Least common multiples

Setup: We're in a Euclidean domain $R$ and have elements $a, b \in R$ .

We've shown that a least common multiple of $a$ and $b$ is any element $e \in R$ such that $⟨ e ⟩ = ⟨ a ⟩ \cap ⟨ b ⟩$ , and we've shown such an element is unique up to multiplication by a unit.

We're now showing that the element $e = \frac{a b}{d}$ is a least common multiple of $a$ and $b$ , where $d$ is a greatest common divisor of $a$ and $b$ .

We've established that $e$ is a common multiple of $a$ and $b$ . Now we'll show it's the least common multiple. First recall that a property of a greatest common divisor is that it can be written as an $R$ -linear combination of $a$ and $b$ . In other words, there are elements $r, s \in R$ such that

d = r a + s b .

Note that since $d$ divides $a$ and $b$ , we can actually "divide" this equation by $d$ to get

1 = r \cdot \frac{a}{d} + s \cdot \frac{b}{d} .

(What we're really doing is writing $a = k d$ and $b = l d$ for some $k, l \in R$ , and then using cancelation on both sides to get $1 = r k + s l$ .) We'll use this equality in a little bit.

Suppose $e^{'}$ is a least common multiple of $a$ and $b$ . Since $e$ is a common multiple and $e^{'}$ is a least common multiple, this means $e = t e^{'}$ for some $t \in R$ . We'll show that $t$ must be a unit, which will prove $e$ is also a least common multiple of $a$ and $b$ . We also know $e^{'} = x a = y b$ for some $x, y \in R$ . Combining this with the equality $e = t e^{'}$ yields

\frac{a b}{d} = e = t e^{'} = t x a and \frac{a b}{d} = e = t e^{'} = t y b .

In the first equality we can cancel $a$ , and in the second equality we can cancel $b$ , to get

\frac{b}{d} = t x and \frac{a}{d} = t y .

Finally we can use our earlier equality to deduce

1 = r \cdot \frac{a}{d} + s \cdot \frac{b}{d} = r \cdot t y + s \cdot t x = t (r y + s x) .

This proves $t$ is a unit in $R$ , with inverse given by $r y + s x$ . Since $e^{'}$ is a least common multiple of $a$ and $b$ , and since $e = t e^{'}$ with $t$ a unit, we've thus proven $e$ is also a least common multiple of $a$ and $b$ .