# Using a basis for a submodule $V$ in order to express $V$ as a direct sum of cyclic groups

The notation I am using is from N. Jacobson Algebra I. I am trying use his method in 3.6 of determining a basis for a sub-module in order to determine the cyclic decomposition of a finitely generated module.

I will go through my reasoning and solution for the problem I am looking for input on if I made any mistakes.

Problem statement: First find a basis for $\mathbb{Z}^{(3)}$ consisting of the set of all $(x,y,z)$ satisfying $2x+6z=0$ and $4x+8y=0$ and determine the structure of the sub-module as a sum of cyclic groups.

First we determine a basis for the submodule by forming the matrix $$\left( \begin{array}{ccc} 2 & 0 & 6\\ 4 & 8 & 0 \end{array} \right)$$ After some elementary row operations we get $$\left( \begin{array}{ccc} 1 & 2 & 0\\ 0 & 2 & -3 \end{array} \right)$$ which tells us our basis must satisfy $x = 2y = -3z$ and $2y + 3z =0$.

Question 1. Does this tell us $<1,2,0>$ and $<0,2,-3>$ form a basis for our sub-module?

Question 2. Since we can generate $x$ from $y$ and $z$ and since $2y + 3z =0$ does it follow that the cyclic subgroup decomposition of the sub-module generated by $x,y,z$ is $\mathbb{Z} \oplus \mathbb{Z_2} \oplus \mathbb{Z_3}$

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You may want to look up the Smith Normal Form, since it provides a mechanical way to solve this kind of problem.

Now, in fact, you've done something wrong; note that $\langle 1,2,0\rangle$ does not satisfy the condition given: you do not have $2x+6z=0$ (you get $2$), nor do you have $4x+8y = 0$ (you get $20$), and neither does $\langle 0,2,-3\rangle$.

Going back to the problem: you want all $\langle x,y,z\rangle$ such that $2x+6z = 4x+8y = 0$. This means that you need $x=-3z$ and $x=-2y$.

If we were working over $\mathbb{Q}$, this would mean that you want the vectors of the form $t\langle 1,-\frac{1}{2},-\frac{1}{3}\rangle$. This doesn't work over $\mathbb{Z}$, though; the only vectors in $\mathbb{Z}^3$ that satisfy the conditions are of the form $\langle 6x, -3x, -2x\rangle$, with $x$ arbitrary.

So a basis for the space is given by $\langle 6,-3,-2\rangle$, and that alone.

As for question $2$: if your original conclusion had been correct, then:

• Since your module is a subgruop of $\mathbb{Z}^3$, it cannot have nontrivial torsion; it must be isomorphic to a certain number of copies of $\mathbb{Z}$.

• Since you thought the module had a basis with two elements, then the decomposition would be $\mathbb{Z}\oplus\mathbb{Z}$.

In fact, since the module is free of rank $1$, it is isomorphic to $\mathbb{Z}$.

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