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I agree that the connection with UFDs is tenuous. Being not a UFD is nether a necessary or a sufficient condition to have powers of a single thing (or a finite

by tprice7 12y ago

I agree that the connection with UFDs is tenuous. Being not a UFD is nether a necessary or a sufficient condition to have powers of a single thing (or a finite number of things) generate all nonzero elements. The ring of integers modulo 7 for example is a UFD, but all nonzero elements are powers of 3. Also, the ring of integers of a number field is only sometimes a UFD, but it is never generated by powers of some finite number of elements.

Chinjut 12y ago
Being a UFD in general is not enough, but being a UFD in which 2, 3, and 5 are (co)prime is certainly sufficient for incommensurability of 3/2, 2, and 5/4 in the relevant sense. I agree that it's rather odd to start discussing UFDs in general, Gaussian integers, etc., just for this incommensurability result, but since unique prime factorization is the key to the whole thing, it's not entirely out of left field. (I'm also not accustomed to considering the ring of integers modulo 7 a UFD, insofar as exponents in prime factorizations are never unique as integers in this context (only as integers modulo 6), but that's just a minor difference in the way we apparently use terminology)
- tprice7 12y ago
  "Being a UFD in general is not enough, but being a UFD in which 2, 3, and 5 are (co)prime is certainly sufficient for incommensurability of 3/2, 2, and 5/4 in the relevant sense." Ok, this seems like kind of a contrived condition though, compared to, say, the condition that 2, 3, and 5 are prime (which also suffices). Something that occurred to me after I made my last post (and possibly what you meant in the second paragraph?): perhaps the author didn't mean to imply that the UFD property is useful for characterizing when you have incommensurability in different rings, but rather that it's relevant because it's used as a step in the proof that these ratios are incommensurable in the case of Z. This makes sense to me but in my opinion it wouldn't hurt if the article were more explicit on this point. I'm not familiar with a commonly-used definition of UFD for which the ring of integers mod 7 is not a UFD. Could you please point me to a reference that contains this alternate definition of UFD you are referring to? The definition I am using is the one in Abstract Algebra by Dummit & Foote, which happens to agree with the definition on wikipedia at the time of this posting.
  - Chinjut 12y ago
    Yes, I think what you're saying in your third paragraph is what I was trying to say in my second paragraph. And I agree that the article could stand to be much clearer on its motivations. Re: the integers mod 7, I had a brainfart; of course the integers mod 7 are a UFD, but trivially so, as they are a field. My apologies!
    - tprice7 12y ago
      Well I'm glad we are in agreement then. No worries about the UFD thing.
jackmaney 12y ago
The ring of integers modulo 7 is also a field. While fields are (trivially) UFDs, the only non-field class of examples of UFDs having every nonzero element as a power of a fixed nonzero nonunit element is a (rank 1) discrete valuation ring.
- tprice7 12y ago
  " the only non-field class of examples of UFDs having every nonzero element as a power of a fixed nonzero nonunit element is a (rank 1) discrete valuation ring." Ok, but the UFD criterion is redundant here because there are no non-fields for which every nonzero element is a power of a fixed nonzero nonunit element. So that doesn't make UFDs any more relevant than any other extra criterion you could add.
  - jackmaney 12y ago
    Well, if you're willing to allow a bit of wiggle room in the form of "up to associates" (which is a qualifier that I forgot to add), there are integral domains in which every nonzero element equals the product of a unit and a power of a fixed nonzero nonunit element. This qualifier isn't too much to ask since "up to associates" is the generalized version of "up to a plus or minus" when factoring integers (which has no real bearing on factorizations). In particular, let F be any field, and consider F[[x]], the ring of formal power series[1] over F. Given any nonzero f(x) in F[[x]], we can "factor out as many x's as we can" to write f(x) = x^n g(x) where n is a nonnegative integer, g(x) is in F[[x]], and the constant term of g(x) is nonzero (here we explicitly define x^0 to be 1). Thus, g(x) is a unit[2] (ie invertible), and every nonzero element of F[[x]] is a unit multiple of a power of x. [1]: http://en.wikipedia.org/wiki/Formal_power_series http://en.wikipedia.org/wiki/Formal_power_series [2]: http://en.wikipedia.org/wiki/Formal_power_series#Inverting_series http://en.wikipedia.org/wiki/Formal_power_series#Inverting_s...
    - tprice7 12y ago
      Ok, but rings for which every nonzero element is a power of a fixed nonzero nonunit element up to associates are exactly discrete valuation rings. So your statement boils down to "The only UFDs that are discrete valuation rings are (rank 1) discrete valuation rings". And I'm not sure what you mean by rank 1 here, but I can deduce that all discrete valuation rings are rank 1 from what you've said, since all discrete valuation rings are UFDs. So what you've said boils down to: "The only UFDs that are discrete valuation rings are discrete valuation rings", which again doesn't make UFDs more relevant than any other condition you might add.

by tprice7 12y ago

Chinjut 12y ago

Being a UFD in general is not enough, but being a UFD in which 2, 3, and 5 are (co)prime is certainly sufficient for incommensurability of 3/2, 2, and 5/4 in the relevant sense. I agree that it's rather odd to start discussing UFDs in general, Gaussian integers, etc., just for this incommensurability result, but since unique prime factorization is the key to the whole thing, it's not entirely out of left field. (I'm also not accustomed to considering the ring of integers modulo 7 a UFD, insofar as exponents in prime factorizations are never unique as integers in this context (only as integers modulo 6), but that's just a minor difference in the way we apparently use terminology)

tprice7 12y ago
"Being a UFD in general is not enough, but being a UFD in which 2, 3, and 5 are (co)prime is certainly sufficient for incommensurability of 3/2, 2, and 5/4 in the relevant sense." Ok, this seems like kind of a contrived condition though, compared to, say, the condition that 2, 3, and 5 are prime (which also suffices). Something that occurred to me after I made my last post (and possibly what you meant in the second paragraph?): perhaps the author didn't mean to imply that the UFD property is useful for characterizing when you have incommensurability in different rings, but rather that it's relevant because it's used as a step in the proof that these ratios are incommensurable in the case of Z. This makes sense to me but in my opinion it wouldn't hurt if the article were more explicit on this point. I'm not familiar with a commonly-used definition of UFD for which the ring of integers mod 7 is not a UFD. Could you please point me to a reference that contains this alternate definition of UFD you are referring to? The definition I am using is the one in Abstract Algebra by Dummit & Foote, which happens to agree with the definition on wikipedia at the time of this posting.
- Chinjut 12y ago
  Yes, I think what you're saying in your third paragraph is what I was trying to say in my second paragraph. And I agree that the article could stand to be much clearer on its motivations. Re: the integers mod 7, I had a brainfart; of course the integers mod 7 are a UFD, but trivially so, as they are a field. My apologies!
  - tprice7 12y ago
    Well I'm glad we are in agreement then. No worries about the UFD thing.

jackmaney 12y ago

The ring of integers modulo 7 is also a field. While fields are (trivially) UFDs, the only non-field class of examples of UFDs having every nonzero element as a power of a fixed nonzero nonunit element is a (rank 1) discrete valuation ring.

tprice7 12y ago
" the only non-field class of examples of UFDs having every nonzero element as a power of a fixed nonzero nonunit element is a (rank 1) discrete valuation ring." Ok, but the UFD criterion is redundant here because there are no non-fields for which every nonzero element is a power of a fixed nonzero nonunit element. So that doesn't make UFDs any more relevant than any other extra criterion you could add.
- jackmaney 12y ago
  Well, if you're willing to allow a bit of wiggle room in the form of "up to associates" (which is a qualifier that I forgot to add), there are integral domains in which every nonzero element equals the product of a unit and a power of a fixed nonzero nonunit element. This qualifier isn't too much to ask since "up to associates" is the generalized version of "up to a plus or minus" when factoring integers (which has no real bearing on factorizations). In particular, let F be any field, and consider F[[x]], the ring of formal power series[1] over F. Given any nonzero f(x) in F[[x]], we can "factor out as many x's as we can" to write f(x) = x^n g(x) where n is a nonnegative integer, g(x) is in F[[x]], and the constant term of g(x) is nonzero (here we explicitly define x^0 to be 1). Thus, g(x) is a unit[2] (ie invertible), and every nonzero element of F[[x]] is a unit multiple of a power of x. [1]: http://en.wikipedia.org/wiki/Formal_power_series http://en.wikipedia.org/wiki/Formal_power_series [2]: http://en.wikipedia.org/wiki/Formal_power_series#Inverting_series http://en.wikipedia.org/wiki/Formal_power_series#Inverting_s...
  - tprice7 12y ago
    Ok, but rings for which every nonzero element is a power of a fixed nonzero nonunit element up to associates are exactly discrete valuation rings. So your statement boils down to "The only UFDs that are discrete valuation rings are (rank 1) discrete valuation rings". And I'm not sure what you mean by rank 1 here, but I can deduce that all discrete valuation rings are rank 1 from what you've said, since all discrete valuation rings are UFDs. So what you've said boils down to: "The only UFDs that are discrete valuation rings are discrete valuation rings", which again doesn't make UFDs more relevant than any other condition you might add.