1
$\begingroup$

Consider the quotient $\mathbb{Z}-\text{module}$

$$M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$$

Prove that $M$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$.

I think the first step is to find a basis of $M$. Clearly $(1,0,0)$ and $(0,0,1)$ are in a basis of $M$. I don't know how to find the third element in that basis.

Also the basis in $\mathbb{Z} \oplus \mathbb{Z}_4$ are $(1,0)$ and $(0,1)$.

  • 4
    Look up *Smith normal form* from your textbook to get the idea behind the algorithm used by the answerers. – 2017-02-19
  • 0
    I found a question here for SNF and finitely generated modules over PID. http://math.stackexchange.com/questions/721523/smith-normal-form-and-classification-of-factor-groups-according-to-the-theorem-o – 2017-02-19

2 Answers 2

1

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Z}{\mathbb{Z}}$You should apply the standard algorithm to the matrix $$ \begin{bmatrix} 3 & 3 & 1\\ 2 & 2 & 2\\ 0 & 0 & 0\\ \end{bmatrix}. $$ First exchange column 1 and 2 to get $$ \begin{bmatrix} 1 & 3 & 3\\ 2 & 2 & 2\\ 0 & 0 & 0\\ \end{bmatrix} $$ then subtract from row 2 twice row 1 to get $$ \begin{bmatrix} 1 & 3 & 3\\ 0 & -4 & -4\\ 0 & 0 & 0\\ \end{bmatrix} $$ then subtract from column 2 thrice column 1 and from column 3 thrice column 1 to get $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & -4 & -4\\ 0 & 0 & 0\\ \end{bmatrix} $$ and finally subtract from column 3 column 2, and multiply row 2 by $-1$ to get $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $$ So the quotient is isomorphic to $$ \Z / \Span{1} \oplus \Z / \Span{4} \oplus \Z / \Span{0} \cong \Z / \Span{4} \oplus \Z . $$

  • 0
    What's the name for this algorithm? I want to see more on this algorithm. – 2017-02-19
  • 0
    It's essentially Gaussian elimination, but performed over a Euclidean domain, or a PID. Look up "finitely generated modules over a PID", for instance in Jacobson, Basic Algebra I. – 2017-02-19
  • 0
    So by "finitely generated modules", you mean $M$ is free here? I can't find a basis for $M$ however. – 2017-02-19
  • 0
    The same algorithm yields a method to calculate the minimal polynomial for a matrix, for instance. – 2017-02-19
  • 0
    No, $M$ is not free. But it is generated by at most three elements, actually two in this case. – 2017-02-19
  • 0
    $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Z}{\mathbb{Z}}$You are thinking of vector spaces. But abelian groups, i.e. $\Z$-modules, need not have bases. For instance if $G = \Set{0, g}$ has order $2$, then no element is linearly independent, as $2 \cdot g = 0$. – 2017-02-19
  • 2
    All of this is hard to learn from comments only. Read a chapter in a good book like (IMHO) Jacobson's, and you'll have a clearer picture. – 2017-02-19
  • 0
    I see that if $M$ is free, then it cannot have a basis of one or three elements. Thus it can only have a basis of two elements. I am trying to figure out how to prove that $M$ cannot have a basis of two elements. – 2017-02-19
2

The standard technique is to apply ($\mathbb{Z}$-invertible) elementary row and column operations to the matrix $$\left[\begin{array}{ccc}3&3&1\\2&2&2\end{array}\right].$$ You will get $$\left[\begin{array}{ccc}1&0&0\\0&4&0\end{array}\right],$$ which means there is a basis of $\mathbb{Z}^3$ such that the module you are factoring out is generated by $(1,0,0)$ and $(0,4,0)$; hence the quotient is $$\{0\}\oplus\mathbb{Z}_4\oplus\mathbb{Z}.$$

  • 0
    Do you mean that $\langle (3,3,1),(2,2,2) \rangle$ is generated by $(1,0,0)$ and $(0,4,0)$? What is the reasoning behind this technique? – 2017-02-19
  • 0
    No. See the comments on previous detailed answer. Row operations don't change the module spanned by the rows, column operations merely change the basis of the free $\mathbb{Z}^3$. – 2017-02-19
  • 0
    Can you slightly explain why $M$ is not free? I see that a basis of $M$, if exists, cannot have more than two elements. – 2017-02-19
  • 0
    Doesn't proving it is isomorphic to $\mathbb{Z}_4\oplus\mathbb{Z}$ ensure that? – 2017-02-20