Euclid's Lemma for Irreducible Elements

Lemma

Let $\left({D, +, \times}\right)$ be a Euclidean domain whose unity is $1$.

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

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

$p \backslash a \times b$

where $\backslash$ means is a divisor of.

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

Generalized Lemma

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

Let $n \in D$ such that:

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

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:

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

Proof

Let $p \backslash a \times b$.

Suppose $p \nmid a$. Then from the definition of irreducible, $p \perp a$.

Thus from Euclid's Lemma for Euclidean Domains 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.

$\blacksquare$

Proof of Generalized 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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle a_1 a_2 \ldots a_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle \left({a_1 a_2 \ldots a_k}\right) \left({a_{k+1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle a_1 a_2 \ldots a_k \lor p \backslash a_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the basis for the induction
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle p\)	\(\backslash\)	\(\displaystyle a_1 \lor p \backslash a_2 \lor \ldots \lor p \backslash a_k \lor p \backslash a_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the induction hypothesis

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

Therefore:

$\displaystyle \forall r \in \N: p \backslash \prod_{i=1}^r a_i \implies \exists i \in \left[{1 . . r}\right]: p \backslash a_i$

$\blacksquare$

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

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.29$: Theorem $56$