Euclid's Lemma for Prime Divisors

Lemma

Let $p$ be a prime number.

Let $a$ and $b$ be integers such that:

$p \divides a b$

where $\divides$ means is a divisor of.

Then $p \divides a$ or $p \divides b$.

General Result

Let $p$ be a prime number.

Let $\ds n = \prod_{i \mathop = 1}^r a_i$.

Then if $p$ divides $n$, it follows that $p$ divides $a_i$ for some $i$ such that $1 \le i \le r$.

That is:

$p \divides a_1 a_2 \ldots a_n \implies p \divides a_1 \lor p \divides a_2 \lor \cdots \lor p \divides a_n$

Corollary

Let $p, p_1, p_2, \ldots, p_n$ be primes such that:

$p \divides \ds \prod_{i \mathop = 1}^n p_i$

Then:

$\exists i \in \closedint 1 n: p = p_i$

Proof 1

We have that the integers form a Euclidean domain.

Then from Irreducible Elements of Ring of Integers we have that the irreducible elements of $\Z$ are the primes and their negatives.

The result then follows directly from Euclid's Lemma for Irreducible Elements.

$\blacksquare$

Proof 2

Let $p \divides a b$.

Suppose $p \nmid a$.

Then from the definition of prime:

$p \perp a$

where $\perp$ indicates that $p$ and $a$ are coprime.

Thus from Euclid's Lemma it follows that:

$p \divides b$

Similarly, if $p \nmid b$ it follows that $p \divides a$.

So:

$p \divides a b \implies p \divides a$ or $p \divides b$

as we needed to show.

$\blacksquare$

Proof 3

Let $p \divides a b$.

Suppose $p \nmid a$.

Then by Proposition $29$ of Book $\text{VII} $: Prime not Divisor implies Coprime:

$p \perp a$

As $p \divides a b$, it follows by definition of divisor:

$\exists e \in \Z: e p = a b$

So by Proposition $19$ of Book $\text{VII} $: Relation of Ratios to Products‎:

$p : a = b : e$

But as $p \perp a$, by:

Proposition $21$ of Book $\text{VII} $: Coprime Numbers form Fraction in Lowest Terms

and:

Proposition $20$ of Book $\text{VII} $: Ratios of Fractions in Lowest Terms

it follows that:

$p \divides b$

Similarly, if $p \perp b$ then $p \divides a$.

Hence the result.

$\blacksquare$

Examples

Example: $6$

Note that Euclid's Lemma for Prime Divisors does not hold for composite numbers.

For example:

$6 \divides 12 = 3 \times 4$

but:

$6 \nmid 3$ and $6 \nmid 4$

Also presented as

Some sources present Euclid's Lemma for Prime Divisors as:

Let $p$ be a prime number.

Let $a$ and $b$ be integers such that:

$a b \equiv 0 \pmod p$

Then either $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$.

Some sources use this property to define a prime number.

Also known as

Some sources give the name of Euclid's Lemma for Prime Divisors as Euclid's First Theorem.

Also see

• Euclid's Lemma

Source of Name

This entry was named for Euclid.

Historical Note

Euclid's Lemma for Prime Divisors was proved by Euclid some time around $\text {300}$$\text { BCE}$.

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Exercise $24.8 \ \text{(a)}$
• 1979: G.H. Hardy and E.M. Wright: An Introduction to the Theory of Numbers (5th ed.) ... (previous) ... (next): $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic: Theorem $3$
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): prime
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): prime