Bézout's identity

(Redirected from Bézout's lemma)

Bézout's identity (also called Bézout's lemma) is a theorem in elementary number theory: let a and b be nonzero integers and let d be their greatest common divisor. Then there exist integers x and y such that

${\displaystyle ax+by=d.}$

• the greatest common divisor d is the smallest positive integer that can be written as ax + by
• every integer of the form ax + by is a multiple of the greatest common divisor d.

The integers x and y are called Bézout coefficients for (a, b); they are not unique. A pair of Bézout coefficients can be computed by the extended Euclidean algorithm. If both a and b are nonzero, the extended Euclidean algorithm produces one of the two pairs such that ${\displaystyle |x|\leq \left|{\frac {b}{d}}\right|}$ and ${\displaystyle |y|\leq \left|{\frac {a}{d}}\right|}$ (equality may occur only if one of a and b is a multiple of the other).

Many other theorems in elementary number theory, such as Euclid's lemma or 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 these domains.

Structure of solutions

When one pair of Bézout coefficients (x, y) has been computed (e.g., using extended Euclidean algorithm), all pairs can be represented in the form

${\displaystyle \left(x+k{\frac {b}{\gcd(a,b)}},\ y-k{\frac {a}{\gcd(a,b)}}\right),}$

where k is an arbitrary integer and the fractions simplify to integers.

Among these pairs of Bézout coefficients, exactly two of them satisfy

${\displaystyle |x|\leq \left|{\frac {b}{\gcd(a,b)}}\right|\quad {\text{and}}\quad |y|\leq \left|{\frac {a}{\gcd(a,b)}}\right|,}$

and equality may occur only if one of a and b divides the other. This relies on a property of Euclidean division: given two integers c and d, if d does not divide c, there is exactly one pair (q,r) such that c = dq + r and 0 < r < |d|, and another one such that c = dq + r and 0 < -r < |d|. The two pairs of small Bézout's coefficients are obtained by choosing k in the above formula for getting either remainder of the division of x by ${\displaystyle b/\gcd(a,b).}$

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

Example

Let a = 12 and b = 42, gcd (12, 42) = 6. Then we have the following Bézout's identities, with the Bézout coefficients written in red for the minimal pairs and in blue for the other ones.

{\displaystyle {\begin{aligned}\vdots \\12&\times \color {blue}{-10}&+\;\;42&\times \color {blue}{3}&=6\\12&\times \color {red}{-3}&+\;\;42&\times \color {red}{1}&=6\\12&\times \color {red}{4}&+\;\;42&\times \color {red}{-1}&=6\\12&\times \color {blue}{11}&+\;\;42&\times \color {blue}{-3}&=6\\12&\times \color {blue}{18}&+\;\;42&\times \color {blue}{-5}&=6\\\vdots \end{aligned}}}

Proof

Given some integers ${\displaystyle a}$ and ${\displaystyle b}$, let ${\displaystyle S={\bigl \{}ax+by|x,y\in \mathbb {Z} {\bigr \}}}$ . Since ${\displaystyle S}$ is trivially nonempty, let ${\displaystyle d=as+bt}$ be the smallest positive integer in ${\displaystyle S}$. Now consider some other ${\displaystyle n\in S}$, and assume by contradiction that ${\displaystyle n}$ is not divisible by ${\displaystyle d}$. Then according to Euclidean division, ${\displaystyle \exists q,r}$ such that:

${\displaystyle n=qd+r,\quad 0

{\displaystyle {\begin{aligned}r&=n-qd\\&=ax+by-q(as+bt)\\&=a(x-qs)+b(y-qt)\end{aligned}}}

But this implies that ${\displaystyle r\in S}$, and since ${\displaystyle 0 this violates the original premise that ${\displaystyle d}$ is the smallest number in ${\displaystyle S}$, thus ${\displaystyle n}$ must be divisible by ${\displaystyle d}$ for all ${\displaystyle n\in S}$. We now show that ${\displaystyle d=\gcd(a,b)}$:

Consider the following specific examples for ${\displaystyle n\in S}$:

{\displaystyle {\begin{aligned}n&=1\cdot a+0\cdot b=a\\n&=0\cdot a+1\cdot b=b\\\end{aligned}}}

Therefore, both ${\displaystyle a}$ and ${\displaystyle b}$ must be divisible by ${\displaystyle d}$.

Now consider a common divisor ${\displaystyle c}$ of ${\displaystyle a}$ and ${\displaystyle b}$:

{\displaystyle {\begin{aligned}a&=pc\\b&=qc\\d&=as+bt\\&=(pc)s+(qc)t\\&=c(ps+qt)\end{aligned}}}

So if ${\displaystyle c}$ divides ${\displaystyle a}$ and ${\displaystyle b}$, then ${\displaystyle c}$ also must divide ${\displaystyle d}$

Therefore, by the definition of the greatest common divisor, ${\displaystyle d=\gcd(a,b)}$ .

Generalizations

For three or more integers

Bézout's identity can be extended to more than two integers: if

${\displaystyle \gcd(a_{1},a_{2},\ldots ,a_{n})=d}$

then there are integers ${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$ such that

${\displaystyle d=a_{1}x_{1}+a_{2}x_{2}+\cdots +a_{n}x_{n}}$

has the following properties:

• d is the smallest positive integer of this form
• every number of this form is a multiple of d

For polynomials

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:

For univariate polynomials f and g with coefficients in a field, there exist polynomials a and b such that af + bg = 1 if and only if f and g have no common root in any algebraically closed field (commonly the field of complex numbers).

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 R is a PID, and a and b are elements of R, and d is a greatest common divisor of a and b, then there are elements x and y in R such that ax + by = d. The reason: the ideal Ra+Rb is principal and indeed is equal to Rd.

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

History

French mathematician Étienne Bézout (1730–1783) proved this identity for polynomials.[2] However, this statement for integers can be found already in the work of another French mathematician, Claude Gaspard Bachet de Méziriac (1581–1638).[3][4][5]