My Math Forum Factor 2 into primes in Z[i].

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 November 20th, 2006, 10:08 PM #1 Newbie   Joined: Nov 2006 Posts: 20 Thanks: 0 Factor 2 into primes in Z[i]. Factor 2 into primes in Z[i]. my answer is 2=(1 + i)(1-i). Do I need to add something more? Please teach me. Thank you very much.
 November 21st, 2006, 03:42 AM #2 Member   Joined: Nov 2006 From: Norway Posts: 33 Thanks: 0 No, you don't. If we define N(a+ib) = (a+ib)(a-ib) = a^2 + b^2, then N is a multiplicative function in Z(i), And, since N(1+i) = 2, 1 + i must be a prime in Z(i), since 2 is a prime in Z. Likewise 1-i.
 November 21st, 2006, 09:26 AM #3 Newbie   Joined: Nov 2006 Posts: 12 Thanks: 0 Factor 2 in Z[i] Sorry, Einar and Alpah: You wrote 2 = (1+i)(1-i). Unfortunately, these are not distinct primes, because 1-i = -i(1+i) and -i is a unit. So the factorisation is 2 = -i(1+i)². Since the discriminant of Q(i) is -4, 2 must ramify in Z[i] so it cannot be the product of 2 distinct primes.

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