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A cyclic ring is a ring (or rng) whose additive group is cyclic.

Two elements of a commutative ring are associates $(\sim)$ iff they divide each other.

An element $d$ of a commutative ring is a $\gcd(a,b)$ iff:

  • $d$ is a common divisor of $a$ and $b$, and
  • any common divisor of $a$ and $b$ divides $d$.

An element $m$ of a commutative ring is an $\mathrm{lcm}(a,b)$ iff:

  • $m$ is a common multiple of $a$ and $b$, and
  • $m$ divides any common multiple of $a$ and $b$.

It looks like the formula

$\gcd(a,b) \cdot \mathrm{lcm}(a,b) \sim a \cdot b$

works for any cyclic ring, even if there is no unity in it, as long as $\gcd(a,b)$ exists.

For example, in $2\mathbb Z_{12}$:

$\gcd(4,8) \sim 4$
$\mathrm{lcm}(4,8) \sim 4$
$\gcd(4,8) \cdot \mathrm{lcm}(4,8) \sim 4 \sim 4 \cdot 8$

There are proofs of the $\gcd \cdot \mathrm{lcm}$ formula for an integral domain:
Prove that $\gcd(M, N)\times \mbox{lcm}(M, N) = M \times N$. and
Transfer Between LCM, GCD for Rings?

How do we show it for an arbitrary cyclic ring?

In an infinite cyclic ring $k \mathbb Z$ a nonzero $\gcd$ exists if only $k = 1$ since any $\gcd$ must divide itself.

I need help with a finite cyclic ring $k \mathbb Z_{kn}$.

Bill Dubuque
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Alex C
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  • How do you define $\gcd(a,b)$ in a finite ring when there is no natural ordering that respects the structure as there usually is in an infinite ring, then it makes sense (because the order exists) to speak of "greatest". – Daniel Donnelly Apr 01 '19 at 20:32
  • Maybe this "universal" definition will work: Define the greatest common divisor of $a, b \in \Bbb{Z}/n\Bbb{Z}$ (if it exists) to be an element $d \in \Bbb{Z}/n \Bbb{Z}$ such that ${da' = a, \ db' = b }$ has a solution $a', b'$ and if $e$ is another such (factors both $a, b$), then $d$ factors through $e$: $d = ed'$. – Daniel Donnelly Apr 01 '19 at 20:38
  • @BananaCats: I was using the following definition from Wikipedia (https://en.wikipedia.org/wiki/Greatest_common_divisor#In_commutative_rings): an element $d$ of a commutative ring is a $GCD(a,b)$ iff $1) d$ is a common divisor of $a$ and $b$, and $2)$ any common divisor of $a$ and $b$ divides $d$. Also, one can use the natural cyclic order on the cyclic rings: https://math.stackexchange.com/q/2213048/427611 – Alex C Apr 01 '19 at 21:04
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    You might be interested in these "Dirichlet" convolution rings that involve GCD / LCM: https://math.stackexchange.com/questions/3169021/can-we-invert-these-analogous-dirichlet-series-for-gcd-lcm-convolution – Daniel Donnelly Apr 01 '19 at 21:21
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    I added the definitions of $gcd$ and $lcm$ into the original question. – Alex C Apr 01 '19 at 21:43

1 Answers1

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  1. Definition. Prime-power cyclic ring is a finite cyclic ring (rng) $p^m \mathbb{Z}_{p^n}$,
    where $p$ is a prime number, and $0 \le m \lt n$;

  2. $a \sim p^k, m \le k \le n$ for any element $a$ of a prime-power cyclic ring $p^m \mathbb{Z}_{p^n}$:

    • $p^k \mathbb{Z}_{p^n}, m \leq k \leq n$, are the only ideals of $p^m \mathbb{Z}_{p^n}$;

  3. $\gcd(a, b) \cdot \mathrm{lcm}(a, b) \sim a \cdot b$ in a prime-power cyclic ring:

    • assuming $gcd(a, b)$ exists, $a \sim p^x, b \sim p^y, m \le x \le y \le n$;
    • then $gcd(a, b) \sim p^{x - m}, lcm(a, b) \sim p^{y + m}$;
    • $\gcd(a, b) \cdot \mathrm{lcm}(a, b) \sim p^{x - y} \cdot p^{y + m} = p^x \cdot p^y \sim a \cdot b$

  4. If $\gcd(a, b) \cdot \mathrm{lcm}(a, b) \sim a \cdot b$ in two rings (rngs) $A$ and $B$,
    then $\gcd(a, b) \cdot \mathrm{lcm}(a, b) \sim a \cdot b$ in the ring $A \times B$:

    • $\gcd((a,b), (c,d)) = (\gcd(a,c), \gcd(b,d))$;
    • $\mathrm{lcm}((a,b), (c,d)) = (\mathrm{lcm}(a,c), \mathrm{lcm}(b,d))$;
    • $(a, b) \sim (c, d) \iff a \sim c \land b \sim d$.

  5. Any finite cyclic ring (rng) $k \mathbb Z_{kn}$ with $k > 0, n > 1$ is a direct sum of prime-power cyclic rings.

Note: in this proof $a \sim b$ in an rng $R$ if $a$ and $b$ generate the same principal ideal $R \cdot a + \mathbb{Z}a$.

Alex C
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