By Stein W.

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**Sample text**

Suppose c ∈ Q is integral over Z, so there is a monic polynomial f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 with ai ∈ Z and f (c) = 0. The ai all lie in the ring of integers OK of the number field K = Q(a0 , a1 , . . an−1 ), and OK is finitely generated as a Z-module, so Z[a0 , . . , an−1 ] is finitely generated as a Z-module. Since f (c) = 0, we can write cn as a Z[a0 , . . , an−1 ]-linear combination of ci for i < n, so the ring Z[a0 , . . , an−1 , c] is also finitely generated as a Z-module.

2 (Discriminant). Suppose a1 , . . , an is any Q-basis of K. The discriminant of a1 , . . , an is Disc(a1 , . . , an ) = Det(Tr(ai aj )1≤i,j≤n ) ∈ Q. The discriminant Disc(O) of an order O in OK is the discriminant of any basis for O. The discriminant dK = Disc(K) of the number field K is the discrimimant of OK . 1. Warning: In Magma Disc(K) is defined to be the discriminant of the polynomial you happened to use to define K, which is (in my opinion) a poor choice and goes against most of the literature.

We will first prove that if p is a prime ideal, then p has an inverse, then we will prove that nonzero integral ideals have inverses, and finally observe that every fractional ideal has an inverse. Suppose p is a nonzero prime ideal of OK . We will show that the OK -module I = {a ∈ K : ap ⊂ OK } 34 CHAPTER 6. UNIQUE FACTORIZATION OF IDEALS is a fractional ideal of OK such that Ip = OK , so that I is an inverse of p. For the rest of the proof, fix a nonzero element b ∈ p. Since I is an OK -module, bI ⊂ OK is an OK ideal, hence I is a fractional ideal.

### A Brief Introduction to Classical and Adelic Algebraic Number Theory by Stein W.

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