By Robin Chapman

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**Extra resources for Algebraic Number Theory: summary of notes [Lecture notes]**

**Example text**

Conversely suppose that I1 ⊇ I2 . Then OK = I1 I1−1 ⊇ I2 I1−1 = J say. Then J is an ideal of OK and I1 J = I1 I2 I −1 = I2 . After all this hard work it is now easy to conclude that factorizations into prime ideals are unique. 3 Let K be a number field, and suppose that I is a nontrivial ideal of OK . If I = P1 P2 · · · Pr = Q1 Q2 · · · Qs (∗) where the Pj and Qk are prime ideals of OK , then r = s and the Qk can be reordered so that Pj = Qj for each j. Proof We use induction on r. Certainly P1 ⊇ I = Q1 Q2 · · · Qs .

J k + 1 k=0 = pβ where β ∈ OK . Comparing norms gives pp−1 = N (λp−1 ) = pp−1 N (β). Consequently N (β) = 1 and β = λp−1 /p is a unit. 7 Let K = Q(ζ) where p is an odd prime number and ζ = exp(2πi/p). Then ∆(1, ζ, ζ 2 , . . , ζ p−2 ) = (−1)(p−1)/2 pp−2 . Proof Call this discriminant ∆. Then ∆ = (−1)p(p−1)/2 f (ζ) where f is the minimum polynomial of ζ. Then f (X) = (X p − 1)/(X − 1), so f (X) = p(X − 1)X p−2 − (X p − 1) (X − 1)2 and so f (ζ) = pζ p−2 /(ζ − 1). Hence N (f (ζ)) = N (p)N (ζ)p−1 /N (ζ − 1) = pp−2 .

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. It will turn out that these are the only ideal classes in K so that |ClK | = 2.