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November 20th, 2006, 10:08 PM   #1
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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.
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November 21st, 2006, 03:42 AM   #2
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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.
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November 21st, 2006, 09:26 AM   #3
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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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