Euclid's Lemma for Irreducible Elements
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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 $a, b \in D$ 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 $56 \ \text{(i)}$