欧几里得引理

可以这样表达这个引理：

如果a|bc，gcd(a,b)=1 那么 a|c。

命题30是这样说的：

如果一个素数整除两个正整数的乘积，那么这个素数可以至少整除这两个正整数中的一个。

如果 p|bc 那么 p|b 或者 p|c。

命题30的证明

设p|ab，但p不是a的因子。于是，可设 rp=ab，其中r|ab。由于p是质数，且不是a的因子，gcd(a,p)=1。这就是说，可以找到两个整数x和y，使得 1=px+ay （贝祖定理）。两边乘以b，可得： b=b(px+ay)

b=bpx+bay.

前面已经说了 rp=ab，因此： b=bpx+rpy

b=p(bx+ry).

所以，p|b。这就是说，p要么整除a，要么整除b，要么都能整除。证毕。

In algebra and number theory, Euclid's lemma is a lemma that captures a fundamental property of prime numbers:

For example, if p = 19, a = 133, b = 143, then ab = 133 × 143 = 19019, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, 133 = 19 × 7.

The lemma first appeared in Euclid's Elements, and is a fundamental result in elementary number theory.

If the premise of the lemma does not hold, that is, if p is a composite number, its consequent may be either true or false. For example, in the case of p = 10, a = 4, b = 15, composite number 10 divides ab = 4 × 15 = 60, but 10 divides neither 4 nor 15.

This property is the key in the proof of the fundamental theorem of arithmetic. It is used to define prime elements, a generalization of prime numbers to arbitrary commutative rings. Euclid's lemma shows that in the integers irreducible elements are also prime elements. The proof uses induction so it does not apply to all integral domains.

Formulations

Euclid's lemma is commonly used in the following equivalent form:

Euclid's lemma can be generalized as follows from prime numbers to any integers.

This is a generalization because a prime number p is coprime with an integer a if and only if p does not divide a.

History

The lemma first appears as proposition 30 in Book VII of Euclid's Elements. It is included in practically every book that covers elementary number theory.

The generalization of the lemma to integers appeared in Jean Prestet's textbook Nouveaux Elémens de Mathématiques in 1689.

In Carl Friedrich Gauss's treatise Disquisitiones Arithmeticae, the statement of the lemma is Euclid's Proposition 14 (Section 2), which he uses to prove the uniqueness of the decomposition product of prime factors of an integer (Theorem 16), admitting the existence as "obvious". From this existence and uniqueness he then deduces the generalization of prime numbers to integers. For this reason, the generalization of Euclid's lemma is sometimes referred to as Gauss's lemma, but some believe this usage is incorrect due to confusion with Gauss's lemma on quadratic residues.

Proofs

The two first subsections, are proofs of the generalized version of Euclid's lemma, namely that: if n divides ab and is coprime with a then it divides b.

The original Euclid's lemma follows immediately, since, if n is prime then it divides a or does not divide a in which case it is coprime with a so per the generalized version it divides b.

Using Bézout's identity

In modern mathematics, a common proof involves Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if x and y are coprime integers (i.e. they share no common divisors other than 1 and −1) there exist integers r and s such that rx+sy=1.

Let a and n be coprime, and assume that n|ab. By Bézout's identity, there are r and s such that rn+sa=1.

Multiply both sides by b: rnb+sab=b.

The first term on the left is divisible by n, and the second term is divisible by ab, which by hypothesis is divisible by n. Therefore their sum, b, is also divisible by n.

By induction

The following proof is inspired by Euclid's version of Euclidean algorithm, which proceeds by using only subtractions.

Suppose that n| ab and that n and a are coprime (that is, their greatest common divisor is 1). One has to prove that n divides b. Since n| ab, there exists an integer q such that nq=ab.

Without loss of generality, one can suppose that n, q, a, and b are positive, since the divisibility relation is independent of the signs of the involved integers.

To prove the theorem by strong induction, we suppose that it has been proved for all smaller values of ab. There are three cases:

If n = a, coprimality implies n = 1, and n divides b trivially.

If n < a, then subtracting nb from both sides gives n(q-b)=(a-n)b. Thus, n divides (a – n) b. Since we assumed that n and a are coprime, it follows that a – n and n must be coprime. (If not, their greatest common divisor d would divide their sum a as well as n, contradicting our assumption.) The conclusion therefore follows by induction hypothesis, since 0 < (a – n) b < ab.

If n > a then subtracting aq from both sides gives (n-a)q=a(b-q). Thus, n - a divides a (b - q). Since (as in the previous case), n – a and a are coprime, and since 0 < b - q < b, then the induction hypothesis implies that n − a divides b − q; that is, b-q=r(n-a) for some integer r. So, (n-a)q=ar(n-a), and, by dividing by n − a, one has q=ar. Therefore, ab=nq=anr, and by dividing by a, one gets b=nr, the desired conclusion.