Euclid's Lemma for Prime Divisors/General Result

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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:

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

Proof 1

As for Euclid's Lemma for Prime Divisors, this can be verified by direct application of general version of Euclid's Lemma for irreducible elements.


Proof 2

Proof by induction:

For all $r \in \N_{>0}$, let $\map P r$ be the proposition:

$\ds p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$

$\map P 1$ is true, as this just says $p \divides a_1 \implies p \divides a_1$.

Basis for the Induction

$\map P 2$ is the case:

$p \divides a_1 a_2 \implies p \divides a_2$ or $p \divides a_2$

which is proved in Euclid's Lemma for Prime Divisors.

This is our basis for the induction.

Induction Hypothesis

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

So this is our induction hypothesis:

$\ds p \divides \prod_{i \mathop = 1}^k a_i \implies \exists i \in \closedint 1 k: p \divides a_i$

Then we need to show:

$\ds p \divides \prod_{i \mathop = 1}^{k + 1} a_i \implies \exists i \in \closedint 1 {k + 1}: p \divides a_i$

Induction Step

This is our induction step:

\(\ds p\) \(\divides\) \(\ds a_1 a_2 \ldots a_{k + 1}\)
\(\ds \leadsto \ \ \) \(\ds p\) \(\divides\) \(\ds \paren {a_1 a_2 \ldots a_k} \paren {a_{k + 1} }\)
\(\ds \leadsto \ \ \) \(\ds p\) \(\divides\) \(\ds a_1 a_2 \ldots a_k \lor p \divides a_{k + 1}\) Basis for the Induction
\(\ds \leadsto \ \ \) \(\ds p\) \(\divides\) \(\ds a_1 \lor p \divides a_2 \lor \ldots \lor p \divides a_k \lor p \divides a_{k + 1}\) Induction Hypothesis

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


$\ds \forall r \in \N: p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$


Proof 3

Let $p \divides n$.

Aiming for a contradiction, suppose:

$\forall i \in \set {1, 2, \ldots, r}: p \nmid a_i$

By Prime not Divisor implies Coprime:

$\forall i \in \set {1, 2, \ldots, r}: p \perp a_i$

By Integer Coprime to all Factors is Coprime to Whole:

$p \perp n$

By definition of coprime:

$p \nmid n$

The result follows by Proof by Contradiction.


Source of Name

This entry was named for Euclid.