15:17  Take your favorite dedekind domain.15:17  mmhm15:17  For instance, consider K a number field15:17  and O_K the ring of integers.15:17  Then take a prime p in Z.15:18  Since Z \subset O_K, p can be considered as an element of O_K, right ?15:18  yes15:18  Ok. p is prime in Z, meaning that the ideal (p) = pZ is a prime ideal of Z.15:18  yep15:18  Consider now this ideal, but in O_K15:18  right15:19  ie the ideal pO_K15:19  yes15:19  It may not be prime anymore15:19  mmhm15:19  So it factors as a product of prime ideals *of O_K*15:20  pO_K = P_1^e_1....P_r^e_r15:20  where P_i are distinct prime ideals of O_K.15:20  yes15:20  You say that p ramifies in O_K (or in K) when there is some e_i which is > 115:21  Example15:21  Take Z[i], the ring of Gauss integers.15:22  It is the ring of integers of the field Q(i).15:22  Take the prime 2 in Z.15:23  (2) = (1 + i) (1 - i) in Z[i] ?15:23  Yes.15:23  But in fact15:23  The ideal (1-i) = (1+i) (as ideals)15:23  So (2) = (1+i)^215:23  And you can prove that (1+i) is a prime ideal in Z[i]15:23  is it because (1 - i)i = i + 1 = 1 + i?15:24  Yes15:24  very cool15:24  Therefore, (2) ramifies in Z[i].15:24  is it prime because the quotient Z[i]/(1 - i) ~= Z is an integral domain? [the quotient tells us to make 1 - i = 0, or to set i = ]15:24  But you can also prove that primes that ramify are not really common15:24  it = (1 - i)15:25  In fact, 2 is the *only* prime that ramifies in Z[i]15:25  More generally, you only have a finite number of primes that ramify15:25  in any O_K?