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In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1. Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?

I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated

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    @Asaf Karagila:$GL_n$is not a ring.2010-08-31

4 Answers 4

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As others have pointed out, the key here is that the determinant of $M^{-1}$ has to be equal to $\frac{1}{det(M)}$ on the one hand, by the properties of the determinant, and also that since $M$ and $M^{-1}$ both have integer coefficients then their determinants must be integers. So you need an integer $e=det(M)$ such that both $e$ and $\frac{1}{e}$ are integers, and the only possibility is $e=\pm 1$.

If you don't consider arguments using the determinant to be "elementary", perhaps the following will do: note that $ad-bc$ must divide each of $a$, $b$, $c$, and $d$. Let $D=ad-bc$; we can then write $a=Da'$, $b=Db'$, $c=Dc'$, $d=Dd'$; then $D = ad-bc=D^2(a'd'-b'c')$, so $D=D^2(a'd'-b'c')$. Therefore, since $D\neq 0$, we must have $D(a'd'-b'c')=1$, so $D|1$, hence $D=ad-bc=\pm 1$, which is what you wanted to show.

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    Fair enough; I was trying to address the part of the question where WWright is asking why can we not have all the given fractions integers with the denominator not equal to $1$ and $-1$. Even if you did not know they came from matrices or inverse matrices, or knew nothing whatsoever about matrices, you can still deduce that the denominator must be either $1$ or $-1$.2010-08-31
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The map $\det:M_2(\mathbb Z)\to\mathbb Z$ is multiplicative, and the determinant of the identity matrix is $1$. It follows from this that the determinant of any inversible element of $M_2(\mathbb Z)$ must be an inversible element of $\mathbb Z$. There are only two such elements in $\mathbb Z$, v.g. $1$ and $-1$.

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The determinant of the inverse is the inverse of the determinant, whence the determinant must be an integral unit.

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HINT $\;\;$ Multiplicative maps preserve units: $\rm\; MN = 1 \;\;\Rightarrow\;\; d(M)\:d(N) = 1$

NOTE $\rm\;\; d(1) = 1\;$ via apply $\rm d$ to $1\cdot 1 = 1\:$ then cancel $\rm d(1)\ne 0,$ valid since $\mathbb Z$ has cancellation.