In the book "Abstract Algebra" by Dummit, the definition of irreducible element in an integral domain $R$ goes like this.

Suppose $r\in R$ is nonzero and is not a unit. Then $r$ is called irreducible in $R$ if whenever $r=ab$ with $a,b\in R$, at least one of $a$ or $b$ must be a unit in $R$. Otherwise $r$ is said to be reducible...

My questions are

  1. why we don't consider the units to be irreducible?

  2. In the definition above, I get confused by the definition of reducible elements: namely, the units are reducible or not?

Thanks in advance..


This is similar to the question of whether $1$ is a prime number. If $1$ were a prime number, then the fundamental theorem of arithmetic would not work because you could add any power of $1$ to a factorization. This generalizes to unique factorization domains, where up to associates an element has a unique factorization into irreducible elements. If units were irreducible, this would not be true.

Just as we say that $1$ is also not composite because it is not the product of two non-unit elements, so we say that units are also not reducible. Thus the answer is that units are neither reducible nor irreducible.

  • 1
    $\begingroup$ Right - perhaps the confusion is in the terminology, where {reducible, irreducible} looks like it ought to be exhaustive, but actually only has two of the three possible categories: {unit, irreducible, reducible}. $\endgroup$
    – Joffan
    Jan 26 '15 at 22:45
  • $\begingroup$ Further, I think it's worth noting that "units" are somehow elements that are "almost" the multiplicative identity. We wouldn't consider 6*1 and 2*3 two distinct factorizations of 6, because the factorization out of 1 is somehow trivial. Similarly, if $u$ is a unit, we wouldn't want to include $u \cdot u^{-1} \cdot r$ as a factorization of $r$. $\endgroup$
    – walkar
    Jan 27 '15 at 5:34

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