The ideal 2 1 5 is not a principal ideal proof of thm

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Unformatted text preview: ID because of the failure of unique factoriza­ P √ √ tion 2 · 3 = (1 + −5)(1 − −5). The ideal (2, 1 + 5) is not a principal ideal. Proof. (of Thm 1.1.IV) Let P be a set of representatives of the irreducible ele­ ments of A, mo dulo units (recall that this means p ∈ A is a prime/irreducible if p is not a unit and p = xy implies x or y is a unit). Then given any x ∈ K ∗ we want to show that x can be uniquely written expressed as � x=u pvp (x) p∈P � 1 2 LECTURE 1 SUPPLEMENTARY NOTES First show existence of factorization. Uniqueness will follow from lemma III of Section 1.1. For existence, we can assume x ∈ A since we can write x = x1 /x2 and express x1 , x2 ∈ A in this form, and divide. So now assume x ∈ A. Then xA is an ideal of A. If it is the entire ring A, then x is a unit and we are done. Else it is contained in a maximal ideal pA for some prime p. Then p|x so write x = x1 p. Again if x1 A = A we are done. Else find a prime dividing x1 and so on. So this gives us a sequence of ele...
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This note was uploaded on 01/23/2014 for the course MATH 18.786 taught by Professor Abhinavkumar during the Spring '10 term at MIT.

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