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Let $V$ be a free $\mathbb{Z}$-module (or a free abelian group) with basis $(v_1$, $v_2$, $v_3)$. Denote by $x$ an arbitrary nonzero element of $V$, say $x=mv_1+nv_2+kv_3 \in V \setminus \{0\}$; here $m,n,k \in \mathbb{Z}$.

It is known that the element $x$ can be extended to a basis iff $\gcd(m,n,k)=1$. I am trying to understand this statement (which of course can be generalized for free modules of an arbitrary finite rank), but first I would like to know how does it work in practice -- given particular integers $m,n,k,$ not all zero, how do I extend the element $x$ to a basis? Further, given a candidate for a basis, how do I verify that it is indeed a basis? Could you give a particular example? If possible, I would like to see matrix interpretation of this problem -- probably this would enable me to get one step closer to this question.

The proof of the above statement (or its generalized version) would be also appreciated, albeit at the moment I am more interested in an example.

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user557
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1 Answers1

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You want to find two vectors which, together with $(m,n,k)$ make a matrix with determinant $\pm 1$. Then the three vectors will form a free basis of ${\mathbb Z}^3$.

One way of doing that is to perform some elementary unimodular column operations on $(m,n,k)$ to transform it to $(1,0,0)$. This you can easily extend to a free basis of ${\mathbb Z}^3$. Then, reverse your column operations on the resulting $3 \times 3$ matrix, to bring the first row back to $(m,n,k)$. The determinant of the new matrix will still be$\pm 1$, so its rows will give you the required free basis.

For example, let $(m,n,k) = (4,6,15)$. The column operations $c_2 \to c_2-c_1$, $c_1\to c_1-2c_2$, $c_3\to c_3-7c_2$, $c_2 \to c_2-2c_3$ in that order, transfors this vector to $(0,0,1)$, which we can extend to the matrix $$\left(\begin{array}{rrr}0&0&1\\0&1&0\\1&0&0\end{array}\right)$$ with determinant $-1$.

Now reversing the column operations on this matrix, we do $c_2\to c_2+2c_3$, $c_3\to c_3+7c_2$, $c_1 \to c_1+2c_2$, $c_2 \to c_2+c_1$, which transforms it to $$\left(\begin{array}{rrr}4&6&15\\2&3&7\\1&1&0\end{array}\right)$$ which still has determinant $-1$, and its rows form a free basis of ${\mathbb Z}^3$.

In practice you might be able to shorten the calculation by extending the first vector to a free basis as soon as you can do this easily. In the exampe above you might have stopped one step earlier and extended $(0,2,1)$ to a free basis.

Derek Holt
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  • Thanks for the answer! A few questions if you don't mind:
    1. Why determining the remaining two elements of a basis is the same as finding two row vectors which, together with the row vector corresponding to the given element, make a matrix with determinant $\pm1$?
    2. What do you mean by a "free basis"?
     – user557 Apr 24 '16 at 12:00
  • By a free basis I just mean a basis. It is a general result about subgroups of finite index in ${\mathbb Z}^n$ generated by $n$ vectors, that the index $k$ is the absolute value of the determinant of the matrix. To see that, note that elementary unimodular row and column operations do not change the absolute value of the determinant, and they transform the matrix to a diagonal matrix (in Smith Normal Form). For a diagonal matrix,the claimed result is clear. – Derek Holt Apr 24 '16 at 12:21
  • Ok, that is clear. Also, it seems that you use implicitly that a free abelian group of rank $n$ is isomorphic to $\mathbb{Z}^n$, but I can't organize my thoughts properly to get in which way. Could you explain it in detail? In particular, column and row operations are supposed to correspond to changes in bases. So to base changes in which group they correspond? – user557 Apr 25 '16 at 12:55
  • Sorry, but I don't think I have time to give more detail, since this is all basic theory of free abelian groups (I have always used ${\mathbb Z}^n$ as the definition of a free abelian group.) Yes, column operations correspond to basis change. For example, if $(x,y,z)$ is a vector w.r.t. basis $(b_1,b_2,b_3)$, then $(x+2y+3z,y,z)$ is the same vector w.r.t. basis $(b_1,b_2-2b_1,b_3-3b_1)$. So that provides an alternative explanation as to why the method works, which does not involve determinants. – Derek Holt Apr 25 '16 at 14:35