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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

Agustí Roig
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WWright
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  • One of your fractions should have a denominator with an a in it; perhaps someone with higher karma than me can edit that. – Arturo Magidin Aug 30 '10 at 19:23
  • @Asaf: Sl_n are those of det 1 typically. It's a simple fact that an integer matrix $A$ is in $Gl_n(\mathbb{Z})$ iff $det A = \pm 1$, though. – Jason DeVito - on hiatus Aug 30 '10 at 19:36
  • @Jason: Take an integer $d\ne 1$ and take $dI_n$ for the scalar matrix of which all the diagonal values are $d$. The determinant is non-zero, therefore it is in $GL_n$, it is an integer matrix, and it is not of determinant $1$. – Asaf Karagila Aug 30 '10 at 23:30
  • @Asaf. It is in $GL_n(\mathbb{R})$, but it is not in $GL_n(\mathbb{Z})$. Said more explicitly, the matrix you are describing is invertible if real (or even rational) values are allowed, but if only integer values are allowed, it's not invertible. – Jason DeVito - on hiatus Aug 31 '10 at 00:02
  • Thanks for the comments, I fixed all the errors, pointed out. I'll try to get better at proof-reading – WWright Aug 31 '10 at 00:57
  • @Jason: Ah, I see your point. You are correct :) – Asaf Karagila Aug 31 '10 at 07:05
  • @Asaf Karagila: GL_n is not a ring. – Qiaochu Yuan Aug 31 '10 at 14:38

4 Answers4

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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.

Arturo Magidin
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    Your argument could be shortened by referring to the coefficients of the inverse matrix provided in the question itself: computing their determinant shows the determinant of the inverse matrix equals 1/(ad-bc) and we're done, because determinants of integral matrices are obviously integral. – whuber Aug 30 '10 at 20:42
  • I assume you are referring to the first part, the one that uses determinants? Yes; using determinants is the nice, elegant, easy, generalizable way of doing it. The second part, on the other hand, does not talk about determinants at all, just about how requiring the coefficients of the inverse (which can be found by solving a system of linear equations over $\mathbb{Q}$ without using determinants at all) to be integers forces the quantity ad-bc to be either 1 or -1. – Arturo Magidin Aug 31 '10 at 00:45
  • Sure your second part talks about determinants, Arturo: it's entirely about the expression D = ad - bc, which clearly is integral whenever a, b, c, d all are. A simple calculation establishes that the corresponding expression for the inverse (as explicitly given in the question) equals D/D^2 = 1/D, whence D must be a unit, QED. – whuber Aug 31 '10 at 14:27
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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 $-1$. – Arturo Magidin Aug 31 '10 at 14:54
  • Brilliant! I love the 2nd argument, treating just the fractions - fascinating how the introduction of new symbols that add no new information make the argument so clear. Where did you learn about this kind of approach? Also, what is meant by the notation “D|1” ? – Rax Adaam Dec 15 '20 at 20:34
  • @RaxAdaam: Notation is standard: $a|b$ means "$a$ divides $b$". – Arturo Magidin Dec 15 '20 at 20:35
  • Lol, sorry, I had never seen a vertical bar used for this, only “a/b” – Rax Adaam Dec 15 '20 at 20:36
  • @RaxAdaam: That usually means the fraction $\frac{a}{b}$, and not that $a$ divides $b$. – Arturo Magidin Dec 15 '20 at 20:47
  • Oh, I'm not sure what that means in this context. When I back-sub your expression, I get $D(a'd' - b'c') = D (\frac{ad-bc}{D^2}) = \frac{D^2}{D^2} = 1$, but that is obviously not what is meant... – Rax Adaam Dec 15 '20 at 20:50
  • @RaxAdaam: That back substitution is nonsense; it just undoes the original substitution. It's like taking $xy=1$, then using that to derive $y=\frac{1}{x}$, and then "back substituting" that into the original equation. All you are doing is spinning your wheels by doing something and then immediately undoing it. Why bother? Why would you do that? – Arturo Magidin Dec 15 '20 at 20:53
  • @RaxAdaam: What "I meant" is exactly what I wrote ten and a half years ago. First you praise it, but then you make statements that show you don't actually understand what you are praising. – Arturo Magidin Dec 15 '20 at 20:54
  • You seem very angry, I was just hoping to understand. Initially I thought I did, and then I realized that clearly I do not. I was hoping you might clarify it instead of put me down and get annoyed with me... I'd also point out that I never questioned what you meant, just that I didn't know what it meant. – Rax Adaam Dec 15 '20 at 20:55
  • @RaxAdaam: I'm not angry, nor am I "very angry". However, one of the things that do annoy me is people who don't know me telling me how I am feeling. So perhaps you might want to avoid that. I am somewhat curious to know why you are pinging me on a 10 year old question, and I am always curious why people lead off with praise for something they do not understand. If you "back substitute", you just undo the substitution, which is a silly thing to do. What's the point of substituting in the first place, if all you are going to do is then do a bunch of operations and undo all the work? (cont) – Arturo Magidin Dec 15 '20 at 20:58
  • @RaxAdaam: (cont). Again: $D$ must divide each of $a,b,c,d$. Writing out what that means, we end up with an equation that tells us that there is an integer $x$ (namely $a'd'-b'c'$) such that $Dx=1$. As both $D$ and $x$ are integers, this means that $D$ must divide $1$. The only integers that divide $1$ are $1$ and $-1$. What I am saying is that there is no hidden meaning or hidden operations in that paragraph. Everything is right there. – Arturo Magidin Dec 15 '20 at 20:59
  • Thanks for finally explaining. N.B. I said you seem very angry because your replies come across as distinctly annoyed, if not aggressive, but I'm not telling how you do feel. I also don't see any contradiction in appreciating an approach that one does not (yet) fully understand (the part that I appreciated is the same part I appreciate now that I follow the last step). Finally, the fact that back-subbing was clearly not the intended direction was the point of my comment (i.e. 'obviously this isn't what was meant, so clearly I'm not understanding'). – Rax Adaam Dec 15 '20 at 22:49
  • @RaxAdaam: I did not "finally explain". I just explained it again in exactly the same way that I did 10 years ago. And having already told me how I feel, you now hide behind your words, thus compounding the problem. Perhaps next time you can try not to tell people how you perceive them to be feeling, either. – Arturo Magidin Dec 15 '20 at 22:55
  • I'm sorry to have bothered you so much, it was not my intention to offend or annoy, I just wanted to understand and was thrown off by your initial response. Thank you for having clarified: I found your comment easier to follow, esp. thanks to the inclusion of $x$ and the emphasis that $a'd'-b'c'$ must be an integer. – Rax Adaam Dec 15 '20 at 23:08
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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 $-1$.

Mariano Suárez-Álvarez
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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.

Pete L. Clark
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Bill Dubuque
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The determinant of the inverse is the inverse of the determinant, whence the determinant must be an integral unit.

whuber
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