Proof. Let $a, b \in R$ be nonzero. Consider the $1 \times 2$ matrix $A = (a\ b)$. Then we see that $u(a\ b)V = (f\ 0)$ with $u \in R$ invertible and $V = (g_{ij})$ an invertible $2 \times 2$ matrix. Then $f = u a g_{11} + u b g_{2 1}$ and $(g_{11}, g_{2 1}) = R$. It follows that $(a, b) = (f)$. An induction argument (omitted) then shows any finitely generated ideal in $R$ is generated by one element. $\square$

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