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Additional info for Algebraic Number Theory: summary of notes [Lecture notes]

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We know that K is norm-Euclidean so that OK = Z[i] is a Euclidean domain. Then each ideal of Z[i] is principal. Let p be a prime number. Then p splits in K whenever p is odd and the congruence a2 ≡ −1 (mod p) is soluble. By elementary number theory this occurs if and only if p ≡ 1 (mod 4). When p ≡ 1 (mod 4) then p = P1 P 2 where P1 = p, a + i and P2 = p, a − i . Here a2 ≡ −1 (mod p). These ideals are principal: P1 = β and as N (P1 ) = p then N (β) = p. Hence 38 p = b2 + c2 where β = b + ci. We have recovered the two-square theorem of elementary number theory: if p is a prime congruent to 1 modulo 4, then p is the sum of two squares of integers.

1 to Λ and a suitable region X . To define X we split into the cases of K real and K imaginary. First suppose that K is imaginary. Let c be any positive number, and let X be the interior of the circle radius c centred at the origin. Note that the interior of a circle is convex; also X is symmetric. 6, area(Λ) = 12 N (J) |∆K |. The area of X is πc2 . 1. Then a = σ ¯ (β) with β ∈ J and |β| < c. Hence 0 < N (β) < c2 . Hence if c2 > 2 N (J) |∆K | = MK N (J) then there exists a nonzero β ∈ J with N (β) < c2 .

Hence 2 + −6 = P2 P5 and √ so [P5 ] = [P2 ]−1 = [P2 ]. Also [Q5 ] = [P5 ]−1 = [P2 ]−1 = [P2 ]. Then 4 + −6 44 has norm 22 and so is the product √of prime ideals of norms√2 and 11. It is either P2 P11 or P2 Q11 , But 4 + −6 ∈ P11 and so 4 + −6 ⊆ P11 . √ Hence 4 + −6 = P2 P11 and so [P11 ] = [P2 ]−1 = [P2 ]. Also [Q11 ] = [P11 ]−1 = [P2 ]−1 = [P2 ]. Summarizing, we have [P2 ] = [P3 ] = [P5 ] = [Q5 ] = [P11 ] = [Q11 ] and [P7 ] = [Q7 ] = [ 1 ]. We also have [P2 ]2 = [ 1 ]. Hence ClK = {[ 1 ], [P2 ]}.

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