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 $M$ be left $R$-module and $A$ ,$B$ are two submodule of $M$ such that $A\times B\cong M$.

Is there submodule $C$ such that $A+C=M$ and $A\cap C=0$?

All my attempts proving the $A$ has a complement in $M$ but it seems to be wrong, but can not find any counter example(s) for it.

share|improve this question
    
What do you mean by $A \times B$? I have never seen this notation before. Do you mean $A+B$? Or $A \oplus B$? –  Aaron Oct 5 '12 at 7:10
    
@Aaron $A\times B$ means the product of two module. some references use the symbol "$\oplus$" for product of two module but I use this symbol"$\times$". –  Babak Miraftab Oct 5 '12 at 7:12
    
Ahh, okay. Finite products are the same as finite direct sums, which is why people generally use direct sum when there are only finitely many modules. But with submodules, there are sums that are not direct, and I was assuming this was some kind of product that was not direct (which confused me). –  Aaron Oct 5 '12 at 7:24
add comment

1 Answer

up vote 2 down vote accepted

This is already false for finite $\mathbb{Z}$-modules, i.e. abelian groups.

Let $M$ be $\mathbb{Z}_4 \times \mathbb{Z}_2$, $A$ be $\langle 2 \rangle \times \langle 0 \rangle$, and $B$ be $\langle 1 \rangle \times \langle 0 \rangle$. Then $M \cong A \times B$, but $A$ is not complemented in $M$.

share|improve this answer
1  
It is probably worth saying why there is no complement. Here is a quick proof. If there were a complement, then we would have a projection map $p:M\to A$ such that elements of $A$ were left fixed. However, this is impossible because $2p(\langle 1\rangle\times \langle 0 \rangle)=0$ (since the subgroup generated by $\langle 2\rangle\times \langle 0\rangle$ is isomorphic to $\mathbb Z_2$) –  Aaron Oct 5 '12 at 9:32
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.