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Let $R$ be a Euclidean domain. Let $a,b,c, d \in R$ with $ad - bc = 1_R = 1$. That is, the matrix:

$$ A = \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) \in \operatorname{Mat}_{2}(R)$$

has determinant one.

Using elementary row and column operations, show that $A$ is equivalent to the identity matrix $I_2 \in \operatorname{Mat}_{2}(R).$

The elementary row/column operations are:

  1. Interchanging two rows or columns,
  2. Adding a multiple of one row or column to another,
  3. Multiplying any row or column by a unit.

This does not seem too difficult at its core, and I have worked about several examples by setting $R = \mathbb{Z}$. However, I am having a difficult time crafting the argument for completely general $R$.

I am familiar with the concept of (and proof of) Smith normal form for matrices over a principal ideal domain. However, what I need to show here is more specific, since the particular diagonalization I seek is the identity matrix.

I do know that the equation $ad - bc = 1$ implies that $\gcd(a,b) = \gcd(c, d) = 1$. And, in an arbitrary Eulcidean domain, we do have the Euclidean algorithm available to us. So, my gut feeling is that this should be the strategy to make the proof work.

For instance, we invoke the Euclidean algorithm to get a sequence of equations:

$$a = q_0b + r_1,$$ $$b = q_1r_1 + r_2,$$ $$r_1 = q_2r_2 + r_3,$$ $$\vdots$$ $$r_{n-1} = q_n \gcd(a,b) + 0.$$

And we can use this sequence to start performing operations on $A$:

$$ A = \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) =\left( \begin{array}{cc} q_0b + r_1 & b \\ c & d \\ \end{array} \right) \sim \left( \begin{array}{cc} r_1 & b \\ c - dq_0 & d \\ \end{array} \right) = \left( \begin{array}{cc} r_1 & q_1r_1 + r_2 \\ c - dq_0 & d \\ \end{array} \right) \sim \left( \begin{array}{cc} r_1 & r_2 \\ c - dq_0 & d - cq_1 + dq_0q_1 \\ \end{array} \right). $$

So we have in a sense "simplified" the first row of $A$, and we could in practice continue this until the first row is of the form:

$$\left( \begin{array}{cc} \gcd(a,b) & 0 \\ \ast & \ast \\ \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ \ast & \ast \\ \end{array} \right). $$

Am I on the right track here? I am worried because the terms in the second row could become quite complicated.

Hints or solutions are greatly appreciated.

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up vote 1 down vote accepted

Hint $\ $ Using row/col operations a la Euclid's algorithm you can reduce to a matrix where $\,a\mid b,c.\,$ Therefore $\,a\mid ad-bc = 1.\,$ Hence wlog $\,a = 1,\,$ so, we can reduce $\,b,c\,$ to $\,0,\,$ so $\,d = 1.$

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