# Is $N_1+N_2\cong N_1\oplus N_2$ iff $N_1\cap N_2=(0)$?

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!

• (I know the question is about commutative rings but) for noncommutative rings it does not work for $N_1$ and $N_2$ even if they are both cyclic. Commented Oct 2, 2023 at 14:57
• There is an obvious surjection $N_1 \oplus N_2 \to N_1 + N_2$, which is injective if and only if $N_1 \cap N_2 = (0)$. Commented Oct 2, 2023 at 15:09
• $M/N \cong M$ implies $N = 0$, sometimes Commented Oct 2, 2023 at 15:12

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, $$N_1 \cap N_2 = (0)$$.