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Anonymous

• CommentRowNumber2.
• CommentAuthorGuest
• CommentTimeMay 20th 2022

For the sentence

An integral domain $R$ is a unique factorization domain (UFD for short) if every non-unit has a factorization $u = r_1 \cdots r_n$ as product of irreducible non-units and this decomposition is unique up to renumbering and rescaling the irreducibles by units.

should the “product of irreducible non-units” be “arbitrary/infinitary product” or “finite product”?

• CommentRowNumber3.
• CommentAuthorJ-B Vienney
• CommentTimeAug 7th 2022
• Precised that $0$ is never irreducible as a consequence of the definition of irreducible.

• Corrected the definition:

The unique factorization condition if for non-zero non-units. With the current definition, there doesn’t exist any UFD. If zero must have a decomposition as a product of irreducibles, then we have $0 = r_{1}...r_{n}$ and thus one of the $r_{i}$ is equal to $0$ because we are in an integral domain. But $0$ is never irreducible, absurd.

• Replaced “irreducible non unit” by “irreducible” because it is redundant.

• Precised that $n$ must be greater than $1$ in the decomposition.

• CommentRowNumber4.
• CommentAuthorJ-B Vienney
• CommentTimeAug 7th 2022

Added a characterization of UFDs. I’ll try to put a proof later.

2. adding redirect for plural unique factorization domains

Anonymous

3. Added fact that the ring of integers $R$ of an algebraic number field is a unique factorization if and only if its Picard group is trivial, and added the reference from which the fact came from:

• John Baez, Hoàng Xuân Sính’s thesis: categorifying group theory (pdf)

Anonymaus