By Robin Chapman

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

**Example text**

Then √ we can take u1 = σ σ ¯ ( m) = ( 12 (1 + m), 12 (1 − m)), and we find that det(U ) = − m. Hence √ √ again area(¯ σ (OK )) = m = ∆. Now consider 4). Then |∆| = 4|m| √ the case where m < 0 and m ≡ 1 (mod 2 identifying x + iy and OK = Z[ m]. Here we need to identify C with R by√ with (x, y). Then u1 = σ ¯ (1) = 1 = (1, 0) and u2 = σ ¯ ( m) = i |m| = (0, |m|) are generators of σ ¯ (O √K ). It is immediate that det(U ) = |m| and √ 1 so so area(¯ σ (OK )) = m = 2 ∆. Finally suppose that m < 0 and m ≡ 1 (mod 4).

Then f1 − f = pf2 where f2 ∈ Z[α]. But f1 (α) = pf2 (α) ∈ p so that P1a1 P2a2 · · · Prar ⊆ p . 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].

It is easy to write down a candidate for the inverse of a fractional ideal I; define I ∗ = {β ∈ K : βI ⊆ OK }. It is clear that I ∗ is an additive subgroup of K and is nonzero since it contains 1/η whenever I = ηJ with J an ideal of OK . Also it is clear that if β ∈ I ∗ and γ ∈ OK then βγ ∈ I ∗ . If δ is a nonzero element of I then δI ∗ ⊆ OK and so (1/δ)I ∗ ⊆ OK . Thus I ∗ is a fractional ideal of K. Also II ∗ ⊆ OK so that II ∗ is an ideal of OK . The hard part is to show that II ∗ = OK . We first prove the invertibility for prime ideals.