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From Dummit and Foot, Abstract Algebra ch10.2:

Defn: Let A,B be submodules of the R-module M. The sum of A and B is the set

$A+B=\{a+b| a\in A, b\in B \}$ it is easily checked that the sum of two submodules A and B is a submodule and is the smallest submodule which contains both A and B.

By the submodule criterion, it follow that $A+B$ is a submodule of M. How can we show that that $A+B$ is the smallest submodule that contains A and B?

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2 Answers

up vote 3 down vote accepted

If $C$ is a submodule containing $A$ and $B$, it is closed under addition, thus for all $a \in A$ and $b \in B$, $a,b \in C$, hence $a+b \in C$ and $A+B \subseteq C$. Therefore $A+B$ "is the smallest" (under inclusion).

Alternatively, you could define $A+B$ as the intersection of all modules containing $A$ and $B$ (intersection of submodules is a submodule, and $\{ a+b \, |\, a \in A, b \in B \}$ must be in the intersection so it's non-empty).

Hope that helps,

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I like the intersection bit. That for me is the meaning of "smallest". –  user38268 Feb 26 '12 at 4:05
    
It actually means "smallest" in a well-defined mathematical sense. If you define a partial order under inclusion over all the modules containing $A$ and $B$, $A+B$ is a minimum. This is a very general idea ; it works for many other algebraic structures ; groups, rings, modules, fields, name it... –  Patrick Da Silva Feb 26 '12 at 4:25
    
Yeah sure, just like how the closure of $A$ is the intersection of all closed sets containing $A$, the intersection of all topologies containing $\tau$ the smallest one containing it, etc... –  user38268 Feb 26 '12 at 4:27
    
Exactly! These are all the same ideas. –  Patrick Da Silva Feb 26 '12 at 4:45
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Modules are close under addition. So any module containing both $A$ and $B$ must contain sums of elements in $A$ and $B$, thus by definition it must contain $A+B$.

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