Image of Sum of Submodules

Question

Statement. Let $R$ be a ring, $M$ be an $R$-module and $(M_i)_{i\in I}$ be a family of $R$-submodules of $M$. Then for every $R$-linear map $f$ defined on $M$, $$f\left(\sum_{i\in I}{M_i}\right)=\sum_{i\in I}{f(M_i)}.$$

Proof. For each $i\in I$, because $M_i\subseteq\sum_{i\in I}{M_i}$, it is clear that $f(M_i)\subseteq f(\sum_{i\in I}{M_i})$ as well. This shows that $$\sum_{i\in I}{f(M_i)}\subseteq f\left(\sum_{i\in I}{M_i}\right).$$

Conversely, let $y\in f(\sum_{i\in I}{M_i})$. Then $y=f(x)$ for some $x\in\sum_{i\in I}{M_i}$. In this case, we can write $$x=x_{i_1}+\cdots+x_{i_k},\quad\text{where}~x_{i_j}\in M_{i_j},~j=1,\ldots,k,~i_1,\ldots,i_k\in I.$$ As a result, \begin{align*} y=f(x)&=f(x_{i_1}+\cdots+x_{i_k}) \\ &=f(x_{i_1})+\cdots+f(x_{i_k})\in\sum_{i\in I}{f(M_i)}. \blacksquare \end{align*}

---

It seems that the above proof is very straightforward, but I still want to verify it here in case there were any counterexamples.

Meanwhile, provided it is correct, I wonder if it follows from some more general results, as I have no impression that any textbook or monographs has stated that explicitly.

Answer 1

Note that the second inclusion can also be proved without taking elements, just as you did with the other one:

For each $i \in I$, $M_i \subseteq f^{-1}(f(M_i)) \subseteq f^{-1}(\sum_{i \in I} f(M_i))$; so, $\sum_{i \in I} M_i \subseteq f^{-1}(\sum_{i \in I} f(M_i))$, which means that $f(\sum_{i \in I} M_i) \subseteq \sum_{i \in I} f(M_i)$.

The thing is that this result only needs the following two facts:

• If $f \colon M \to N$ is a linear map, then for each submodule $A$ of $M$ and each submodule $B$ of $N$, $$ f(A) \subseteq B \iff A \subseteq f^{-1}(B). $$
• The sum of a family of submodules of a module is the smallest submodule containing each member of the family.

In other words:

• If $f \colon M \to N$ is a linear map, then the direct and inverse image functions forms a (monotone) Galois connection between the posets of submodules of $M$ and $N$.
• In the poset of submodules of a module, the supremum is given by the sum of submodules.

The general result that you're looking for is:

In a (monotone) Galois connection, the lower adjoint preserves all supremum that exists in its domain.

You already have the proof of this.