# Euclidean Domain is GCD Domain

## Theorem

Let $\struct {D, +, \times}$ be a Euclidean domain.

Then any two elements $a, b \in D$ have a greatest common divisor $d$ such that:

- $d \divides a \land d \divides b$
- $x \divides a \land x \divides b \implies x \divides d$

and $d$ is written $\gcd \set {a, b}$.

For any $a, b \in D$:

- $\exists s, t \in D: s a + t b = d$

Any two greatest common divisors of any $a, b$ are associates.

## Proof

Let $a, b \in D$.

Let $U \subseteq D$ be the set of all elements $h a + k b$ of $D$ where $h, k \in D$.

Then $U$ is an ideal of $D$.

Note that $U = \ideal a + \ideal b$ where $\ideal a$ and $\ideal b$ are Principal Ideal.

By Sum of Ideals is Ideal, $U$ is an ideal.

By Euclidean Domain is Principal Ideal Domain, $U$ is a principal ideal, $\ideal d$ say.

As $a, b \in U$ it follows that $d$ is a divisor of $a$ and $b$, that is:

- $d \divides a \land d \divides b$

Since $d$ itself is in $U$, we have:

- $\exists s, t \in D: s a + t b = d$

By Common Divisor in Integral Domain Divides Linear Combination:

- $x \divides a \land x \divides b \implies x \divides d$

So $d$ is a greatest common divisor of $a$ and $b$.

If $d$ and $d'$ are both greatest common divisors of $a$ and $b$, then $d \divides a \land d \divides b$ and so $d \divides d'$.

Similarly $d' \divides d$.

So $d$ and $d'$ are associates.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 28$. Highest Common Factor: Theorem $54$