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Unformatted text preview: 386 8. MODULES M Œp, where p r is the largest power of p dividing the period m of M . See
Exercise 8.5.8. M Œp D f0g if p does not divide m.
We will show that M is the (internal) direct sum of the submodules
M Œp for p appearing in an irreducible factorization of a period of M ..
Theorem 8.5.9. (Primary decomposition theorem) Let M be a ﬁnitely generated torsion module over a principal ideal domain R, let m be a period
of M with irreducible factorization m D p1 1 p2 2 ps s . Then
M Š M Œp1 ˚ ˚ M Œpk : m
Proof. For each index i let ri D m=pi i ; that is, ri is the product of all
the irreducible factors of m that are relatively prime to pi . For all x 2 M ,
we have ri x 2 M Œpi , because pi i .ri x/ D mx D 0. Furthermore, if
x 2 M Œpj for some j ¤ i , then ri x D 0, because pj j divides ri .
The greatest common divisor of fr1 ; : : : ; rs g is 1. Therefore, there
exist t1 ; : : : ; ts in R such that t1 r1 C C ts rs D 1. Hence for any x 2 M ,
x D 1x D t1 r1 x C
C ts rs x 2 M Œp1 C M Œp2 C
C M Œps . Thus
G D M Œp1 C C M Œps .
Suppose that xj 2 M Œpj for 1 Ä j Ä s and j xj D 0. Fix an index
i . Since ri xj D 0 for j ¤ i , we have
0 D ri .
xj / D
ri xj D ri xi :
j j Because ri is relatively prime to the period of each nonzero element of
M Œpi , it follows that xi D 0. Thus by Proposition 8.1.27
M D M Œp1 ˚ ˚ M Œps :
I Corollary 8.5.10. Let y 2 M and write y D y1 C y2 C
yj 2 M Œpj for each j . Then yj 2 Ry . C ys , where Proof. If rj and tj are as in the proof of the theorem, then yj D rj tj y . I
The primary decomposition and the Chinese remainder theorem. For
the remainder of this subsection, we study the primary decomposition of
a cyclic torsion module over a principal ideal domain R. The primary
decomposition of a cyclic torsion module is closely related to the Chinese
remainder theorem, as we shall explain. ...
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- Fall '08