4 ms·

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

by GFK_of_xmaspast 10y ago

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

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