Bézout's identity

In mathematics, Bézout's identity (also called Bézout's lemma), named after Étienne Bézout who proved it for polynomials, is the following theorem:

Here the greatest common divisor of and is taken to be . The integers and are called Bézout coefficients for ; they are not unique. A pair of Bézout coefficients can be computed by the extended Euclidean algorithm, and this pair is, in the case of integers one of the two pairs such that and ; equality occurs only if one of and is a multiple of the other.

As an example, the greatest common divisor of 15 and 69 is 3, and 3 can be written as a combination of 15 and 69 as , with Bézout coefficients −9 and 2.

Many other theorems in elementary number theory, such as Euclid's lemma or the Chinese remainder theorem, result from Bézout's identity.

A Bézout domain is an integral domain in which Bézout's identity holds. In particular, Bézout's identity holds in principal ideal domains. Every theorem that results from Bézout's identity is thus true in all principal ideal domains.

Structure of solutions

If and are not both zero and one pair of Bézout coefficients has been computed (for example, using the extended Euclidean algorithm), all pairs can be represented in the form
$\left(x-k\frac,\ y+k\frac\right),$
where is an arbitrary integer, is the greatest common divisor of and , and the fractions simplify to integers.

If and are both nonzero and none of them divides the other, then exactly two of the pairs of Bézout coefficients satisfy
$, x, < \left , \frac\right , \quad \text\quad , y, < \left , \frac\right , .$ If and are both positive, one has $x>0$ and $y<0$ for one of these pairs, and $x<0$ and $y>0$ for the other. If is a divisor of (including the case $b=0$), then one pair of Bézout coefficients is .

This relies on a property of Euclidean division: given two non-zero integers and , if does not divide , there is exactly one pair such that and , and another one such that and .

The two pairs of small Bézout's coefficients are obtained from the given one by choosing for in the above formula either of the two integers next to .

The extended Euclidean algorithm always produces one of these two minimal pairs.

Example

Let and , then . Then the following Bézout's identities are had, with the Bézout coefficients written in red for the minimal pairs and in blue for the other ones.

$\begin \vdots \\ 12 &\times () & + \;\; 42 &\times \color &= 6 \\ 12 &\times () & + \;\;42 &\times \color &= 6 \\ 12 &\times \color & + \;\;42 &\times() &= 6 \\ 12 &\times \color & + \;\;42 &\times () &= 6 \\ 12 &\times \color & + \;\;42 &\times () &= 6 \\ \vdots \end$

If is the original pair of Bézout coefficients, then yields the minimal pairs via , respectively ; that is, , and .

Existence proof

Given any nonzero integers and , let . The set is nonempty since it contains either or (with and ). Since is a nonempty set of positive integers, it has a minimum element , by the well-ordering principle. To prove that is the greatest common divisor of and , it must be proven that is a common divisor of and , and that for any other common divisor , one has .

The Euclidean division of by may be written as
a=dq+r\quad\text\quad 0\le r
The remainder is in , because
$\begin r & = a - qd \\ & = a - q(as+bt)\\ & = a(1-qs) - bqt. \end$
Thus is of the form , and hence . However, , and is the smallest positive integer in : the remainder can therefore not be in , making necessarily 0. This implies that is a divisor of . Similarly is also a divisor of , and therefore is a common divisor of and .

Now, let be any common divisor of and ; that is, there exist and such that and . One has thus
$\begin d&=as + bt\\ & =cus+cvt\\ &=c(us+vt). \end$
That is, is a divisor of . Since , this implies .

Generalizations

For three or more integers

Bézout's identity can be extended to more than two integers: if
$\gcd(a_1, a_2, \ldots, a_n) = d$
then there are integers $x_1, x_2, \ldots, x_n$ such that
$d = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n$
has the following properties:
* is the smallest positive integer of this form
* every number of this form is a multiple of

For polynomials

Bézout's identity does not always hold for polynomials. For example, when working in the polynomial ring of integers: the greatest common divisor of and is ''x'', but there does not exist any integer-coefficient polynomials and satisfying .

However, Bézout's identity works for univariate polynomials over a field exactly in the same ways as for integers. In particular the Bézout's coefficients and the greatest common divisor may be computed with the extended Euclidean algorithm.

As the common roots of two polynomials are the roots of their greatest common divisor, Bézout's identity and fundamental theorem of algebra imply the following result:

The generalization of this result to any number of polynomials and indeterminates is Hilbert's Nullstellensatz.

For principal ideal domains

As noted in the introduction, Bézout's identity works not only in the ring of integers, but also in any other principal ideal domain (PID).
That is, if is a PID, and and are elements of , and is a greatest common divisor of and ,
then there are elements and in such that . The reason is that the ideal is principal and equal to .

An integral domain in which Bézout's identity holds is called a Bézout domain.

History and attribution

The French mathematician Étienne Bézout (1730–1783) proved this identity for polynomials. The statement for integers can be found already in the work of an earlier French mathematician, Claude Gaspard Bachet de Méziriac (1581–1638).
Andrew Granville traced the association of Bézout's name with the identity to Bourbaki, arguing that it is a misattribution since the identity is implicit in Euclid's ''Elements''.

See also

* , an analogue of Bézout's identity for homogeneous polynomials in three indeterminates
*
*
*

Notes

External links

Online calculator
for Bézout's identity.
*

{{DEFAULTSORT:Bezouts Identity

Articles containing proofs
Diophantine equations
Lemmas in number theory