Let $K$ be a principal ideal ring. How to prove that for any $ x= (x_1, x_2)^t \in K^2 $ there exists a matrix $G \in SL_2(K)$ such that $Gx = (\gcd(x_1, x_2),0)^t $ ?
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$\begingroup$ Do you mean greatest common divisor ? Its not quite clear. $\endgroup$– Rene SchipperusJun 23, 2014 at 12:22
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1$\begingroup$ @ReneSchipperus Yes. $\endgroup$– Max MalyshJun 23, 2014 at 12:25
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1$\begingroup$ Remark that the matrix appears in the bottom right corner of the table constructed by the Extended Euclidean algorithm - see here. $\endgroup$– Bill DubuqueJun 23, 2014 at 14:19
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If it's the greatest common divisor $d$ of $x_{1}$ and $x_{2}$ you mean, and you assume that $x_{1}$ and $x_{2}$ are not both zero, so that $d \ne 0$, then there are $a, b \in K$ such that $a x_{1} + b x_{2} = d$, and now take $$ G = \begin{bmatrix} a & b\\ -x_{2}/d & x_{1}/d \end{bmatrix} $$
answered Jun 23, 2014 at 12:26
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1$\begingroup$ Of course if $x_{1} = x_{2} = 0$, any $G$ will do. $\endgroup$ Jun 23, 2014 at 12:56