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I was reading PID—"In a PID every non-zero, non-unit element can be written as product of irreducibles."

So here Uniqueness of the product is not mentioned so I think the product may or may not be unique. But when the product becomes unique up to associates then such a PID is called a UFD but this does not mean that every PID is a UFD, rather it means every UFD is a PID, am I wrong somewhere logically?

This came to me since when I was referring the Proof of PID implies UFD, there starting from PID it proved that the product is unique up to associates and hence proved UFD?

glS
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BAYMAX
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  • In a PID if $p$ is irreducible and $c \not \in (p)$ then $(p,c) = (1)$. This means that $(p,qb) = (p,pb,qb) = (p,b(p,q)) = (p,b)$ so that if $a$ is a product of irreducible all different from $p$ then $a \not \in (p)$, and that $p$ appears in some factorization of $a$ in irreducible if and only if $a \in (p)$ – reuns Apr 16 '17 at 02:46
  • Is $(p)$ the ideal generated by $p$ ? – BAYMAX Apr 16 '17 at 02:48
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    @BAYMAX yes. The above notation is standard, AFAIK. – Xam Apr 16 '17 at 03:09
  • related: https://math.stackexchange.com/q/697204/173147 – glS Nov 13 '21 at 14:18

2 Answers2

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Every PID is a UFD. Not every UFD is a PID.

Example: A ring $R$ is a unique factorization domain if and only if the polynomial ring $R[X]$ is one. But $R[X]$ is a principal ideal domain if and only if $R$ is a field. So, $\mathbb{Z}[X]$ is an example of a unique factorization domain which is not a principal ideal domain.

The statement "In a PID every non-zero, non-unit element can be written as product of irreducibles" is true, but it is not the definition of a principal ideal domain. Nor is it the definition of a unique factorization domain: as you pointed out, it does not account for uniqueness.

Note that in an integral domain, every prime element is irreducible, but the converse is false.

Here are two equivalent definitions of unique factorization domain:

1 . An integral domain $R$ in which every nonzero nonunit is a product of prime elements,

2 . An integral domain $R$ in which every nonzero nonunit is a product of irreducible elements, and this product is unique up to associates.

In the proof you were reading, I guess what they are trying to say is the following: in a principal ideal domain $R$, it is true that irreducible is the same thing as prime. So once you show that every nonzero nonunit can be written as a product of irreducibles, it follows that $R$ is a unique factorization domain.

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D_S
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    If PID $\implies $ UFD, consider a case like- an element from PID such that it is written as a product of irreducible elements whose representation is not unique i.e. there exists another representation now, can this belong to UFD?.I think no, as in UFD elements are such that they are written as product of irreducibles which are unique upto associates.?and we have shown a element in a PID which doesnot belong to UFD . – BAYMAX Apr 16 '17 at 02:41
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    That won't happen though. Once you write an element in a PID as a product of irreducibles, that product will be unique, as a consequence of the fact that every irreducible is prime. – D_S Apr 16 '17 at 03:27
  • Little confused,So what additional condition does UFD have that PID lacks? – BAYMAX Apr 16 '17 at 03:39
  • You've got it backwards. It is principal ideal domains which have something that unique factorization domains in general lack. – D_S Apr 16 '17 at 03:41
  • :) something is ?. – BAYMAX Apr 16 '17 at 03:53
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    They lack the property that every ideal is principal. – D_S Apr 16 '17 at 04:21
  • ok.that is Ok,so in UFD there exista an ideal which is not principal. – BAYMAX Apr 16 '17 at 04:26
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In a PID every non-zero,non-unit element can be written as product of irreducibles".

This statement claims that a PID is an atomic domain, i.e., a domain where every non-zero, non unit admit a factorization into irreducibles (aka atoms, hence the name atomic).

As you noted, being a UFD requires the uniqueness of such a factorization and this leads to another definition: an AP-domain is a domain where every atom is prime.

So we have the following characterization of UFDs: $$\text{UFD}\iff \text{Atomic domain}+\text{AP-domain}.$$

Therefore you can prove that a PID is a UFD showing that is both an atomic and an AP-domain. For further reference you can check these notes, specially the sections two and four.

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Xam
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