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By Robin Chapman

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But we have seen these ideals have the same norm, so they are equal. 37 In fact this result is true even when OK = Z[α] as long as p |OK : Z[α]|. We shall not prove this generalization. √ Example Let K = Q( m) be a quadratic field where √ m is a squarefree integer with m ≡ 1 (mod 4). Then OK = Z[α] where α = m has minimum polynomial X 2 − m. To determine the prime ideal factorization of p , where p is a prime number, in OK , we must factorize X 2 − m in Fp [X]. There are three possibilities: 1.

We have already seen that I and J are nonprincipal. We can now √ write this as [I] = [OK ] and [J] = [OK ]. But we 2 −6 and J 2 = 3 . Thus [I]2 = [I][J] = [J]2 = also have I = 2 , IJ = [OK ] in ClK . Thus [J] = [I]−1 = [I], that √ is, the ideals I and √ J lie in the −1 −1 −2 −6 2 = −6/2 . Hence same ideal class. Indeed IJ = IJJ = √ J = ( −6/2)I. We can confirm this by calculating √ √ √ √ √ −6 −6 I= 2, −6 = −6, −3 = 3, −6 = J. 2 2 So we have at least two elements, [OK ] and [I], in ClK . As [I]2 = [OK ] then [I] has order 2 in the class-group.

Thus γv = Av where v is the column vector with entries the βj and A is the matrix with entries the ajk . Thus γ is an eigenvalue of A which is a matrix with integer entries. Thus γ is an algebraic integer and so γ ∈ K ∩ B = OK . We now show that prime ideals are invertible. 2 Let K be a number field and let P be a prime ideal of OK . Then there is a fractional ideal J of K with P J = 1 . Proof We let P ∗ = {β ∈ K : βP ⊆ OK }. Then P ∗ is a fractional ideal of K, OK ⊆ P ∗ and P ⊆ P P ∗ ⊆ OK . By the maximality of the prime ideal P , either P P ∗ = P or P P ∗ = OK .

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