Euclid's Lemma for Prime Divisors

Lemma

Let $p$ be a prime number.

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

$p \backslash a b$

where $\backslash$ means is a divisor of.

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

Some sources use this property to define a prime number.

Generalized Lemma

Let $p$ be a prime number.

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

If $p$ divides $n$, then $p$ divides $a_i$ for some $i$ such that $1 \le i \le r$.

That is:

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

Corollary

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

$\displaystyle p \backslash \prod_{i=1}^n p_i$

Then:

$\exists i \in \left[{1 . . n}\right]: p = p_i$

Proof

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.

$\blacksquare$

Alternative Proof

Alternatively we can prove this result directly, without needing to refer to results from abstract algebra.

Let $p \backslash a b$.

Suppose $p \nmid a$. Then from the definition of prime, $p \perp a$.

Thus from Euclid's Lemma it follows that $p \backslash b$.

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

So:

$p \backslash a b \implies p \backslash a$ or $p \backslash b$

as we needed to show.

$\blacksquare$

Proof of Generalized Lemma

As for the main lemma, this can be verified by direct application of generalized version of Euclid's Lemma for irreducible elements.

Alternatively, we can adopt a proof by induction:

For all $r \in \N^*$, let $P \left({r}\right)$ be the proposition:

$\displaystyle p \backslash \prod_{i=1}^r a_i \implies \exists i \in \left[{1 . . r}\right]: p \backslash a_i$.

$P(1)$ is true, as this just says $p \backslash a_1 \implies p \backslash a_1$.

Basis for the Induction

$P(2)$ is the case:

$p \backslash a_1 a_2 \implies p \backslash a_2$ or $p \backslash a_2$

which is the main lemma as proved above.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle p \backslash \prod_{i=1}^k a_i \implies \exists i \in \left[{1 . . k}\right]: p \backslash a_i$

Then we need to show:

$\displaystyle p \backslash \prod_{i=1}^{k+1} a_i \implies \exists i \in \left[{1 . . {k+1}}\right]: p \backslash a_i$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle a_1 a_2 \ldots a_{k+1}\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle \left({a_1 a_2 \ldots a_k}\right) \left({a_{k+1} }\right)\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle a_1 a_2 \ldots a_k \lor p \backslash a_{k+1}\)	\(\displaystyle \)	from the basis for the induction
	\(\displaystyle \implies\)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle a_1 \lor p \backslash a_2 \lor \ldots \lor p \backslash a_k \lor p \backslash a_{k+1}\)	\(\displaystyle \)	from the induction hypothesis

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall r \in \N: p \backslash \prod_{i=1}^r a_i \implies \exists i \in \left[{1 . . r}\right]: p \backslash a_i$

$\blacksquare$

Proof of Corollary

From the main result, $p \backslash p_i$ for some $i$.

But by definition of prime, the only divisors of $p_i$ are $1$ and $p_i$ itself.

As $1$ is not prime, it follows that $p = p_i$.

Hence the result.

$\blacksquare$

Source of Name

This entry was named for Euclid.

Sources

• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 0.1$: Definition
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 0.1$: Theorem $3$
• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 3.12$: Theorem $21$
• George E. Andrews: Number Theory (1971): $\S 2.2$: Corollary $2.3, \ 2.4$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 23$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 12.5, \ \S 12.6$