Euclid's Lemma for Irreducible Elements/General Result

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Lemma

Let $\struct {D, +, \times}$ be a Euclidean domain whose unity is $1$.

Let $p$ be an irreducible element of $D$.

Let $n \in D$ such that:

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

where $a_i \in D$ for all $i: 1 \le i \le r$.

If $p$ divides $n$, then $p$ divides $a_i$ for some $i$.


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

Proof by induction:

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

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


$\map P 1$ 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 Irreducible Elements.

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:

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


Then we need to show:

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


Induction Step

This is our induction step:

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

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


Therefore:

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

$\blacksquare$


Source of Name

This entry was named for Euclid.


Also see


Sources