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Let $A$ be a commutative ring with $1$, $M$ an $A$-module and $N_1,N_2$ two submodules. It is easy to see that $$N_1\cap N_2=(0)\implies N_1+N_2\cong N_1\oplus N_2.$$ I am trying to understand if the converse is true given that $N_1$ and $N_2$ are finitely generated.

If we don't make this assumption, then it's possible to produce a counterexample. Consider, for example, the $\mathbb{Z}$-module $M=\prod_{n\in\mathbb{N}}\mathbb{Z}$ with component-wise addition and scalar multiplication, and let $N_1=\mathbb{Z}\times0\times0\times\cdots$ and $N_2=M$. Then, $N_1+N_2=M\cong N_1\oplus N_2$ but clearly, $N_1\cap N_2\neq(0)$, so in this case, the converse implication is not true. However, I haven't been able to find a counterexample in the case when both $N_1$ and $N_2$ are finitely generated. Any help would be greatly appreciated!

user26857
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Jon
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1 Answers1

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Any finitely generated module over a commutative ring is Hopfian (i.e., all surjective endomorphisms are automorphisms).

So, suppose that $N_1 \oplus N_2 \cong N_1 + N_2$, and $N_1$ and $N_2$ are finitely generated. Then, the obvious surjection $N_1 \oplus N_2 \to N_1 + N_2$ induces a surjective endomorphism of $N_1 + N_2$. The union of a generating set for $N_1$ and one for $N_2$ is one for $N_1 + N_2$, so $N_1 + N_2$ is also finitely generated and thus Hopfian. So, the surjective endomorphism must be an automorphism, which means that the obvious surjection $N_1 \oplus N_2 \to N_1 + N_2$ must be an isomorphism. So, $N_1 \cap N_2 = (0)$.

Geoffrey Trang
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