4 ms·
A nice discussion, but a bit of a nitpick: in Z[sqrt(-5)], 2, 3, 1+sqrt(-5), and 1-sqrt(-5) are actually irreducible, not prime. In an integral domain D, a non
by jackmaney 10y ago
A nice discussion, but a bit of a nitpick: in Z[sqrt(-5)], 2, 3, 1+sqrt(-5), and 1-sqrt(-5) are actually irreducible, not prime.
In an integral domain D, a nonzero element x is called irreducible if x is not a unit and whenever x = ab (for a, b in D), then one of a or b is a unit.
On the other hand, a nonunit element x in D is called prime if for all a, b in D, if x divides ab then x divides a or x divides b (by "x divides ab", I mean that there's some element--call it y--in D such that xy = ab).
In Z (or any unique factorization domain[1]), these concepts coincide. In Z[sqrt(-5)], however, there are irreducible elements that are not prime. In particular, 2 is irreducible in Z[sqrt(-5)], but it isn't prime, since 2 divides (1+sqrt(-5))*(1-sqrt(-5)), but 2 divides neither 1+sqrt(-5) nor 1-sqrt(-5).
[1]: https://en.wikipedia.org/wiki/Unique_factorization_domain https://en.wikipedia.org/wiki/Unique_factorization_domain
- n4r9 10y agoThis is brought up at the top of the comments section, along with Gowers' response and a neat comment from someone called Fabian: >It seems it is the thoughtless combination of the definitions of “prime” and “irreducible” that makes the theorem appear obvious.