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When defining irreducible integers, we restrict our attention $Z = \{n \in \mathbb{Z} \; | \; |n| > 1\}$. Then we say that for some $p \in Z$, $p$ is irreducible if for some $a,b \in \mathbb{Z}$: $$p = ab \implies |a| = 1 \;\vee\; |b| = 1$$

Why do we exclude $1$ and $-1$ from the definition? Wouldn't they fit the condition?

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Generally it's because we want to talk about unique factorization into the product of irreducible elements. If we allowed units to be irreducible, we'd lose uniqueness (since you could include or not include any amount of 1's)

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We really want unique prime factorization. And we also really want irreducible iff prime in the integers. So we had better not make 1 irreducible!

If 1 were a prime, the number of possible prime factors in a given integer is unbounded, as $1^n$ divides anything.

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  • $\begingroup$ Jinx :). Yeah almost simulpost. $\endgroup$
    – Alan
    Sep 23 '15 at 3:03
  • $\begingroup$ He, that's funny. $\endgroup$
    – Zach Stone
    Sep 23 '15 at 3:05

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