Bézout's identity in a Euclidean domain

Statement

Let $R$ be a Euclidean domain and let $a, b \in R$. Then $a$ and $b$ have a greatest common divisor $d$, and there are $s, t \in R$ with $$d = s a + t b.$$

Remarks

The Euclidean algorithm computes $d$. The extended Euclidean algorithm also computes $s$ and $t$, by back substitution. In $\mathbb{Z}$: $\gcd(252, 198) = 18 = 4 \cdot 252 - 5 \cdot 198$. The same identity holds in every principal ideal domain, because the ideal $(a, b)$ is principal. If $d \neq 0$, the coefficients are not unique: $s + k b/d$ and $t - k a/d$ also work for each $k \in R$.

Explanations

L1 human

The gcd of two elements is a combination of them with coefficients in the ring. For example, $\gcd(4, 6) = 2 = 2 \cdot 4 - 1 \cdot 6$. This is the main tool to solve linear equations $s a + t b = c$ in a Euclidean domain.