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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 px 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.
 Spring '10
 AbhinavKumar
 Algebra, Number Theory, Fractions

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