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In this case the above analysis does not quite work since the Chinese remainder theorem does not apply. However, it suggests strongly that 2 pOK = p, g(α) , and this we can check directly. We check the case where d ≡ 2, 3 (mod 4) and p = 2; the other cases are similar. We have that f¯(x) = x2 − d is a square in Fp [x]. One checks easily that this implies that p divides d (since p = 2), so we can take g(x) = x. Thus the claim is that 2 pOK = p, α . To check this, we simply compute p, α p, α = p2 , pα, α2 .

At the moment we have a submodule c of K such that γc ⊆ R, which isn’t quite good enough. Note, however, that b ⊆ c since α ∈ a. Thus γb ⊆ γc ⊆ R. We will show that γb ⊆ b. So, take an arbitrary element β ∈ b. We want to show that γβ ∈ b. To do this we will show that for all α ∈ a, we have γβα ∈ α ; this will imply that γβ ∈ b by the definition of b. ) So fix α ∈ a. Then βα ∈ α by definition of b, so we can write βα = αδ for some δ ∈ R. Now, visibly δ ∈ c, so γδ ∈ γc ⊆ R. So, finally, γβα = (γδ)α ∈ α .

Then there exists γ ∈ K such that γ ∈ / R and γa ⊆ R. 3. UNIQUE FACTORIZATION OF IDEALS IN DEDEKIND DOMAINS 45 The lemma just says that a is significantly distinct from R in the sense that there is some non-integral element of K we can multiply it by which will not cause the ideal to become non-integral. R itself certainly does not have this property, for example. Proof. Fix any non-zero α in a. 4 the principal ideal α contains some product of non-zero prime ideals; choose (not necessarily distinct) primes p1 , .

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Algebraic Number Theory [Lecture notes] by Tom Weston

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