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I would want to make an example of a matrix $M \in GL_3(\mathbb{Z}_7)$ such that $\langle M; +, \cdot \rangle \simeq GF(7^3)$ and the multiplicative order of $M$ is $3$.

Any hints how to do that with a computer algebra system would be appreciated.

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    I seem to have left out that what always holds is $\langle M,+,\cdot\rangle\simeq GF(p)[x]/\langle \chi_M(x)\rangle,$ where $\chi_M(x)$ is the characteristic polynomial of $M$.2011-11-03

2 Answers 2

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While you figure out what the question is supposed to be, here is some advice on using GAP to find an answer:

The naive way of looking for such a matrix is just to walk through the elements of GL(3,7):

First( GL(3,7), m -> Order(m) = 3 and Size(Algebra(GF(7),[m])) = 7^3 ); 

However, this runs into a performance problem as GL(3,7) is quite large:

gap> Size(GL(3,7)); 33784128 

Instead, we use the fact that GL(3,7) has relatively few and relatively well understood conjugacy classes:

gap> ccreps := List( ConjugacyClasses( GL( 3, 7 ) ), Representative );; gap> Size( ccreps ); 336 

Now we can examine those 336 conjugacy class representatives to see if any satisfy your question:

gap> ms := Filtered( ccreps, m -> Order(m) = 3 and > Size(Algebra(GF(7),[m])) = 7^3 );; Size( ms ); 1 gap> Perform( ms, Display );  4 . .  . 1 .  . . 2 

So we see there is only one matrix of order 3 in GL(3,7), up to conjugacy, such that $\langle M,+,\cdot\rangle$ has $7^3$ elements. Unfortunately, it is isomorphic to $GF(7)^3$ not $GF(7^3)$.

It is quite easy to find an M that generates a copy of $GF(7^3)$: one simply looks for a matrix whose order divides $7^3-1$ but not $7-1$. In other words, we just want a matrix of order 57:

gap> ms := Filtered( ccreps, m -> Order(m) = 57 );; gap> Size( ms ); 12 

We get 12 conjugacy classes, but in fact these form a single rational class:

gap> ForAll( ms, m -> ForAny(PrimeResidues(57), > k -> IsConjugate(GL(3,7),m,m^k))); true 

Now we want to check the algebras just to make sure:

gap> as:=List( ms, m -> Algebra(GF(7),[m]));; gap> ForAll( as, IsSimple ); true gap> ForAll( as, IsCommutative ); true gap> ForAll( as, a -> Size(a) = 7^3 ); true 
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    Thank you for the answer! I appreciate the hints including advanced GAP commands.2011-11-02
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Actually you can do the generalized version (described in pol's comment) with Mathematica, too. The task at hand is to find a 3x3 matrix $M$ with entries in $GF(p)=\mathbf{Z}/p\mathbf{Z}$ such that $M$ has multiplicative order $q$, and that the algebra it generates is isomorphic to $GF(p^3)$. The multiplicative group of $GF(p^3)$ is cyclic of order $p^3-1$, so for such an $M$ to exist, it is necessary that $q\mid p^3-1$. For the algebra $GF(p)[M]$ to be a field, it is necessary that the minimal polynomial of $M$ is cubic, and that it does not split into lower degree factors over $GF(p)$. The minimal polynomial of a matrix is always a factor of its characteristic polynomial $\chi_M(x)$, so we only need $\chi_M(x)$ to be irreducible over $GF(p)$. Then $\langle M,+,\cdot\rangle=GF(p)[x]\langle \chi_M(x)\rangle$, which is isomorphic to $GF(p^3)$.

But the constraint on the order of $M$ forces the minimal polynomial to be a factor of $x^q-1$. So we need $x^q-1$ to have an irreducible factor of degree three. Note that when $q\mid p-1$, then the polynomial $x^q-1$ splits into a product of linear factors over $GF(p)$, so we have to avoid that.

Following Jack's suggestion ($57=3\cdot19$) let's try with $p=7$, $q=19$ with Mathematica

In[1]:=Factor[x^19-1,Modulus->7] Out[1]=(6+x)(6+2x+x^3)(6+3x+3x^2+x^3)(6+x+4x^2+x^3)(6+4x+4x^2+x^3)(6+5x^2+x^3)(6+3x+6x^2+x^3) 

So we have no less than six cubic factors (it was actually possible to tell this in advance, but let's skip that). Let's use the first: $6+2x+x^3$. We use a standard recipe of companion matrix to produce a matrix with a given minimal polynomial, and pick $ M=\pmatrix{0&0&-6\cr 1&0&-2\cr 0&1&0\cr}. $ It is straightforward to verify that $M^{19}\equiv I_3 \pmod 7$.

As a final check we see that $q=3, p=7$ does not work:

In[2]:=Factor[x^3-1,Modulus->7] Out[2]=(3+x)(5+x)(6+x) 

confirming the findings from my comments and Jack's answer that the third roots of unity in a field of characteristic seven are $4=-3$, $5=-2$, and $1=-6$.