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For $A=\begin{pmatrix} a_1 & b_1 \\ c_1&d_1 \end{pmatrix}, B=\begin{pmatrix} a_2 & b_2 \\ c_2&d_2 \end{pmatrix}\in M_2(\mathbb Z)$, define

$A*B:=a_1L_1BR_1+b_1L_1BR_2+c_1L_2BR_1+d_1L_2BR_2$, where

$L_1=I_2=\begin{pmatrix} 1 & 0 \\ 0&1 \end{pmatrix}, L_2=\begin{pmatrix} 0 & 1 \\ 1&0 \end{pmatrix}, R_1=\begin{pmatrix} 0 & 1 \\ -1&-1 \end{pmatrix}, R_2=-I_2-R_1$.

Is there an elegant/clever way to see that $(M_2(\mathbb Z),+,*)$ forms a commutative ring with unity, where $+$ is the usual matrix addition, and $*$ acts as the multiplication?

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closed as off-topic by Steven Landsburg, Andreas Thom, YCor, Mike Miller, Ben Linowitz Oct 29 '18 at 0:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Steven Landsburg, Andreas Thom, YCor, Mike Miller, Ben Linowitz
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Where does this come from? $\endgroup$ – Igor Rivin Oct 27 '18 at 23:17
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    $\begingroup$ @IgorRivin: To tell you the truth, this is just what my professor mentioned ... I'm supposed to find a structure ... $\endgroup$ – user521337 Oct 27 '18 at 23:42
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    $\begingroup$ If your professor wants you to find a structure, shouldn't you be trying to do this rather than asking other people to do it? Or have I misunderstood your meaning? $\endgroup$ – Yemon Choi Oct 28 '18 at 0:34
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    $\begingroup$ So, the professor just disguised a standard law (retrieved by Matt's solution) using a change of coordinates, and made an exercise out of it (check associativity, etc). Solving this kind of exercises has nothing to do with research, and is not the purpose of this site. (It's also not the purpose of MathSE to solve such exercises, when no effort is made by the OP to show one's progress.) $\endgroup$ – YCor Oct 28 '18 at 12:40
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    $\begingroup$ @YCor, I agree that solving the professor’s exercise of verifying associativity, etc, is not a research problem. But the question was to exhibit an isomorphism to an easily intelligible structure, and I did not find that so obvious. $\endgroup$ – Matt F. Oct 28 '18 at 16:29
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There is an isomorphism $f:\mathbb{C\times C}\rightarrow (M_2(\mathbb{R}),+,*)$ given by

$$f((1,0))=\left(\begin{array}{cc} -1 & -1\\-1 & -1 \end{array}\right)\!/2,\ \ \ f((i,0))=\left(\begin{array}{cc} 1 & -1\\1 & -1 \end{array}\right)\!/2\sqrt{3}, $$

$$f((0,1))=\left(\begin{array}{cc} -1 & -1\\1 & 1 \end{array}\right)\!/2,\ \ \ f((0,i))=\left(\begin{array}{cc} 1 & -1\\-1 & 1 \end{array}\right)\!/2\sqrt{3} $$

which I found by finding the identity for *, then finding two idempotents which add up to the identity, then finding two elements whose squares are the negatives of those idempotents.

So there is an isomorphism $g:(M_2(\mathbb{Z}),+,*) \rightarrow \mathbb{Z[\omega]\times Z[\omega]}$ given by

$$g\left(\left(\begin{array}{cc} 1 & 0\\0 & 0 \end{array}\right)\right) = \left(\omega,\ \omega\right) ,\ \ \ g\left(\left(\begin{array}{cc} 0 & 1\\0 & 0 \end{array}\right)\right) = \left(\omega^2,\omega^2\right) ,\\ g\left(\left(\begin{array}{cc} 0 & 0\\1 & 0 \end{array}\right)\right) = \left(\omega, -\omega\right) ,\ \ \ g\left(\left(\begin{array}{cc} 0 & 0\\0 & 1 \end{array}\right)\right) = \left(\omega^2,-\omega^2\right) $$ Here $\omega=(-1+\sqrt{3i})/2$, and there are copies of $\mathbb{Z}$ in the image because $1=-(\omega+\omega^2)$.

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  • $\begingroup$ The multiplication and addition of $(\omega\mathbb Z +\omega^2\mathbb Z)\times (\omega\mathbb Z +\omega^2\mathbb Z)$ is the usual co-ordinate wise right ? So the operations on $M_2(\mathbb Z)$ is just the transport of that structure by the bijection $g$ right ? $\endgroup$ – user521337 Oct 28 '18 at 17:12
  • $\begingroup$ Yes, the bijection is with (the subset of) CxC with coordinatewise addition and multiplication. $\endgroup$ – Matt F. Oct 28 '18 at 19:03

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