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I am exploring whether the following assertion holds true in integral domains: $\gcd(\text{multiples}(a,b))= \text{lcm}(a, b)$. Let us make this formal below.

Consider two elements $a$ and $b$ in an integral domain. Define the LCM of $a$ and $b$ and also define the GCD of all multiples of $a$ and $b$. Specifically:

  • Let $l = \gcd(\{m : a \mid m \text{ and } b \mid m\})$, where $m$ is any common multiple of $a$ and $b$.
  • Let $s = \text{lcm}(a, b)$.

We describe $l$ and $s$ in terms of the following conditions:

  • Condition (i): $l$ divides every $m$ where $a \mid m$ and $b \mid m$ (This seems identical to the condition for $s$).
  • Condition (ii): For any $\tilde{l}$ that divides every $m$ where $a \mid m$ and $b \mid m$, $l \mid \tilde{l}$.

For the LCM:

  • Condition (a): $a$ divides $s$ and $b$ divides $s$.
  • Condition (b): $s$ divides every $m$ where $a \mid m$ and $b \mid m$ (This matches Condition (i) for $l$).

The identity between Condition (i) for $l$ and Condition (b) for $s$ is clear, but I am uncertain if Conditions (i) and (ii) together imply Condition (a). If they do, then we could conclude that $l = s$, proving the property in integral domains.

Could anyone help clarify if this implication is valid or provide a proof or counterexample? Any input or detailed insight into this matter would be greatly appreciated.

Bill Dubuque
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Martin Geller
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  • Yes, $,{\rm lcm}(a,b),$ is the gcd of all common multiples of $,a,b.$ Indeed we have the descent: if $,m,m'$ are divisibly incomparable common multiples then $,\gcd(m,m'),$ is a smaller common multiple - see my comments here. Dually, the gcd is the lcm of all common divisors. – Bill Dubuque Jun 17 '24 at 16:23
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    Note also that gcds and lcms generally don't exist in a domain. Are you working in a gcd domain? – Bill Dubuque Jun 17 '24 at 16:35
  • I guess the statement would hold vacuously true if $gcd's$ and $lcm$'s didn't exist in a given integral domain; and, yes, I am working in a PID – Martin Geller Jun 17 '24 at 16:37
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    In a PID it follows immediately by the definition = universal property of the ideal sum, viz. $$\gcd(A,B) = A+B = \cap_C{C\supseteq A, B)= {\rm lcm}_C{C\mid A,B}\qquad$$ – Bill Dubuque Jun 17 '24 at 16:42
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    When you say "multiples" (of $a$ and $b$) it seems like you always mean "common multiples". All in all your question is hard to parse; it would be easier if you just asked something of the form: if some element satisfies such and such conditions, does it always also satisfy this condition? – Marc van Leeuwen Jun 19 '24 at 08:12
  • I disagree. Using common multiples is to abuse notation. For $A\subset R$ where $R$ is a ring, define $\text{multiples}(A):= {r\in R : a|r : \forall a \in A}$. The non-common multiples of the elements of $a$ can simply be expressed as $\cup_{a\in A}\text{multiples}(a)$. The conditions were just my first approach to prove them, which is almost by definition not going to be the most streamlined: they had nothing to do with my question per se, which is succintly expressed in the question title. For a streamlined approach, see my own answer below – Martin Geller Jun 20 '24 at 09:09

3 Answers3

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We know that $\text{lcm}(a,b)\in\text{divisors}(\text{multiples}(a,b))$. Hence $\text{lcm}(a,b)| \text{gcd}(\text{multiples}(a,b))$. On the other hand, $\text{lcm}(a,b) \in \text{multiples}(a,b)$. So any divisor of the multiples of $a$ and $b$, and in particular $\text{gcd}(\text{multiples}(a,b))$, divides $\text{lcm}(a,b)$. Hence $\text{lcm}(a,b) \sim \text{gcd}(\text{multiples}(a,b))$, and since the $\text{gcd}$ is unique up to the equivalence relation $\sim$, $\text{lcm}(a,b)$ is a $\text{gcd}(\text{multiples}(a,b))$, which is the correct statement that we should have been trying to prove.

Martin Geller
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Clearer: $ $ if a set $M = \{m_1,m_2,\ldots\}$ contains a $ $ $\rm\color{#c00}{common}$ $ $ divisor $\,\color{#c00}{m_1\mid m_2,\,m_3,\,\ldots}\,$ then $\,\gcd\,\{m_i\} \sim m_1,\ $ since $\,m_1\,$ is also $ $ (divisibly) $ $ $\rm\color{#0a0}{greatest}$ $ $ by $\,d\mid \color{#0a0}{m_1},m_2,\ldots \Rightarrow d\mid \color{#0a0}{m_1}$.

OP is special case $\,M =\,$ all common multiples of $\,a,b\,$ and $\,m_1 = {\rm lcm}(a,b),\,$ which divides every common multiple $\,m_i\,$ of $\,a,b\,$ by the lcm definition / universal property.

Or we can apply gcd mod reduction to reduce: $m_i\to m_i\bmod m_1 = 0\,$ for all $\,i>1,\,$ i.e.

$$(m_1,m_2,m_3,\ldots)\, =\, (m_1,\,\underbrace{m_2\bmod m_1}_{0},\,\underbrace{m_3\bmod m_1}_{0},\,\ldots) \,=\, (m_1)\qquad\qquad$$

Bill Dubuque
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Define $a\leq b$ iff $a | b$. If your ring is nice enough, then $\leq$ is a partial order on the equivalence classes $x\sim y$ iff $x = c\cdot y$ for some invertible $c$.

I claim your question has not much to do with rings.

Then $\gcd = \inf$ and $\text{lcm} = \sup$ in this order. You are asking if $\inf M = \sup \{a, b\}$ where $M$ is the set of all upper bounds of $\{a, b\}$, assuming both sides exist. Recall that definition of $\sup\{a, b\}$ is that of $\min M$, so existence of $\sup \{a, b\}$ implies existence of $\inf M$ and even something stronger, and the two sides are equal - by definition.

Jakobian
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