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$gcd cdot lcm$ for cyclic rings

Prove that $gcd(M, N)times mbox{lcm}(M, N) = M times N$.Transfer Between LCM, GCD for Rings?Are these about determinant true for commutative rings?Rings isomorphic to $mathbb{Z}_6timesmathbb{Z}_{10}$Prove if If $m in Z^+$, $a|m$, and $b|m$, then $mbox{lcm}(a,b) leq m$.Prove that $G times H$ is cyclic $iff$ $gcd(m,n) =1$$Goplus H$ is cyclic iff finite groups $G$ and $H$ are cyclic and $gcd(|G|,|H|)=1$$gcd(|H_{1}|,|H_{2}|,cdots,|H_{n}|) > 1$ implies $H_{1} times H_{2} times cdots times H_{n} = G$ not cyclicStrict cyclic orderUniqueness of least common multiple up to associatesRing theory: well-definedness of associates without commutativityConnection between GCD and LCM of two numbers

0

$begingroup$

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

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

It looks like the formula $gcd(a,b) cdot lcm(a,b) sim a cdot b$ works for any cyclic ring, even if there is no unity in it, as long as $gcd(a,b)$ exists:

In $2mathbb Z$:

$gcd(4,8) sim 2$
$lcm(4,8) sim 16$
$gcd(4,8) cdot lcm(4,8) sim 4 cdot 8 sim 32$

In $2mathbb Z_{12}$:

$gcd(4,8) sim 4$
$lcm(4,8) sim 4$
$gcd(4,8) cdot lcm(4,8) sim 4 cdot 8 sim 4$

There are proofs of the $gcd(a,b) cdot lcm(a,b)$ formula for an integral domain:
Prove that $gcd(M, N)times mbox{lcm}(M, N) = M times N$.
https://math.stackexchange.com/a/717775/427611

How do we show it for an arbitrary cyclic ring?

For an infinite cyclic ring $kmathbb Z$ the proof looks simple:

if there is a $gcd(a,b)$ in $kmathbb Z$, then

$gcd'(a,b) sim k cdot gcd(a,b)$
$lcm(a,b) sim k cdot lcm'(a,b)$

where $gcd'(a,b)$ and $lcm'(a,b)$ are the corresponding values in $mathbb Z$.

I need help with a finite cyclic ring $kmathbb Z_{kn}$.

abstract-algebra ring-theory modular-arithmetic greatest-common-divisor least-common-multiple

share|cite|improve this question

asked 16 hours ago

Alex CAlex C

7518

$endgroup$

add a comment | 

0

$begingroup$

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

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

It looks like the formula $gcd(a,b) cdot lcm(a,b) sim a cdot b$ works for any cyclic ring, even if there is no unity in it, as long as $gcd(a,b)$ exists:

In $2mathbb Z$:

$gcd(4,8) sim 2$
$lcm(4,8) sim 16$
$gcd(4,8) cdot lcm(4,8) sim 4 cdot 8 sim 32$

In $2mathbb Z_{12}$:

$gcd(4,8) sim 4$
$lcm(4,8) sim 4$
$gcd(4,8) cdot lcm(4,8) sim 4 cdot 8 sim 4$

There are proofs of the $gcd(a,b) cdot lcm(a,b)$ formula for an integral domain:
Prove that $gcd(M, N)times mbox{lcm}(M, N) = M times N$.
https://math.stackexchange.com/a/717775/427611

How do we show it for an arbitrary cyclic ring?

For an infinite cyclic ring $kmathbb Z$ the proof looks simple:

if there is a $gcd(a,b)$ in $kmathbb Z$, then

$gcd'(a,b) sim k cdot gcd(a,b)$
$lcm(a,b) sim k cdot lcm'(a,b)$

where $gcd'(a,b)$ and $lcm'(a,b)$ are the corresponding values in $mathbb Z$.

I need help with a finite cyclic ring $kmathbb Z_{kn}$.

abstract-algebra ring-theory modular-arithmetic greatest-common-divisor least-common-multiple

share|cite|improve this question

asked 16 hours ago

Alex CAlex C

7518

$endgroup$

add a comment | 

$begingroup$

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

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

It looks like the formula $gcd(a,b) cdot lcm(a,b) sim a cdot b$ works for any cyclic ring, even if there is no unity in it, as long as $gcd(a,b)$ exists:

In $2mathbb Z$:

$gcd(4,8) sim 2$
$lcm(4,8) sim 16$
$gcd(4,8) cdot lcm(4,8) sim 4 cdot 8 sim 32$

In $2mathbb Z_{12}$:

$gcd(4,8) sim 4$
$lcm(4,8) sim 4$
$gcd(4,8) cdot lcm(4,8) sim 4 cdot 8 sim 4$

There are proofs of the $gcd(a,b) cdot lcm(a,b)$ formula for an integral domain:
Prove that $gcd(M, N)times mbox{lcm}(M, N) = M times N$.
https://math.stackexchange.com/a/717775/427611

How do we show it for an arbitrary cyclic ring?

For an infinite cyclic ring $kmathbb Z$ the proof looks simple:

if there is a $gcd(a,b)$ in $kmathbb Z$, then

$gcd'(a,b) sim k cdot gcd(a,b)$
$lcm(a,b) sim k cdot lcm'(a,b)$

where $gcd'(a,b)$ and $lcm'(a,b)$ are the corresponding values in $mathbb Z$.

I need help with a finite cyclic ring $kmathbb Z_{kn}$.

abstract-algebra ring-theory modular-arithmetic greatest-common-divisor least-common-multiple

share|cite|improve this question

asked 16 hours ago

Alex CAlex C

7518

$endgroup$

share|cite|improve this question

asked 16 hours ago

Alex CAlex C

7518

share|cite|improve this question

asked 16 hours ago

Alex CAlex C

7518

asked 16 hours ago

Alex CAlex C

7518

asked 16 hours ago

Alex CAlex C

7518