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...

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    $\begingroup$ A good place to start would be: which submodules of $\mathbb{Z}\oplus\mathbb{Z}$ do you know to not be direct summands? Then figure out what $A$ and $B$ ought to be. $\endgroup$ – Zev Chonoles Sep 6 '11 at 5:25
  • $\begingroup$ Yes I am having trouble determining what proper submodules of $\mathbb{Z} \oplus \mathbb{Z}$ that are not direct summands. Should I just take some generating set $<{a_1,b_1}>$? $\endgroup$ – user7980 Sep 6 '11 at 5:32
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    $\begingroup$ @user7980 Any submodule is going to be free of rank $\leq 2$, so that's certainly the same task. One way of thinking about this is that if $A$ has a complement, so that $K = A \oplus C$ for some submodule $C \subset K$, then $K/A \approx C$. So find quotients that can't possibly be submodules. $\endgroup$ – Dylan Moreland Sep 6 '11 at 5:37

$\newcommand\ZZ{\mathbb Z}$Every subgroup of $\ZZ\oplus\ZZ$ generated by an element of the form $(x,y)$ with $x$,~$y\in\ZZ$ coprime is a direct summand. Using this, it is easy to give examples of subgroups which are direct summands. For example, it follows from this that the subgroups $A$ and $B$ generated by $(2,3)$ and by $(2,5)$ are direct summands of $\ZZ\oplus\ZZ$.

Now, it is easy to check that $$ \left( \begin{array}{cc} -1 & 1 \\ -3 & 2 \end{array} \right) \left( \begin{array}{cc} 2 & 2 \\ 3 & 5 \end{array} \right) \left( \begin{array}{cc} 1 & -3 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array} \right). $$ (I found this factorization using a Smith normal form implementation) As a consequence of this, we see that the image of the map $$\left( \begin{array}{cc} 2 & 2 \\ 3 & 5 \end{array} \right):\ZZ^2\to\ZZ^2$$ has index $4$ in its codomain, so that the subgroup $A+B$ is not a summand.


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.

  • $\begingroup$ Is there a similar result for $\mathbb{Z}^n$? That is a vector $(a_1, ... ,a_n) \in \mathbb{Z}^n$ spans a direct summand if and only if it is primitive. And also if there are $k < n$ linearly independent vectors when will they span a direct summand of $\mathbb{Z}^n$? - Thanks $\endgroup$ – R_D May 22 '15 at 9:15

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