Bézout's Lemma/Euclidean Domain

From ProofWiki
Jump to: navigation, search

Theorem

Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$ and whose unity is $1$.

Let $\nu: D \setminus \set 0 \to \N$ be the Euclidean valuation on $D$.

Let $a, b \in D$ such that $a$ and $b$ are not both equal to $0$.


Let $\gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$.

Then:

$\exists x, y \in D: a \times x + b \times y = \gcd \set {a, b}$

such that $\gcd \set {a, b}$ is the element of $D$ such that:

$\forall c = a \times x + b \times y \in D: \map \nu {\gcd \set {a, b} } \le \map \nu c$


Proof

We are given that $a, b \in D$ such that $a$ and $b$ are not both equal to $0$.

Without loss of generality, suppose specifically that $b \ne 0$.

Let $S \subseteq D$ be the set defined as:

$S = \set {x \in D_{\ne 0}: x = m \times a + n \times b: m, n \in D}$

where $D_{\ne 0}$ denotes $D \setminus 0$.

Setting $m = 0$ and $n = 1$, for example, it is noted that $b \in S$.

Therefore $S \ne \O$.


By definition, $\nu$ has the properties:

$(1): \quad \forall a, b \in D, b \ne 0: \exists q, r \in D$ such that $\map \nu r < \map \nu b$, or $r = 0$, such that:
$a = q \times b + r$
$(2): \quad \forall a, b \in D, b \ne 0$:
$\map \nu a \le \map \nu {a \times b}$

Let $\nu \sqbrk S$ denote the image of $S$ under $\nu$.

We have that:

$\nu \sqbrk S \subseteq \N$

Hence by the Well-Ordering Principle $\nu \sqbrk S$ has a smallest element.

Let $d \in S$ be such that $\map \nu d$ is that smallest element of $\nu \sqbrk S$.

By definition of $S$, we have that:

$d = u \times a + v \times b$

for some $u, v \in D$.


Let $x \in S$.

By $(2)$ above:

$x = q \times d + r$

such that either:

$\map \nu r < \map \nu d$

or:

$r = 0$


Aiming for a contradiction, suppose $r \ne 0$.

Then:

\(\, \displaystyle \exists m, n \in D: \, \) \(\displaystyle x\) \(=\) \(\displaystyle m \times a + n \times b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r\) \(=\) \(\displaystyle x - q \times d\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {m \times a + n \times b} - q \paren {u \times a + v \times b}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {m - q \times u} a + \paren {n - q \times v} b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle \paren {r \in S} \land \paren {\map \nu r < \map \nu d}\)

which contradicts the choice of $d$ as the element of $S$ such that $\map \nu d$ is the smallest element of $\nu \sqbrk S$.


Therefore:

$\forall x \in S: x = q \times d$

for some $q \in D$.

That is:

$\forall x \in S: d \divides x$

where $\divides$ denotes divisibility.


In particular:

$d \divides a = 1 \times a + 0 \times b$
$d \divides b = 0 \times a + 1 \times b$

Thus:

$d \divides a \land d \divides b \implies \map \nu 1 \le \map \nu d \le \map \nu {\gcd \set {a, b} }$


However, note that as $\gcd \set {a, b}$ also divides $a$ and $b$ (by definition), we have:

\(\displaystyle \gcd \set {a, b}\) \(\divides\) \(\displaystyle \paren {u \times a + v \times b} = d\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \gcd \set {a, b}\) \(\divides\) \(\displaystyle d\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \nu {\gcd \set {a, b} }\) \(\le\) \(\displaystyle \map \nu d\)


Since $d$ is the element of $S$ such that $\map \nu d$ is the smallest element of $\nu \sqbrk S$:

$\gcd \set {a, b} = d = u \times a + v \times b$

$\blacksquare$


Source of Name

This entry was named for Étienne Bézout.


Historical Note

There are sources which suggest that Bézout's Lemma was first noticed by Claude Gaspard Bachet de Méziriac.

Étienne Bézout's contribution was to prove a more general result, for polynomials.