We will call a submodule $A$ a direct summand of $K$ if there exists a submodule $B$ such that $A \oplus B = K$. I think this is a question that can be formulated in terms of rank of a proper free sumbmodules but I am not sure how to ask it.
Consider the $\mathbb{Z}$-module $\mathbb{Z} \oplus \mathbb{Z}$. Is there an example of two submodules of $A,B$ of $\mathbb{Z} \oplus \mathbb{Z}$ such that $A$ and $B$ are direct summands of $\mathbb{Z} \oplus \mathbb{Z}$ but $A+B$ is not a direct summand of $\mathbb{Z} \oplus \mathbb{Z}$?
I first thought that $\mathbb{Z}\oplus 0$ and $ 0 \oplus \mathbb{Z}$ was an example until I realized every module is a direct summand of itself…
Best Answer
Here's a slightly more high-brow way to say what Mariano is saying. A vector $(a,b) \in \mathbb{Z}^2$ spans a direct summand if and only if it is primitive, i.e. not divisible by any integer other than $\pm 1$. However, two linearly independent vectors $(a,b)$ and $(c,d)$ span a direct summand of $\mathbb{Z}^2$ (which necessarily must be $\mathbb{Z}^2$ itself) if and only if the determinant of the matrix $\left( \begin{array}{cc} a & c\\b & d \end{array}\right)$ is $\pm 1$. This is a very special property. If you write down two random primitive vectors, you will probably get a huge determinant.