I want to show that $\mathbb Q$ as an abelian group or as an $\mathbb Z$- module is indecomposable. An $R$-module $M$ is said to be indecomposable if it cannot be written as a direct sum of its non-trivial(not $0$ or $M$ itself) submodule. Any hint how to proceed?
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this question is years old, but nevertheless this other question is the same as it: Why is the additive group of rational numbers indecomposable? – Ben Millwood Dec 20 '21 at 16:49
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A direct sum $M \oplus N$ always has a pair of submodules that intersect only at the identity, namely $\{ (m,0) : m \in M\}$ and $\{(0,n):n\in N\}$.
But any two nontrivial subgroups of $\mathbb Q$ have infinite intersection.
Ben Millwood
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It's easier than it sounds. Can you prove it for cyclic subgroups? – Ben Millwood Oct 15 '15 at 16:57
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Suppose $M_1$ be the subgroup generated by $\frac{p_1}{q_1}$ and $M_2$ be the submodule generated by $\frac{p_2}{q_2}$ then for any natural number $n$, $np_1p_2$ is in both $M_1$ and $M_2$. So, the intersection of $M_1$ and $M_2$ is non-empty. From here I can easily see your last statement in the answer. Again very nice answer from you. – Oct 16 '15 at 03:59
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Theorem 3.1.5[Cohen Macaulay rings-Bruns and Herzog] Let $R$ be a ring and $I$ an $R$ module. If $R$ is a principal domain and $I$ is divisible, then $I$ is injective.
By theorem $\mathbb{Q}$ is an injective $\mathbb{Z}$ module
Theorem 3.2.6[Cohen Macaulay rings-Bruns and Herzog] Let $R$ be a Noethering ring.
$(a)$ For all $p \in spec(R)$ the module $E(R/p)$ is indecomposable (Where $E(R/p)$ is injective hull of $R/p$)
put $R=\mathbb{Z}$ and $M=\mathbb{Q}$ , since $0 \in spec(R)$. $E(R/<0>)=\mathbb{Q}=M$ ( Injective hull of integral domain is field of fractions see)
user177523
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