# Direct summands of $\mathbb Z \oplus \mathbb Z$

Consider the $\mathbb Z$-module $\mathbb Z \oplus \mathbb Z$.

I have seen two assertions about the direct summands of $\mathbb Z \oplus \mathbb Z$ but I have trouble with these:

1. $(a,b)\in\mathbb Z \oplus \mathbb Z$ spans a direct summand if and only if it is primitive, that is, $(a,b)=1$.

2. Two linearly independent vectors $(a,b)$ and $(c,d)$ span a direct summand of $\mathbb Z \oplus \mathbb Z$ if and only if the determinant of the matrix $\left( \begin{array}{cc} a & c\\b & d \end{array}\right)$ is $±1$.

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what is the meaning of primitive here? –  Babak Miraftab Sep 30 '12 at 17:40
I would guess primitive to mean gcd(a,b)=1 without knowing more about the problem. –  Chris Leary Sep 30 '12 at 17:49

The following steps lead to a solution for (2):

1. The condition that $(a,b)$ and $(c,d)$ be a basis for $\Bbb{Z} \oplus \Bbb{Z}$ is that the the linear system $$\left(\begin{array}{cc} a& b \\ c & d \end{array}\right)\left(\begin{array}{c}x \\ y \end{array}\right) = \left(\begin{array}{c}e \\ f \end{array}\right)$$ has a unique solution for every $(e,f) \in \Bbb{Z} \oplus \Bbb{Z}$.

2. This happens if and only if the matrix $A = \left( \begin{array}{cc} a & b \\ c & d \end{array}\right)$ is invertible over $\Bbb{Z}$.

3. This happens if and only if $\det A = ad-bc$ is non-zero and divides all of $a,b,c$ and $d$.

4. Conclude.

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The first part isn't really right -- being $\mathbb{Z}$-linearly independent doesn't imply that they actually generate all of $\mathbb{Z}\oplus\mathbb{Z}$. (E.g. (2,0) and (0,2).) The second part is the right condition but isn't equivalent to the first part. –  Harry Altman Dec 3 '12 at 18:53
@HarryAltman I have corrected my answer. 1) was wrong previously. –  user38268 Dec 4 '12 at 22:25

The matrix $A=\left(\begin{matrix}a&c\\b&d\end{matrix}\right)$ has the property that it transfers the coordinates relative to $(a,b)$ and $(c,d)$ to standard coordinates: $x\cdot (a,b)+y\cdot(c,d) = u\cdot(1,0)+v\cdot (0,1)$ with $\left(\begin{matrix}u\\v\end{matrix}\right)=A\left(\begin{matrix}x\\y\end{matrix}\right)$. There must also be a matrix $B$ in theopposite direction, i.e. such that $\left(\begin{matrix}x\\y\end{matrix}\right)=B\left(\begin{matrix}u\\v\end{matrix}\right)$. What does that tell about $AB$? What does that imply for $\det(A)$ and $\det(B)$?

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