5 ms·

I know I'm going up against a brilliant mathematician and Fields medalist here, but I find this article to be unenlightening. It seems that Gowers has glossed o

by nilkn 10y ago

I know I'm going up against a brilliant mathematician and Fields medalist here, but I find this article to be unenlightening. It seems that Gowers has glossed over something about the integers that's built incredibly deeply into our intuition about them when he talks about Z[sqrt(-5)]: > These numbers have various properties in common with the integers: you can add them and multiply them, there are identities for both addition and multiplication, and every number has an additive inverse. And as with integers, if you divide one by another, you don’t always get a third, so the notion of divisibility makes sense too. That means that we could if we wanted try to define a notion of a “prime” number of the form a+b\sqrt{-5}. He skipped over the fact that the positive integers are well-ordered by < (and in fact < also respects the arithmetic operations on the positive integers as well). I just can't take the comparison seriously without this being explicitly discussed, because the ordering of the integers is incredibly fundamental to human intuition about them.

GFK_of_xmaspast 10y ago
Here's a stumper then: why do Z[sqrt(-1)] and Z[sqrt(-3)] have unique factorization but Z[sqrt(-5)] doesn't?
- nilkn 10y ago
  Did you mean to write Z[sqrt(-2)] instead of Z[sqrt(-3)]?
  - GFK_of_xmaspast 10y ago
    https://en.wikipedia.org/wiki/Eisenstein_integer https://en.wikipedia.org/wiki/Eisenstein_integer
    - nilkn 10y ago
      But Z[sqrt(-3)] is not the ring of Eisenstein integers. Regardless, the fact that Z[sqrt(-n)] is a UFD for n = 1,2 but not for n >= 3 is indeed quite counter-intuitive. But Gowers isn't asking whether it's intuitive for Z[sqrt(-n)]. He's asking about the positive integers, and the positive integers have a very special well-ordering which respects the arithmetic operations and for which humans have quite a powerful intuition. I maintain that Z[sqrt(-n)] is more different from Z than Gowers is letting on and that if you actually look at all the differences, rather doing the opposite and trying to make it seem similar to Z, you'll quickly realize why people feel that the FTA is not very bizarre for Z but it is for Z[sqrt(-n)].
- Double_Cast 10y ago
  Do mathematicians actually have an answer for this? Or is it considered an open question.
  - jackmaney 10y ago
    There's a pretty good handle on it, in the form of the ideal class group[1]. [1]: https://en.wikipedia.org/wiki/Ideal_class_group https://en.wikipedia.org/wiki/Ideal_class_group
- mrkgnao 10y ago
  https://en.wikipedia.org/wiki/Quadratic_integer#Definition https://en.wikipedia.org/wiki/Quadratic_integer#Definition This might help.
- jackmaney 10y ago
  Z[sqrt(-3)] does not have unique factorization. However, Z[w] does, where w = (-1 + sqrt(-3))/2 denotes a primitive third root of unity. In particular, 2 * 2 = (-1 + sqrt(-3)) * (-1 - sqrt(-3)) gives two irreducible factorizations of 4 in Z[sqrt(-3)] (you can use norm arguments in Z[w] to show irreducibility). Note that the right hand side can also be written as (2w) * (2w^2) however, since w is not in Z[sqrt(-3)], the two factorizations above are distinct in Z[sqrt(-3)]. They become the same factorization in Z[w]. Edit: Another way to think about why Z[sqrt(-3)] doesn't have unique factorization is because it isn't integrally closed[1]--that is, there is a monic polynomial (ie a polynomial with a leading coefficient of 1) with coefficients in Z[sqrt(-3)] that doesn't have roots in Z[sqrt(-3)]. In particular, since w^2 + w + 1 == 0, w is a root of the polynomial x^2 + x + 1, which is monic over Z[sqrt(-3)]. It turns out that any unique factorization domain[2] is integrally closed. Since Z[sqrt(-3)] is not integrally closed, it is not a UFD. [1]: https://en.wikipedia.org/wiki/Integrally_closed_domain https://en.wikipedia.org/wiki/Integrally_closed_domain [2]: https://en.wikipedia.org/wiki/Unique_factorization_domain https://en.wikipedia.org/wiki/Unique_factorization_domain
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archgoon 10y ago
The Gaussian integers are well ordered as well. That '<' respects arithmetic operations isn't an aside; it's the core feature.
- nilkn 10y ago
  I did mention that part for a reason. And of course if one accepts the axiom of choice then every set can be well-ordered, but that would not force every ring to be a UFD.
  - archgoon 10y ago
    You mentioned it as as an aside. The well ordering property is completely irrelevant.
    - nilkn 10y ago
      I appreciate that you're trying to clarify my post, although I did not intend my use of parentheses to imply that the enclosed fact was not important. But I'm unclear on what your ultimate point is. It's certainly not true that well ordering is "completely irrelevant", and I think you must know that if you have the confidence to make such a bold statement, which makes me wonder why you made it in the first place. It plays a very prominent role in many proofs of these facts for the positive integers. It alone is not sufficient to establish unique factorization, but I never claimed it was. Are you agreeing with me? Disagreeing with me? What sort of response are you expecting? I'd like to have a productive discussion about this, but you're giving me a single bread crumb to go off of here.
Chinjut 10y ago
Yes, but the only way in which that ordering compels the integers to be a unique factorization domain is through the fact that it allows us the Euclidean algorithm; put another way, through the fact that, for any bunch of integers, their smallest positive combination (in the sense of adding or subtracting various multiples of them together) divides all of them. But this fact about the integers actually doesn't seem to be obvious to many people! If I were to ask "What's the smallest positive value of the form 98X - 60Y?", how many people would say "Psh, obviously it's 2"? If I were to pick three particular primes p, q, and r, and ask what the smallest positive value I could get by adding and subtracting copies of p * q, p * r, and q * r together was, how many people would see right off the bat "Oh, it has to be possible to get 1 itself, as that is their only common factor"? Not many, I suspect. And if your magic integers-with-< intuition doesn't even give you that, then I don't think it's in any sound way giving you uniqueness of prime factorization. I think you're just back-rationalizing your sense that uniqueness of prime factorization must be obvious because you've never seen it fail and heard people talk about it a lot and so on, instead of having access to genuine grounds for certainty.
- nilkn 10y ago
  > I think you're just back-rationalizing your sense that uniqueness of prime factorization must be obvious because you've never seen it fail and heard people talk about it a lot and so on, instead of having access to genuine grounds for certainty. I don't think it's fair for you to make assumptions like this about me. I have a degree in mathematics and have a fair amount of experience with this stuff. I certainly do not have the same level of experience as Gowers, but the odds are very high that you don't either. > Yes, but the only way in which that ordering compels the integers to be a unique factorization domain is through the fact that it allows us the Euclidean algorithm; put another way, through the fact that, for any bunch of integers, their smallest positive combination (in the sense of adding or subtracting various multiples of them together) divides all of them. What do you mean by "only"? You'll need to use something roughly equivalent to this, but you certainly don't have to use it stated in this precise form. For instance, a proof has already been given in this very comment thread which does not require the Euclidean algorithm in the form in which you've stated it. It requires only a much more intuitive statement about modular arithmetic, which can be proven relatively easily using the properties I've mentioned (as was done by makomk). If you're still not satisfied, I encourage you to read this thread: http://math.stackexchange.com/questions/385967/origin-of-well-ordering-proof-of-uniqueness-in-the-ftoarithmetic http://math.stackexchange.com/questions/385967/origin-of-wel... --- The other questions you've posed seem irrelevant to me. Any mathematical fact can be spun so as to make it seem difficult or counter-intuitive. For instance, I might tell you that a very simple and elementary proof of the infinitude of primes makes it obvious that n^8+4n^7+8n^6+10n^5+9n^4+6n^3+3n^2+n has at least three prime factors for all positive integers n. But how obvious is that to you?
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by nilkn 10y ago

GFK_of_xmaspast 10y ago

Here's a stumper then: why do Z[sqrt(-1)] and Z[sqrt(-3)] have unique factorization but Z[sqrt(-5)] doesn't?

nilkn 10y ago
Did you mean to write Z[sqrt(-2)] instead of Z[sqrt(-3)]?
- GFK_of_xmaspast 10y ago
  https://en.wikipedia.org/wiki/Eisenstein_integer https://en.wikipedia.org/wiki/Eisenstein_integer
  - nilkn 10y ago
    But Z[sqrt(-3)] is not the ring of Eisenstein integers. Regardless, the fact that Z[sqrt(-n)] is a UFD for n = 1,2 but not for n >= 3 is indeed quite counter-intuitive. But Gowers isn't asking whether it's intuitive for Z[sqrt(-n)]. He's asking about the positive integers, and the positive integers have a very special well-ordering which respects the arithmetic operations and for which humans have quite a powerful intuition. I maintain that Z[sqrt(-n)] is more different from Z than Gowers is letting on and that if you actually look at all the differences, rather doing the opposite and trying to make it seem similar to Z, you'll quickly realize why people feel that the FTA is not very bizarre for Z but it is for Z[sqrt(-n)].
Double_Cast 10y ago
Do mathematicians actually have an answer for this? Or is it considered an open question.
- jackmaney 10y ago
  There's a pretty good handle on it, in the form of the ideal class group[1]. [1]: https://en.wikipedia.org/wiki/Ideal_class_group https://en.wikipedia.org/wiki/Ideal_class_group
mrkgnao 10y ago
https://en.wikipedia.org/wiki/Quadratic_integer#Definition https://en.wikipedia.org/wiki/Quadratic_integer#Definition This might help.
jackmaney 10y ago
Z[sqrt(-3)] does not have unique factorization. However, Z[w] does, where w = (-1 + sqrt(-3))/2 denotes a primitive third root of unity. In particular, 2 * 2 = (-1 + sqrt(-3)) * (-1 - sqrt(-3)) gives two irreducible factorizations of 4 in Z[sqrt(-3)] (you can use norm arguments in Z[w] to show irreducibility). Note that the right hand side can also be written as (2w) * (2w^2) however, since w is not in Z[sqrt(-3)], the two factorizations above are distinct in Z[sqrt(-3)]. They become the same factorization in Z[w]. Edit: Another way to think about why Z[sqrt(-3)] doesn't have unique factorization is because it isn't integrally closed[1]--that is, there is a monic polynomial (ie a polynomial with a leading coefficient of 1) with coefficients in Z[sqrt(-3)] that doesn't have roots in Z[sqrt(-3)]. In particular, since w^2 + w + 1 == 0, w is a root of the polynomial x^2 + x + 1, which is monic over Z[sqrt(-3)]. It turns out that any unique factorization domain[2] is integrally closed. Since Z[sqrt(-3)] is not integrally closed, it is not a UFD. [1]: https://en.wikipedia.org/wiki/Integrally_closed_domain https://en.wikipedia.org/wiki/Integrally_closed_domain [2]: https://en.wikipedia.org/wiki/Unique_factorization_domain https://en.wikipedia.org/wiki/Unique_factorization_domain

deleted 10y ago

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archgoon 10y ago

The Gaussian integers are well ordered as well. That '<' respects arithmetic operations isn't an aside; it's the core feature.

nilkn 10y ago
I did mention that part for a reason. And of course if one accepts the axiom of choice then every set can be well-ordered, but that would not force every ring to be a UFD.
- archgoon 10y ago
  You mentioned it as as an aside. The well ordering property is completely irrelevant.
  - nilkn 10y ago
    I appreciate that you're trying to clarify my post, although I did not intend my use of parentheses to imply that the enclosed fact was not important. But I'm unclear on what your ultimate point is. It's certainly not true that well ordering is "completely irrelevant", and I think you must know that if you have the confidence to make such a bold statement, which makes me wonder why you made it in the first place. It plays a very prominent role in many proofs of these facts for the positive integers. It alone is not sufficient to establish unique factorization, but I never claimed it was. Are you agreeing with me? Disagreeing with me? What sort of response are you expecting? I'd like to have a productive discussion about this, but you're giving me a single bread crumb to go off of here.

Chinjut 10y ago

Yes, but the only way in which that ordering compels the integers to be a unique factorization domain is through the fact that it allows us the Euclidean algorithm; put another way, through the fact that, for any bunch of integers, their smallest positive combination (in the sense of adding or subtracting various multiples of them together) divides all of them. But this fact about the integers actually doesn't seem to be obvious to many people! If I were to ask "What's the smallest positive value of the form 98X - 60Y?", how many people would say "Psh, obviously it's 2"? If I were to pick three particular primes p, q, and r, and ask what the smallest positive value I could get by adding and subtracting copies of p * q, p * r, and q * r together was, how many people would see right off the bat "Oh, it has to be possible to get 1 itself, as that is their only common factor"? Not many, I suspect. And if your magic integers-with-< intuition doesn't even give you that, then I don't think it's in any sound way giving you uniqueness of prime factorization. I think you're just back-rationalizing your sense that uniqueness of prime factorization must be obvious because you've never seen it fail and heard people talk about it a lot and so on, instead of having access to genuine grounds for certainty.

nilkn 10y ago
> I think you're just back-rationalizing your sense that uniqueness of prime factorization must be obvious because you've never seen it fail and heard people talk about it a lot and so on, instead of having access to genuine grounds for certainty. I don't think it's fair for you to make assumptions like this about me. I have a degree in mathematics and have a fair amount of experience with this stuff. I certainly do not have the same level of experience as Gowers, but the odds are very high that you don't either. > Yes, but the only way in which that ordering compels the integers to be a unique factorization domain is through the fact that it allows us the Euclidean algorithm; put another way, through the fact that, for any bunch of integers, their smallest positive combination (in the sense of adding or subtracting various multiples of them together) divides all of them. What do you mean by "only"? You'll need to use something roughly equivalent to this, but you certainly don't have to use it stated in this precise form. For instance, a proof has already been given in this very comment thread which does not require the Euclidean algorithm in the form in which you've stated it. It requires only a much more intuitive statement about modular arithmetic, which can be proven relatively easily using the properties I've mentioned (as was done by makomk). If you're still not satisfied, I encourage you to read this thread: http://math.stackexchange.com/questions/385967/origin-of-well-ordering-proof-of-uniqueness-in-the-ftoarithmetic http://math.stackexchange.com/questions/385967/origin-of-wel... --- The other questions you've posed seem irrelevant to me. Any mathematical fact can be spun so as to make it seem difficult or counter-intuitive. For instance, I might tell you that a very simple and elementary proof of the infinitude of primes makes it obvious that n^8+4n^7+8n^6+10n^5+9n^4+6n^3+3n^2+n has at least three prime factors for all positive integers n. But how obvious is that to you?

deleted 10y ago

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