By Robin Chapman

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An ideal I of OK is maximal if I is nontrivial but the only ideals J of OK with I ⊆ J are J = I and J = OK . 4 Let K be a number field. An ideal I of OK is prime if and only if it is maximal. Proof First suppose that I is maximal. Let β, γ ∈ OK with βγ ∈ I and β∈ / I. To show that I is prime it suffices to show that γ ∈ I. Let J = I + β . Then J is an ideal and I ⊆ J, but I = J since β ∈ J. By maximality of I, J = OK . Hence 1 ∈ J so 1 = η + δβ where η ∈ I and δ ∈ OK . Then 1 ≡ δβ (mod I). Consequently, γ = 1γ ≡ δβγ ≡ 0 (mod I), as βγ ∈ I.

Hence area((a + F) ∩ X ) area(X ) = a∈Λ where only a finite number of terms in the sum are nonzero. Consider the set (a + F) ∩ A. Its elements are a + t where t ∈ F and t ∈ {x − a : x ∈ X } = X − a using the obvious notation. Hence (a + F) ∩ A = a + (F ∩ (X − a)) and clearly area((a + F) ∩ A) = area(F ∩ (X − a)). Thus area(F ∩ (X − a)). area(X ) = a∈Λ As area(X ) > area(F) by hypothesis, the sets F ∩ (X − a)) cannot all be disjoint, for then area(F ∩ (X − a)) = area a∈Λ F ∩ (X − a) ≤ area(F) a∈Λ as this union is a subset of F.

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.