# Euclid's lemma

In algebra and number theory, Euclid's lemma is a lemma that captures a fundamental property of prime numbers, namely:[note 1]

Euclid's lemma — If a prime p divides the product ab of two integers a and b, then p must divide at least one of those integers a or b.

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.

If the premise of the lemma does not hold, i.e., 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.[note 2] 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

Let $p$ be a prime number, and assume $p$ divides the product of two integers $a$ and $b$ . In symbols, this is written $p\mid ab$ . Its negation, $p$ does not divide $ab$ , is written $p\nmid ab$ . Then $p\mid a$ or $p\mid b$ (or both). Equivalent statements are:

• If $p\nmid a$ and $p\nmid b$ , then $p\nmid ab$ .
• If $p\nmid a$ and $p\mid ab$ , then $p\mid b$ .

Euclid's lemma can be generalized from prime numbers to any integers:

Theorem — If $n\mid ab$ , and $n$ is relatively prime to $a$ , then $n\mid b$ .

This is a generalization because if $n$ is prime, either

• $n\mid a$ or
• $n$ is relatively prime to $a$ . In this second possibility, $n\nmid a$ so $n\mid b$ .

## 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 1681.

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, it is coprime with a if and only if it does not divide a.

### 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 relatively prime, 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 $n\mid ab$ and that n and a are coprime (that is, their greatest common divisor is 1). One has to prove that n divides b. Since $n\mid ab,$ there is a 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 from the signs of the involved integers.

For proving this by strong induction, we suppose that the result has been proved for all positive lower values of ab.

There are three cases:

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

If n < a, one has

$n(q-b)=(a-n)b.$ The positive integers an and n are coprime: their greatest common divisor d must divide their sum, and thus divides both n and a. It results that d = 1, by the coprimality hypothesis. So, the conclusion follows from the induction hypothesis, since 0 < (an) b < ab.

Similarly, if n > a one has

$(n-a)q=a(b-q),$ and the same argument shows that na and a are coprime. Therefore, one has 0 < a (bq) < ab, and the induction hypothesis implies that na divides bq; that is, $b-q=r(n-a)$ for some integer. So, $(n-a)q=ar(n-a),$ and, by dividing by na, one has $q=ar.$ Therefore, $ab=nq=anr,$ and by dividing by a, one gets $b=nr,$ the desired result.

### Proof of Elements

Euclid's lemma is proved at the Proposition 30 in Book VII of Euclid's Elements. The original proof is difficult to understand as is, so we quote the commentary from Euclid (1956, pp. 319–332).

Proposition 19
If four numbers be proportional, the number produced from the first and fourth is equal to the number produced from the second and third; and, if the number produced from the first and fourth be equal to that produced from the second and third, the four numbers are proportional.[note 3]
Proposition 20
The least numbers of those that have the same ratio with them measures those that have the same ratio the same number of times—the greater the greater and the less the less.[note 4]
Proposition 21
Numbers prime to one another are the least of those that have the same ratio with them.[note 5]
Proposition 29
Any prime number is prime to any number it does not measure.[note 6]
Proposition 30
If two numbers, by multiplying one another, make the same number, and any prime number measures the product, it also measures one of the original numbers.[note 7]
Proof of 30
If c, a prime number, measure ab, c measures either a or b.
Suppose c does not measure a.
Therefore c, a are prime to one another. VII. 29
Suppose abmc.
Therefore c : ab : m. VII. 19
Hence ［VII. 20, 21bnc, where n is some integer.
Therefore c measures b.
Similarly, if c does not measure b, c measures a.
Therefore c measures one or other of the two numbers a, b.
Q.E.D.