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Let M and N be left R-Modules, is it possible to construct an example of a sub-module of $M \oplus N$ that is not a direct sum of a submodule of M and a submodule of N?

I don't know a whole lot of Modules so I was trying to think of ideals. Maybe some polynomials in two variables x,y?

Anyone think this is on track?

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Try the direct sum of $\mathbb{R}$-modules $\mathbb{R} \oplus \mathbb{R}$. – Chris Eagle Sep 11 '11 at 0:15
up vote 7 down vote accepted

Hint: Let $R=N=M=\mathbb{Z}$. There are many $(x,y)\in \mathbb{Z}\oplus \mathbb{Z}$ such that the submodule generated by $(x,y)$ is not the direct sum of submodules of $\mathbb{Z}$. Can you find some?

This works in much greater generality than $\mathbb{Z}$, of course.

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since $ \mathbb{Z} $ only has $ n \mathbb{Z} $ as submodules (they're the only subgroups) wouldn't any module generated by (n,m) be the direct sum of $ n \mathbb{Z} $ and $ m \mathbb{Z} $? I guess if I use two points though like $ (2,3),(5,7) $ maybe they generate a submodule that can't be the direct sum of any submodules of $ \mathbb{Z} $ – user9352 Sep 11 '11 at 1:46
Compare $$n\mathbb{Z}\oplus m\mathbb{Z}=\{(a,b)\in\mathbb{Z}\oplus\mathbb{Z}\mid n\text{ divides } a, m\text{ divides } b\}$$ and $$\langle(n,m)\rangle = \{(nt,mt)\in\mathbb{Z}\oplus\mathbb{Z}\mid t\in\mathbb{Z}\},$$ they are not the same. – Zev Chonoles Sep 11 '11 at 1:48
@user9352: If you replace $\mathbb Z$ by $\mathbb R$ in Zev's answer, your question becomes: are there lines in the plane $\mathbb R^2$ which are neither vertical nor horizontal? – Pierre-Yves Gaillard Sep 11 '11 at 4:59
ok I think i see, the direct sum is bigger like if I use (2,3) then (2,6) is in the direct sum $2\mathbb{Z} \oplus 3\mathbb{Z}$ but not the submodule generated by (2,3) – user9352 Sep 11 '11 at 13:38
What about a bigger sub-module, would it be possible to get a submodule of a direct sum that is not contained in the direct sum of two submodules? – user9352 Sep 12 '11 at 1:54

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