By Stein W.

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**Extra resources for A Brief Introduction to Classical and Adelic Algebraic Number Theory**

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As an abelian group OK is free of rank equal to the degree [K : Q] of K, and I is of finite index in OK , so I can be generated as an abelian group, hence as an ideal, by [K : Q] generators. The following proposition asserts something much better, namely that I can be generated as an ideal in OK by at most two elements. 7. Suppose I is a fractional ideal in the ring OK of integers of a number field. Then there exist a, b ∈ K such that I = (a, b). Proof. If I = (0), then I is generated by 1 element and we are done.

Every ideal is contained in a maximal ideal, so I is contained in a nonzero prime ideal p. 6 we can cancel I from both sides of this equation to see that p−1 = OK , a contradiction. Thus I is strictly contained in Ip−1 , so by our maximality assumption on I there are maximal ideals p1 , . . , pn such that Ip−1 = p1 · · · pn . Then I = p · p1 · · · pn , a contradiction. Thus every ideal can be written as a product of primes. Suppose p1 · · · pn = q1 · · · qm . If no qi is contained in p1 , then for each i there is an ai ∈ qi such that ai ∈ p1 .

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

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