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Artin says that every element in an integral domain can be expressed uniquely as a product of irreducible elements. This article says irreducible elements are non-units which are not the product of two non-units.

  1. Is $1$ a unit? $1*1=1$, hence $1$ is the inverse of itself.
  2. Say $a$ is an irreducible. $a=b*c$, where $b$ and $c$ are both units, but not the inverses of each other. Is this scenario possible?
  3. What is the motivation behind not allowing a unit to be an irreducible factor of another element? I somehow gather that then the factorization of an element won't be unique (i.e. $a=b*c*d=u*u^{-1}*b*c*d$), but this seems to be a triviality. Why can't $a=u*b*c$, where $u$ is a unit?

Thanks in advance!

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Factorization is not a "trivial" property or thing to consider. – anon Aug 6 '13 at 5:07
It seems unlikely that Artin claimed that every element can be expressed uniquely as a product of irreducibles, as this is famously false in the ring $\mathbb{Z}[\sqrt{-5}]$. Nor is it necessarily true that any element can be expressed as such a product at all; this requires some kind of finiteness condition such as being noetherian. So check your hypotheses. – Ryan Reich Aug 6 '13 at 5:22
up vote 2 down vote accepted

1) No, $1$ is not irreducible. By definition, irreducible elements cannot be unit. Since $1$ is unit, $1$ is not irreducible.

2) This scenario is impossible. If $b$ and $c$ are units, then their product $bc$ is also a unit, so it cannot be irreducible.

3) You are right. The reason units are not allowed to be irreducible is exactly the reason you point out. For example, in $\mathbb{Z}$ if we allow $1$ or $-1$ (which are the only units in $\mathbb{Z}$) to appear in factorization, then prime factorization of an integer would not be unique: since we can write $15=(-1)\cdot 3\cdot 1\cdot 5\cdot (-1)$ or $15=3\cdot 5\cdot (-1)\cdot (-1)$ or in many other ways.

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@Prism- sorry I'm a little confused; you've answered the first question as "No", and then gone on to say $1$ is a unit. So is it a unit or not? – Ayush Khaitan Aug 6 '13 at 5:13
@AyushKhaitan: Sorry, I meant that "No, $1$ is not irreducible". It is always the case that $1$ is a unit! – Prism Aug 6 '13 at 5:14
As for question 2, what if $a$ is irreducible, and $a=b*c$, where only $b$ is a unit and $c$ is not? Is this scenario possible? Thanks! – Ayush Khaitan Aug 6 '13 at 5:14
@AyushKhaitan: Yes, definitely! For example, $a=1*a$ (here $b=1$, $c=a$, and $a$ is arbitrary irreducible element). But there can be more non-trivial examples. – Prism Aug 6 '13 at 5:16
@prism- Thanks! – Ayush Khaitan Aug 6 '13 at 5:19

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