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Unformatted text preview: 306 6. RINGS product of the denominators of the coefﬁcients of '.x/. Factoring g.x/ as
'.x/ D .d=b/f .x/;
where f .x/ is primitive in RŒx. This decomposition is unique up to units
in R. In fact, if
.d1 =b1 /f1 .x/ D .d2 =b2 /f2 .x/;
where f1 and f2 are primitive in RŒx, then d1 b2 f1 .x/ D d2 b1 f2 .x/. By
the uniqueness of the decomposition 6.6.1 for RŒx, there exists a unit u in
R such that d1 b2 D ud2 b1 . Thus d1 =b1 D ud2 =b2 .
Example 6.6.8. Take R D Z.
7=10 C 14=5x C 21=20x 3 D .7=20/.2 C 8x C 3x 3 /;
where 2 C 8x C 3x 3 is primitive in ZŒx.
Lemma 6.6.9. (Gauss’s lemma). Let R be a unique factorization domain
with ﬁeld of fractions F .
(a) The product of two primitive elements of RŒx is primitive.
(b) Suppose f .x/ 2 RŒx. Then f .x/ has a factorization f .x/ D
'.x/ .x/ in F Œx with deg.'/; deg. / 1 if, and only if, f .x/
has such a factorization in RŒx.
Proof. Suppose that f .x/ D
ai x i and g.x/ D
bj x j are primitive
in RŒx. Suppose p is irreducible in R. There is a ﬁrst index r such
that p does not divide ar and a ﬁrst index s such thatP does not divide
r Cs in f .x/g.x/ is a b C
i <r ai br Cs i C
P. The coefﬁcient of x
ar Cs j bj . By assumption, all the summands are divisible by p ,
except for ar bs , which is not. So the coefﬁcient of x r Cs in fg.x/ is not
divisible by p . It follows that f .x/g.x/ is also primitive. This proves part
Suppose that f .x/ has the factorization f .x/ D '.x/ .x/ in F Œx
with deg.'/; deg. /
1. Write f .x/ D ef1 .x/, '.x/ D .a=b/'1 .x/
and .x/ D .c=d / 1 .x/, where f1 .x/, '1 .x/, and 1 .x/ are primitive
in RŒx. Then f .x/ D ef1 .x/ D .ac=bd /'1 .x/ 1 .x/. By part (a),
the product '1 .x/ 1 .x/ is primitive in RŒx. By the uniqueness of such
decompositions, it follows that .ac=bd / D eu, where u is a unit in R, so
f .x/ factors as f .x/ D ue'1 .x/ 1 .x/ in RŒx.
Corollary 6.6.10. If a polynomial in ZŒx has a proper factorization in
QŒx, then it has a proper factorization in ZŒx. ...
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