May 11 '24 at 03:09

1 Answers1

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Your intuition seems right, but it may be easier to use what we have in $\mathbb{Z}$ rather than try to copy-paste it onto a new ring. This will avoid concerns about new factors (e.g. presence of zerodivisors) from becoming relevant. I have given one solution below as an example of working in $\mathbb{Z}$ and then applying that to $R$.

I would guess that the recommended solution is recommended for the same reason: to avoid thinking about too many moving parts.

First, consider the following:

Fact: If $f|g$, say $g =fn$, then $a^f-1$ divides $a^g-1$. This is because we can write:

$$(\sum_{i=0}^{n-1}a^{if})(a^f-1) = a^g-1$$

We will use this fact twice below.

Write $d = xn + ym$ via Bezout’s Lemma. Then:

$$a^d-1 = a^{xn+ym}-1 = a^{xn}(a^{ym}-1) + a^{xn}-1$$

and by the above fact, as $a^{xn}-1$ and $a^{ym}-1$ are respectively in $(a^n-1)$ and $(a^m-1)$, we have that $a^d-1$ is in their sum.

However, we already have that $a^d-1$ divides both of them, so applying the above fact we are done.

  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque May 11 '24 at 02:54
  • Got it - thank you and sorry for this being a dupe I looked at other threads but they seemed to be more specific eg to integer rings or polynomial rings so I thought there is value in posing the more general question but perhaps it is superfluous in any case – giorgio May 11 '24 at 03:44