Is the field of real numbers a unique factorization domain?

1.9k Views Asked by At

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 $\mathbb{R}$?

1

There are 1 best solutions below

0
On

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.