current community

your communities

more stack exchange communities

Stack Exchange
tour help
Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 6 down vote favorite
2

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 $A\equiv M \pmod m$?

share|improve this question
    
As you can see from my answer, one can ask $M\in \mathcal{M}_n(\mathbb Z)$ with $\det M=1$. –  user26857 14 hours ago
    
Nice, I just accepted the answer. –  Nathan Portland 9 hours ago
add comment

1 Answer

up vote 3 down vote accepted

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 $k+xd_2\cdots d_n+m^{n-1}y=0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.