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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.

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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

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