Euclid's Lemma for Irreducible Elements

Lemma

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

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

Let $a, b \in D$ such that:

$p \divides a \times b$

where $\divides$ means is a divisor of.

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

General Result

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

Let $n \in D$ such that:

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

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

Let $p \divides a \times b$.

Suppose $p \nmid a$.

Then from the definition of irreducible:

$p \perp a$

Thus from Euclid's Lemma for Euclidean Domains:

$p \divides b$

Similarly, if $p \nmid b$:

$p \divides a$

So:

$p \divides a b \implies p \divides a$

or:

$p \divides b$

as we needed to show.

$\blacksquare$

Source of Name

This entry was named for Euclid.

Also see

• Euclid's Lemma for Prime Divisors, for the usual statement of this result, which is this lemma as applied specifically to the integers.

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Theorem $56 \ \text{(i)}$