Every field $\mathbb{F}$, with the norm function $\phi(x) = 1, \forall x \in \mathbb{F}$ is a Euclidean domain. Every Euclidean domain is a unique factorization domain.

So, it means that $\mathbb{R}$ is a UFD?

What are the irreducible elements of $\mathbb{R}$?

  • 3
    $\begingroup$ Every field is a UFD. Every non-zero element has the trivial factorization. $\endgroup$
    – pisco
    Commented Nov 13, 2017 at 11:03

1 Answer 1


The (most common) definition of UFD requires that every non-zero non-unit element has a unique factorisation into irreducible elements. A field has no non-zero non-unit elements, so that condition is vacuously satisfied.

No irreducible (non-unit) elements are needed, because we have no non-unit elements that need factorisations.


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