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 $1 \le i \le r$.

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
Let $p \backslash a b$.

Suppose $p \nmid a$. Then from Prime Not Divisor then Coprime, $p \perp a$.

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.

Proof of Generalised Lemma
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:

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.