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My understanding is that irreducible elements cannot be written as the product of two non-units. Therefore, in $\mathbb{Z}_7$, the ring/field of integers modulo 7, because every nonzero element is a unit, every element must be irreducible.

My understanding is that an element $p \in \mathbb{Z}_7$ is prime iff: $$p|ab \Rightarrow p|a \text{ or } p|b$$ In $\mathbb{Z}_7$, every element is prime because p|a is true of any nonzero elements $a,b,p \in \mathbb{Z}_7$

Is my explanation true?

Similarly, this would work for any finite field.

Do these constitute unique factorisation domains (UFDs)? Would it not be vacuously true, since there are no nonzero non-units that "every non zero non-unit cannot be written uniquely as the product of non-irreducible elements"? I guess I am looking for more intuition as to why the integers modulo p, where p is prime, constitute a UFD given that all nonzero elements are both irreducible and prime.

Also: I tried to find an example of a non-UFD. I tried $\mathbb{Z}_6$ and concluded that it is NOT a UFD because you can't write elements like 2,3,4 (which are nonzero non-units) as the product of irreducibles (the irreducibles in this set being 1 and 5) at all. Is this true? Are there examples of a set of integers modulo n that is NOT a UFD because there are 2 or more ways to write a number as the product of irreducible elements, or do most/all rings that are not UFDs just have elements that can't be written at all as the product of irreducibles? Sorry for long question - just very interested!

Andrew
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  • $\Bbb Z_6$ is not even an integral domain. For an interesting example of a non-UFD consider $\Bbb Z[\sqrt{-6}]$. – Angina Seng Oct 16 '18 at 05:39
  • Sorry, I asked the question poorly. I meant to say are there any examples of non-UFDs amongst the sets of integers modulo n, in which you can write an element in more than one way as the product of irreducible elements. I want to find these examples as I want to teach someone about this who doesn't know about complex numbers. – Andrew Oct 16 '18 at 05:41
  • A finite integral domain is necessarily a field. – Angina Seng Oct 16 '18 at 05:42
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    Is that because if "ab=ac and a nonzero implies b=c" means that a must be invertible? – Andrew Oct 16 '18 at 05:44
  • Generally factorization theory is more complicated in non-domains, e.g. $\rm:x = (3+2x)(2-3x)\in \Bbb Z_6[x].:$ Basic notions such as associate and irreducible bifurcate into a few inequivalent notions, e.g. see here. – Bill Dubuque Oct 16 '18 at 15:41

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