Euclid's Lemma for Euclidean Domains
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Theorem
Let $\left({D, +, \times}\right)$ be a Euclidean domain whose unity is $1$.
Let $a \backslash b$ denote that $a$ is a divisor of $b$.
Let $a, b, c \in D$.
If $a \backslash b \times c$, where $a$ and $b$ are relatively prime, then $a \backslash c$.
Proof
Let $\left({D, +, \times}\right)$ be a Euclidean domain whose unity is $1$.
Let $a, b, c \in D$.
$a \perp b$ 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 $a \backslash a \times c$ and $a \backslash b \times c$, it is clear that $a \backslash \left({c \times a \times x + c \times b \times y}\right)$.
However, $c \times a \times x + c \times b \times y = c \left({a \times x + b \times y}\right) = c \times 1 = c$.
Therefore, $a \backslash c$.
$\blacksquare$
Source of Name
This entry was named for Euclid.
Also see
- Euclid's Lemma, for the usual statement of this result, which is this lemma as applied specifically to the integers.
Sources
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.29$: Theorem $55$
