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Unformatted text preview: 298 6. RINGS coefﬁcients whose constant term is an integer. R is an integral domain, as
it is a subring of QŒx. The units in R are ˙1. No rational multiple of x
is irreducible, because x D 2. 1 x/ is a proper factorization. If x is written
as a product of several elements in R, exactly one of these elements must
be a rational multiple of x , while the remaining factors must be integers.
Therefore, x has no factorization by irreducibles. Factorization in Principal Ideal Domains
We are going to show that a principal ideal domain is a unique factorization domain. The proof has two parts: First we show that every nonzero,
nonunit element of a PID has at least one factorization by irreducibles.
Then we show that an element cannot have two essentially different factorizations by irreducibles.
Lemma 6.5.15. Suppose that R is a principal ideal domain. If a1 R Â
a2 R Â a3 R
is an increasing sequence of ideals in R, then there exists
an n 2 N such that [m 1 am R D an R. Proof. Let I D [n an R. Then I is an ideal of R, by Proposition 6.2.21.
Since R is a PID, there exists an element b 2 I such that I D bR. Since
b 2 I , there exists an n such that b 2 an R. It follows that bR Â an R Â
I D bR; so I D an R.
I Lemma 6.5.16. Let R be a commutative ring with multiplicative identity
element, and let a; b 2 R. Then ajb , bR Â aR. Moreover, a is a proper
factor of b , bR aR R.
¤ Proof. Exercise 6.5.13. ¤ I Lemma 6.5.17. Let R be a principal ideal domain. Then every nonzero
element of R that is not a unit has at least one factorization by irreducible
elements. Proof. Let R be a principal ideal domain. Suppose that R has a nonzero
element a that is not a unit and has no factorization by irreducible elements. Then a itself cannot be irreducible, so a admits a proper factorization a D bc . At least one of b and c does not admit a factorization by ...
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- Fall '08