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.
- Is $1$ a unit? $1*1=1$, hence $1$ is the inverse of itself.
- 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?
- 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!
Best Answer
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.