my problem here is:
Find all units in the ring $M_2(\mathbb{Z})$
I've done it before but now it's a matrix and I'm lost. First I think on the determinant because I saw some other wrote about it, but I'm not sure. Any help?
Thank you
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$\begingroup$ Determinants was a good way to go. How can you tell from a determinant whether the matric is invertible? $\endgroup$– ArthurCommented Jan 16, 2018 at 13:23
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$\begingroup$ If a matrix is invertible, then $det(M)\neq 0$ if we let $M$ be the matrix. Am I right? And we also know that we can get identity matrix when $M$ and $M^{-1}$ is "multiplied" together. I hope it makes sense. $\endgroup$– Amir HassanCommented Jan 16, 2018 at 13:26
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$\begingroup$ Yes, but $\det(M)\det(M^{-1})=\det{I}=1$ and $\det(M), \det(M^{-1})$ are integers. What can they be? $\endgroup$– user491874Commented Jan 16, 2018 at 13:29
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$\begingroup$ I believe that it can be $\pm 1$ $\endgroup$– Amir HassanCommented Jan 16, 2018 at 13:29
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$\begingroup$ Exactly! This is why this is true: $M$ is invertible in $M_2(\mathbb Z)$ if and only if $\det(M)=\pm 1$. Try to prove it. $\endgroup$– user491874Commented Jan 16, 2018 at 13:30
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1 Answer
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More generally, for any commutative ring $R$, an $n\times n$ matrix $M$ lies in $\mathbf{GL}(n,R)$ if and only if $\det M\in R^{\times}$ (the units of $R$).
The units of $\mathbf Z$ are indeed only $1$ and $-1)$, simply because if $|u|>1$, you cannot find another integer $v$ such that $uv=1$ since $|uv|=|u||v|>|v|$.
