Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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 $B\cong D$

share|improve this answer
    
that was very elegant and clear, thanks –  Leon Lampret Sep 18 '11 at 23:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.