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.
Best Answer
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.