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Let $m,n\geqslant 2$, and $A\in \mathcal{M}_n(\mathbb Z)$ such that $\det A \equiv 1 \pmod m$.

Does it (necessarily) exist $M\in \mathrm{GL}_n(\mathbb Z)$ such that $A\equiv M \pmod m$?

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  • $\begingroup$ As you can see from my answer, one can ask $M\in \mathcal{M}_n(\mathbb Z)$ with $\det M=1$. $\endgroup$
    – user26857
    Jul 15, 2014 at 5:55

1 Answer 1

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By using the Smith Normal Form of $A$ one can assume that $A$ is diagonal with $d_1,\dots,d_n$ on the the main diagonal and $d_1\cdots d_n=1+mk$.
Then take

$$ M=\begin{pmatrix} d_1+mx & m & 0 & 0 &\dots & 0 & 0\\ 0 & d_2 & m & 0 &\dots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \dots & d_{n-1} & m\\ (-1)^{n+1}my & 0 & 0 & 0 & \dots & 0 & d_n \end{pmatrix}.$$ Clearly $$\det M=(d_1+mx)d_2\cdots d_n+m^ny=1+m(k+xd_2\cdots d_n+m^{n-1}y),$$ and since $\gcd(d_2\cdots d_n,m)=1$ one can find $x,y\in\mathbb Z$ such that $k+xd_2\cdots d_n+m^{n-1}y=0$.

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