# Finitely generated modules over PID

Let $A$, $B$, $C$, and $D$ be finitely generated modules over a PID $R$ such that $A\oplus$ $B$ $\cong$ $C\oplus$ $D$ and $A\oplus$ $D$ $\cong$ $C\oplus$ $B$ . Prove that $A$ $\cong$ $C$ and $B$ $\cong $$D. The only tool I have is the theorem about finitely generated modules, but I don't quite see the connection. Please Help. Thanks. - ## 1 Answer According to the theorem about finitely generated modules, any finitely generated module over a PID R can be expressed as$$ R^r \oplus R/(p_1^{k_1}) \oplus R/(p_2^{k_2}) \oplus \cdots \oplus R/(p_n^{k_n})$$where$p_1^{k_1},\ldots,p_n^{k_n}$are prime powers in$R$. The$R^r$is called the free part, and the other summands are elementary divisors. According to the theorem, these summands are unique up to permutation. Let$p^k$be a prime power in$R$. Let$a$,$b$,$c$, and$d$denote the numbers of times that$R/(p^k)$appears as an elementary divisor of$A$,$B$,$C$, and$D$, respectively. Since$A\oplus B \cong C \oplus D$, the uniqueness part of the theorem tells us that$a+b=c+d$. Similarly, since$A\oplus D = C\oplus B$, we know that$a+d=c+b$. Since$a$,$b$,$c$, and$d$are nonnegative, it follows that$a=c$and$b=d$. This holds for all$p^k$, and a similar argument works for the rank of the free parts, which proves that$A\cong C$and$B\cong D\$

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that was very elegant and clear, thanks – Leon Sep 18 '11 at 23:58