# Finding a necessary and sufficient condition for a free abelian group of rank $1$

Let $$V=\left\{\begin{pmatrix}\lambda\\ \mu\\ \nu\end{pmatrix}\colon \lambda,\mu,\nu\in\mathbb Z\right\}$$, so $$V$$ is a free abelian gorup of rank $$3$$.

Suppose that $$a=\begin{pmatrix}\alpha_1\\\alpha_2\\\alpha_3\end{pmatrix}$$ and $$b=\begin{pmatrix}\beta_1\\\beta_2\\\beta_3\end{pmatrix}$$ are in $$V$$. Let $$U$$ be the subgroup of $$V$$ generated by $$a$$ and $$b$$. Assume that $$U$$ has rank $$2$$.

State and prove a necessary and sufficient condition involving the three determinants $$\begin{vmatrix}\alpha_1&\beta_1\\\alpha_2&\beta_2\end{vmatrix}$$, $$\begin{vmatrix}\alpha_1&\beta_1\\\alpha_3&\beta_3\end{vmatrix}$$, and $$\begin{vmatrix}\alpha_2&\beta_2\\\alpha_3&\beta_3\end{vmatrix}$$, insuring that $$V/U$$ is free abelian of rank $$1$$.

My attempt:

If $$V/U$$ is a free abelian of rank $$1$$, then there exist two invertible matrices $$A\in M_3(\mathbb Z)$$ and $$B\in M_2(\mathbb Z)$$ such that $$A[a\ b]B=\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}$$. But how am I supposed to find a necessary and sufficient condition involving the three determinants?

Can someone give me a hint? Thank you.

Addendum: The minors may not be preserved under elementary transformations.

• Where is this from? – fourier1234 Aug 7 '19 at 14:35

Here is an outline for one possible way to think about this.

Also I notice that I have answered the transpose of your question. Just take the transpose of all the matrices I have written down. The ideas are still the same.

As you have realised yourself, the group theory part of the question is just realising that you need the Smith Normal Form to be $$\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}.$$

It seems that you are happy with this and this is a standard exercise. If you are not happy with this, see the excellent answer here by Derek Holt.

The tricky part, is the following.

Find a necessary and sufficient condition involving the minor determinants $$\begin{vmatrix} a_1&a_2 \\ b_1&b_2 \end{vmatrix}, \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}, \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}$$ for the matrix $$T =\begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\end{bmatrix}$$ to have Smith Normal Form as above.

We will need the following Theorems about existence and uniqueness of the Smith Normal Form. These Theorems are 5.8 and 5.11 in the typed notes here.

Theorem Existence of Smith Normal Form

Let $$R$$ be an Euclidean domain. Then every $$A \in M_{m \times n}(R)$$ is equivalent to a diagonal matrix of the form

$$\begin{bmatrix}f_1 & \\ & f_2 \\ & & \ddots \\ & & & f_{r} \\ & & & & 0 \\ & & & & & \ddots \end{bmatrix}$$

where $$f_{1} \mid f_{2} \mid \dots \mid f_{r-1} \mid f_{r}$$ .

And we need

Theorem Uniqueness of Smith Normal Form

Let $$R$$ be a Euclidean domain and let $$A \in M_{ m \times n} (R)$$ and let $$S = \begin{bmatrix}f_1 & \\ & f_2 \\ & & \ddots \\ & & & f_{r} \\ & & & & 0 \\ & & & & & \ddots \end{bmatrix}$$ be a Smith normal form of $$A$$. Then the gcd of $$k \times k$$ sub-determinants of $$A = \begin{cases} f_{1}f_{2} \dots f_{k}, \; 1\leq k \leq r \\ 0, r Therefore the elements $$f_{1}, f_{2}, \dots, f_{r}$$ are unique up to multiplication by units.

We will prove the following

Claim The matrix $$T$$ has Smith normal form $$\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}$$ if and only if the greatest common divisor of the three minor determinants $$\begin{vmatrix} a_1&a_2 \\ b_1&b_2 \end{vmatrix}, \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}, \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}$$ is $$1$$.

Proof. First suppose that the matrix $$T$$ has Smith normal form $$\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}$$. Then by the Uniqueness Theorem above, we know that the greatest common divisor of $$a_1, a_2, a_3, b_1, b_2, b_3$$ is $$1$$ (since the first element on the diagonal is $$f_{1} = 1$$). Then since the second element on the diagonal is $$f_{2}=1$$, we know that the greatest common divisor of the three $$2 \times 2$$ Determinants $$\begin{vmatrix} a_1&a_2 \\ b_1&b_2 \end{vmatrix}, \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}, \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}$$ is equal to $$f_{1} f_{2} = 1$$.

Conversely suppose that the gcd of the determinants of the $$2 \times 2$$ Submatrices is $$1$$. Notice that this implies that the gcd $$g$$ of $$(a_1, a_2, a_3, b_1, b_2,b_3)=1$$ since this number divides the determinants of all the $$2 \times 2$$ minors. By the Existence Theorem of the Smith Normal Form, we know that $$T$$ has Smith Normal Form

$$\begin{bmatrix} f_1 & 0 & 0 \\ 0 & f_2 & 0 \end{bmatrix}.$$ By the Uniqueness of Smith Normal form and the observations about $$gcd$$ above, we see that $$f_{1}=1$$ and $$f_{2}=1$$.

• Are you sure that "the minor determinants have their values preserved" claimed in the paragraph below $$N=\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}.$$ What about adding the last column to the second column and consider the minor formed by the first column and the second? – Bach Aug 10 '19 at 1:27
• Sorry @Bach you are absolutely right. My original solution was wrong. I hope I have fixed things now and I have given references for my claims. – fourier1234 Aug 11 '19 at 17:47