Euclid's Lemma for Euclidean Domains

Theorem
Let $\left({D, +, \times}\right)$ be a Euclidean domain whose unity is $1$.

Let $a \mathop \backslash b$ denote that $a$ is a divisor of $b$.

Let $a, b, c \in D$.

If $a \mathop \backslash b \times c$, where $a$ and $b$ are relatively prime, then $a \mathop \backslash c$.

Proof
We have $a \perp b$ from the definition of relatively prime.

That is, $\gcd \left\{{a, b}\right\} = 1$.

From Bézout's lemma, we may write $a \times x + b \times y = 1$ for some $x, y \in D$.

Upon multiplication by $c$, we see that $c = c \times \left({a \times x + b \times y}\right) = c \times a \times x + c \times b \times y$.

Since $a \mathop \backslash a \times c$ and $a \mathop \backslash b \times c$, it is clear that $a \mathop \backslash \left({c \times a \times x + c \times b \times y}\right)$.

However, $c \times a \times x + c \times b \times y = c \times \left({a \times x + b \times y}\right) = c \times 1 = c$.

Therefore, $a \mathop \backslash c$.

Also see

 * Euclid's Lemma for Irreducible Elements


 * Euclid's Lemma, for the usual statement of this result, which is this lemma as applied specifically to the integers.