# Bézout's Lemma/Principal Ideal Domain

## Theorem

Let $\struct {D, +, \circ}$ be a principal ideal domain.

Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a set of non-zero elements of $D$.

Let $y$ be a greatest common divisor of $S$.

Then $y$ is expressible in the form:

- $y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$

where $d_1, d_2, \dotsc, d_n \in D$.

## Proof

From Finite Set of Elements in Principal Ideal Domain has GCD we have that at least one such greatest common divisor exists.

So, let $y$ be a greatest common divisor of $S$.

Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.

From Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal:

- $J = \ideal x$

for some $x \in D$, where $\ideal x$ denotes the principal ideal generated by $x$.

From Finite Set of Elements in Principal Ideal Domain has GCD, $x$ is a greatest common divisor of $S$.

From Greatest Common Divisors in Principal Ideal Domain are Associates, $y$ is an associate of $x$.

By definition of associate:

- $\ideal y = \ideal x$

Therefore:

- $y \in J$

and so by definition, $y$ is expressible in the form:

- $y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$

where $d_1, d_2, \dotsc, d_n \in D$.

$\blacksquare$

## Source of Name

This entry was named for Étienne Bézout.

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 62.4$ Factorization in an integral domain: $\text{(iii)}$