An abelian group with a basis of $n$ elements is called a free abelian group of rank $n$. If $G$ is free abelian of rank $n$ and $\{x_1, ... ,x_n\},\{y_1, ... ,y_n\}$ are both bases, then there exist integers $a_{ij}, b_{ij}$, such that $$y_i=\sum_ja_{ij}x_j,\; x_i=\sum_jb_{ij}y_j,$$

If we consider the Appropriate Matrices, $A=(a_{ij}), \; \;B=(b_{ij})$,it follows that $AB = I_n$, the identity matrix.

Now consider the next lemma.

Every subgroup $H$ of a free abelian group $G$ of rank $n$ is free of rank $s \leq n$. Moreover there exists a basis $u_1, ... , u_n$ for $G$ and positive integers $\alpha_1, ... ,\alpha_s$ such that $\alpha_1 u_1, ... , \alpha_s u_s$ is a basis for $H$.

This is Theorem 1.16. in the book Algebraic-Number Theory by Ian Stewart and David Tall, on page 29. The first part of the proof is -

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In the proof it is written that -

$u_1, w_2, ... , w_n$ is another basis for $G$. (The appropriate matrix is clearly unimodular.)


If $G$ is rank $n$, pick any basis $w_1, ... , w_n$ of $G$ for $h \in G$ is of the form, $h=h_1w_1+ \cdots h_nw_n$, then what is the appropriate matrix of basis $w_1, ... , w_n$ and what is the appropriate matrix of $u_1, w_2, ... , w_n$ obtained from for $u_1=w_1+ q_2w_2 \cdots +q_nw_n$ ?


Let $G$ be a free abelian group of rank $n$ with basis ${x_1, ... ,x_n}$. Suppose $(a_{ij})$ is an $n \times n$ matrix with integer entries. Then the elements $y_i= \sum_ja_{ij}x_j$ form a basis of $G$ if and only if $(a_{ij})$ is unimodular (Lemma 1.15. on page 28), so the vector of basis, $(y_i)=\begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \\ \vdots \\ y_{n} \end{bmatrix} = \begin{bmatrix} a_{11}&a_{12}&\cdots && a_{1n}\\ a_{21}&a_{22}&&&\vdots\\ a_{31} & a_{32} & &a_{3(n-1)}&a_{3n}\\ \vdots & \vdots& & & a_{(n-1)n}\\ a_{n1} & a_{n2} & \cdots &a_{n(n-1)}& a_{nn} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{n} \end{bmatrix}$

If $u_1, w_2, ... , w_n$ is another basis for $G$, then the vector of basis, say $w_i$ is,

$$\begin{bmatrix} u_{1} \\ w_{2} \\ w_{3} \\ \vdots \\ w_{n} \end{bmatrix}$$

We want $a_{ij}$ for which $w_i = \sum a_{ij}x_j$.

Comparing with $y_i= \sum_ja_{ij}x_j$, we have $y_i = w_i$ for all $i$ and $x_1 = u_1, x_i = w_i$ for $i \geq 2$.

From $u_1=w_1+q_2 w_2+ \cdots q_n w_n$, if the associated matrix we get is -

$$(c_{ij})=\begin{bmatrix}1&0&\cdots && 0\\ q_2 &1&\ddots&&\vdots\\ q_3 & 0 & \ddots&0&0\\ \vdots & \vdots& \ddots & \ddots & 0\\ q_n & 0 & \cdots &0& 1 \end{bmatrix},$$

The determinant of a lower triangular matrix is the product of the diagonal entries, in this case this is $1$, thus the matrix is unimodular, by the definition of unimodular (see page 28). To verify $(c_{ij})$ is the appropriate matrix we do the following calculation:

$$(c_{ij})(x_j)=\begin{bmatrix}1&0&\cdots && 0\\ q_2 &1&\ddots&&\vdots\\ q_3 & 0 & \ddots&0&0\\ \vdots & \vdots& \ddots & \ddots & 0\\ q_n & 0 & \cdots &0& 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{n} \end{bmatrix} = \begin{bmatrix} 1 \cdot x_{1} \\ q_2\cdot x_{1}+1 \cdot x_{2} \\ q_3\cdot x_{1}+1 \cdot x_{3} \\ \vdots \\ q_n\cdot x_{1}+1 \cdot x_{n} \end{bmatrix} =\begin{bmatrix} u_{1} \\ w_{2} \\ w_{3} \\ \vdots \\ w_{n} \end{bmatrix} $$

$\therefore u_1=w_1+q_2 w_2+ \cdots q_n w_n \implies x_1 = x_1(q_2+q_3+ \cdots q_n) + x_2+x_3 \cdots x_n$

$\implies 0 = x_1(-1+q_2+q_3+ \cdots q_n) + x_2+x_3 \cdots x_n$.

If $(-1+q_2+q_3+ \cdots q_n)=0 $ then the set $\{ x_2, x_3 \cdots x_n \}$ is not a basis by the definition of basis since $0 = x_2+x_3 \cdots x_n$ means one of the element of $\{ x_2, x_3 \cdots x_n \}$ is depended on others. Thus the super-set $\{ x_1, x_2, x_3 \cdots x_n \}$ can not be basis either.

If $(-1+q_2+q_3+ \cdots q_n)\neq 0 $ then the set $\{ x_1, x_2, x_3 \cdots x_n \}$ is not a basis by the definition of basis since $0 = x_1+x_2+x_3 \cdots x_n$ means one of the element of $\{x_1, x_2, x_3 \cdots x_n \}$ is depended on others.

Thus, $(c_{ij})$ can not be the appropriate matrix because then $\{ x_1, x_2, x_3 \cdots x_n \}$ is not a basis which is against the hypothesis.


1 Answer 1


The basis $w_1,w_2,\dots,w_n$ is changed to a basis $u_1,w_2,\dots,w_n$ where $$ u_1 = w_1 + q_2 w_2 + \dots + q_n w_n. $$ This might be written as $$ \begin{bmatrix} u_1 \\ w_2 \\ \vdots \\ w_n\end{bmatrix} = \begin{bmatrix} 1 & q_2 & \dots & q_n \\ & 1 \\ & & \ddots \\ & & & 1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ \vdots \\ w_n\end{bmatrix}. $$ The other way around is obtained by inverting this matrix or noting that $$ w_1 = u_1 - q_2 w_2 - \dots - q_n w_n. $$ We obtain $$ \begin{bmatrix} w_1 \\ w_2 \\ \vdots \\ w_n\end{bmatrix} = \begin{bmatrix} 1 & -q_2 & \dots & -q_n \\ & 1 \\ & & \ddots \\ & & & 1 \end{bmatrix} \begin{bmatrix} u_1 \\ w_2 \\ \vdots \\ w_n\end{bmatrix}. $$ These two are the appropriate unimodular matrices for the basis changes between $w_1,w_2,\dots,w_n$ and $u_1,w_2,\dots,w_n$.

It seems you mixed up rows and columns when setting up the matrix.

  • $\begingroup$ I have been told, see this answer: math.stackexchange.com/a/3805165/613063 did he make a mistake? $\endgroup$ Oct 5, 2020 at 18:30
  • $\begingroup$ Yes, he mixed up rows and columns. Do you see anything wrong with my answer? Why are you posting a new question with a bounty when you already made a post including this question before? $\endgroup$
    – Christoph
    Oct 5, 2020 at 18:41
  • $\begingroup$ That was an old and resolved question (slightly different than this, this question is more specific and based on that post), when I found the inconsistency the member stopped answering, what else can I do! $\endgroup$ Oct 5, 2020 at 18:48
  • $\begingroup$ So, is your question answered now? $\endgroup$
    – Christoph
    Oct 5, 2020 at 19:03
  • $\begingroup$ And then you also posted yet another question on the same topic here math.stackexchange.com/questions/3851572/… — Please consider posting a single question only once and when multiple questions are related, add links to the other questions so that people don't give answers that are already given on other questions. $\endgroup$
    – Christoph
    Oct 5, 2020 at 19:05

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