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Suppose $A$ is a $m\times n$ ($n\geq m$) matrix on the ring $\mathbb Z$ of integers and the greatest common divisor of its $m\times m$ minor determinants is $1$. Prove that there is a $n\times m$ matrix $C$ on $\mathbb Z$ such that $AC=I$, where $I$ is the unit $m\times m$ matrix.

Thanks.

Stefan Hansen
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Li Xinghe
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1 Answers1

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This question is relevant.

Since $f:\mathbb Z^n \longrightarrow \mathbb Z^m, \ v\mapsto Av$ is surjective there are $m$ vectors $c_i$ in $\mathbb Z^n$ s.t. $f(c_i)=e_i=(0,0,\ldots,0,1,0,\ldots,0)^t , \ \ \forall i=1,2,\ldots,m$.

Take as $C$ the matrix with columns $c_i$.

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