By Stein W.

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**Sample text**

Given K, find some order O ⊂ K, and compute d = Disc(O). Factor d, and use the factorization to write d = s · f 2 , where f 2 is the largest square that divides d. Then the index of O in OK is a divisor of f , and we (tediously) can enumerate all rings R with O ⊂ R ⊂ K and [R : O] | f , until we find the largest one all of whose elements are integral. 4. Consider the ring OK = √ Z[(1 + 5)/2] of integers of K = Q( 5). The discriminant of the basis 1, a = (1 + 5)/2 is 2 1 = 5. 1 3 √ √ √ Let O = Z[ 5] be the order generated by 5.

It thus suffices to prove that if p1 , . . , pr are distinct prime ideals of OK and e1 , . . , er are positive integers, then r ψ : OK / r penn n=1 → (OK /penn ) n=1 is an isomorphism. Let ϕ : OK → ⊕rn=1 (OK /penn ) be the natural map induced by reduction mod penn . 1 is equal to r en en n=1 pn , so ψ is injective. Note that the projection OK → OK /pn of ϕ onto each factor is obviously surjective, so it suffices to show that the element (1, 0, . . , 0) is in the image of ϕ (and the similar elements for the other factors).

Let p be a nonzero prime ideal of OK , and let n ≥ 0 be an integer. Then pn /pn+1 ∼ = OK /p as OK -modules. Proof. ) Since pn = pn+1 (by unique factorization), we can fix an element b ∈ pn such that b ∈ pn+1 . Let ϕ : OK → pn /pn+1 be the OK -module morphism defined by ϕ(a) = ab. The kernel of ϕ is p since clearly ϕ(p) = 0 and if ϕ(a) = 0 then ab ∈ pn+1 , so pn+1 | (a)(b), so p | (a), since pn+1 does not divide (b). Thus ϕ induces an injective OK -module homomorphism OK /p → pn /pn+1 . 3. Suppose c ∈ pn .