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.

Generalised 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 $1 \le i \le r$.

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 by Induction
For all $r \in \N^*$, let $P \left({r}\right)$ be the proposition:
 * $\displaystyle p \backslash \prod_{i=1}^r a_i \implies \exists i \in \left[{1 \, . \, . \, r}\right]: p \backslash a_i$.

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

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:

Consider any product of $k+1$ integers which is divisible by $p$:
 * $p \backslash a_1 a_2 \cdots a_k a_{k+1}$

Either $p$ divides $a_{k+1}$ or it does not.

If $p$ divides $a_{k+1}$ we have finished.

If $p$ does not divide $a_{k+1}$, then from GCD with Prime we have:
 * $\gcd \left\{{p, a_{k+1}}\right\} = 1$

... that is, $p \perp a_{k+1}$.

So we have $p \backslash \left({a_1 a_2 \cdots a_k}\right) a_{k+1}$ and $p \perp a_{k+1}$.

Hence by Euclid's Lemma we have that $p \backslash a_1 a_2 \cdots a_k$.

But then by the induction hypothesis, $p \backslash a_i$ for some $i$ such that $1 \le i \le k$.

So either $p \backslash a_i$ for some $i$ such that $1 \le i \le k$, or $p \backslash a_{k+1}$.

That is, $P \left({k+1}\right)$ holds.

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

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

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.