Can we find element of order $q^2-1$ in $text{GL}_2(mathbb{F}_q)$?
$begingroup$
How to find element of order $q^2-1$ in $text{GL}_2(mathbb{F}_q)$? I am hoping to find field $mathbb F_{q^2}$ as subalgebra of $2times 2$ matrices over field $mathbb F_q$ where $q$ is power of prime number. I was trying with element $pmatrix {n&1 \ 1&0}$ but it works only for $q=2,3,4,8,16$.
It is suggested that this is duplicate of question
$GL_n(mathbb F_q)$ has an element of order $q^n-1$
But I don't know how to find $mathbb F_{q^2}$ in $M_2(mathbb F_q)$.
Here is test in GAP for above matrix - $n$ is generator of the field multiplicative group.
gap> List([2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41],
> k->Order([[Z(k),1],[1,0]]*Z(k)^0));
[ 3, 8, 5, 12, 16, 9, 20, 24, 28, 17, 16, 40, 22, 52, 56, 20, 64, 31, 76, 40 ]
To give my motivation - I want to prove that algebra $M_2(mathbb F_q)$ can be represented as ${a+bj}$ for $a,b$ belonging to $mathbb F_{q^2}$ and multiplication given by Cayley-Dickson
$$(a+bj)(c+dj)=ac+bar db + (da+bbar c)j,$$
where $j$ is matrix satisfying $bar j=-j$.
abstract-algebra group-theory finite-groups finite-fields general-linear-group
$endgroup$
add a comment |
$begingroup$
How to find element of order $q^2-1$ in $text{GL}_2(mathbb{F}_q)$? I am hoping to find field $mathbb F_{q^2}$ as subalgebra of $2times 2$ matrices over field $mathbb F_q$ where $q$ is power of prime number. I was trying with element $pmatrix {n&1 \ 1&0}$ but it works only for $q=2,3,4,8,16$.
It is suggested that this is duplicate of question
$GL_n(mathbb F_q)$ has an element of order $q^n-1$
But I don't know how to find $mathbb F_{q^2}$ in $M_2(mathbb F_q)$.
Here is test in GAP for above matrix - $n$ is generator of the field multiplicative group.
gap> List([2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41],
> k->Order([[Z(k),1],[1,0]]*Z(k)^0));
[ 3, 8, 5, 12, 16, 9, 20, 24, 28, 17, 16, 40, 22, 52, 56, 20, 64, 31, 76, 40 ]
To give my motivation - I want to prove that algebra $M_2(mathbb F_q)$ can be represented as ${a+bj}$ for $a,b$ belonging to $mathbb F_{q^2}$ and multiplication given by Cayley-Dickson
$$(a+bj)(c+dj)=ac+bar db + (da+bbar c)j,$$
where $j$ is matrix satisfying $bar j=-j$.
abstract-algebra group-theory finite-groups finite-fields general-linear-group
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 9 at 1:33
add a comment |
$begingroup$
How to find element of order $q^2-1$ in $text{GL}_2(mathbb{F}_q)$? I am hoping to find field $mathbb F_{q^2}$ as subalgebra of $2times 2$ matrices over field $mathbb F_q$ where $q$ is power of prime number. I was trying with element $pmatrix {n&1 \ 1&0}$ but it works only for $q=2,3,4,8,16$.
It is suggested that this is duplicate of question
$GL_n(mathbb F_q)$ has an element of order $q^n-1$
But I don't know how to find $mathbb F_{q^2}$ in $M_2(mathbb F_q)$.
Here is test in GAP for above matrix - $n$ is generator of the field multiplicative group.
gap> List([2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41],
> k->Order([[Z(k),1],[1,0]]*Z(k)^0));
[ 3, 8, 5, 12, 16, 9, 20, 24, 28, 17, 16, 40, 22, 52, 56, 20, 64, 31, 76, 40 ]
To give my motivation - I want to prove that algebra $M_2(mathbb F_q)$ can be represented as ${a+bj}$ for $a,b$ belonging to $mathbb F_{q^2}$ and multiplication given by Cayley-Dickson
$$(a+bj)(c+dj)=ac+bar db + (da+bbar c)j,$$
where $j$ is matrix satisfying $bar j=-j$.
abstract-algebra group-theory finite-groups finite-fields general-linear-group
$endgroup$
How to find element of order $q^2-1$ in $text{GL}_2(mathbb{F}_q)$? I am hoping to find field $mathbb F_{q^2}$ as subalgebra of $2times 2$ matrices over field $mathbb F_q$ where $q$ is power of prime number. I was trying with element $pmatrix {n&1 \ 1&0}$ but it works only for $q=2,3,4,8,16$.
It is suggested that this is duplicate of question
$GL_n(mathbb F_q)$ has an element of order $q^n-1$
But I don't know how to find $mathbb F_{q^2}$ in $M_2(mathbb F_q)$.
Here is test in GAP for above matrix - $n$ is generator of the field multiplicative group.
gap> List([2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41],
> k->Order([[Z(k),1],[1,0]]*Z(k)^0));
[ 3, 8, 5, 12, 16, 9, 20, 24, 28, 17, 16, 40, 22, 52, 56, 20, 64, 31, 76, 40 ]
To give my motivation - I want to prove that algebra $M_2(mathbb F_q)$ can be represented as ${a+bj}$ for $a,b$ belonging to $mathbb F_{q^2}$ and multiplication given by Cayley-Dickson
$$(a+bj)(c+dj)=ac+bar db + (da+bbar c)j,$$
where $j$ is matrix satisfying $bar j=-j$.
abstract-algebra group-theory finite-groups finite-fields general-linear-group
abstract-algebra group-theory finite-groups finite-fields general-linear-group
edited Jan 13 at 21:55
Marek Mitros
asked Jan 6 at 16:57
Marek MitrosMarek Mitros
357212
357212
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Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 9 at 1:33
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 9 at 1:33
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 9 at 1:33
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 9 at 1:33
add a comment |
3 Answers
3
active
oldest
votes
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Let $x^2+ax+binmathbb{F}_q[x]$ be an irreducible polynomial one of whose roots is a generator of the cyclic group $mathbb{F}_{q^2}^times$ of order $q^2-1$. Define
$$A:=begin{bmatrix}0&-b\1&-aend{bmatrix},.$$
Show that the order of $A$ is precisely $q^2-1$.
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$begingroup$
I would prove by considering $A$ as matrix over $F_{q^2}$. We can diagonalize it and the roots of this polynomial are on the diagonal. It is remaining to show that both roots of polynomial are the same order and they are conjugate. If $t$ is root of the polynomial then $pmatrix{-b\t}$ is eigenvector with eigen value $t$.
$endgroup$
– Marek Mitros
Jan 8 at 14:31
2
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You only need to know that one root has exactly order $q^2-1$. That is why I placed the condition that one of the roots of $x^2+ax+b$ generates $mathbb{F}_{q^2}^times$.
$endgroup$
– Batominovski
Jan 8 at 22:14
add a comment |
$begingroup$
The main question (the part with Cayley-Dickson) follows from the Skolem-Noether theorem.
Let's denote $K=Bbb{F}_q, L=Bbb{F}_{q^2}$, and for all $zin L$ let $z^q=overline{z}$ be the Frobenius automorphism. As the other answers have explained there exist ways of embedding $L$ into $M_2(K)$. Let's fix one such embedding $phi:Lto M_2(K)$. Then $zmapsto phi(overline{z})$ is another embedding of $L$, so by Skolem-Noether there exists an invertible matrix $uin M_2(K)$ such that $uphi(z)=phi(overline{z})u$ for all $zin L$. Identifying $L$ with $phi(L)$ allows us to rewrite this as
$$uz=overline{z}uquadtext{for all $zin L$.}qquad(*)$$
A priori there is no reason to think that $u^2$ would be equal to the identity. However, we can always arrange that to be the case. Namely, if $cin L$ is arbitrary, it follows that $cz=zc$ for all $zin L$ implying that $u'=uc$ can take the role of $u$ in equation $(*)$. Actually, the double centralizer theorem tells us that any two matrices satisfying $(*)$ are gotten from each other by multiplication by an element of $L$.
Anyway, applying $(*)$ twice tells us that
$$u^2z=u(uz)=u(overline{z}u)=(uoverline{z})u=(zu)u=zu^2$$
for all $zin L$. In other words $u^2$ commutes with all the matrices in $L$. Trivially $u^2$ commutes with $u$. The sum of subspaces $L+uL$ must be direct, and hence all of $M_2(K)$. We have just shown that $u^2$ is in the center of $M_2(K)$. Therefore $u^2$ is a scalar matrix, and we can think of it as an element $u^2=alphain K$.
But, for all $cin L$
$$(uc)^2=u(cu)c=u^2overline{c}c=u^2N(c),$$
where $N:Lto K, zmapsto zoverline{z}$ is the norm map. In the case of finite fields the norm map is surjective, so we can find an element $c_0$ such that $N(c_0)=1/alpha$.
Therefore
$$
j=uc_0
$$
satisfies the relation $j^2=1$ as well as the relation $jz=overline{z}j$ for all $zin L$
as prescribed.
$endgroup$
$begingroup$
It should not be too difficult to find the matrix $j$ given an embedding $phi$. Different choices of $phi$ lead to different matrices $j$. They seem to form a conjugacy class in $GL_2(K)$.
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– Jyrki Lahtonen
Jan 8 at 18:52
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nice answer. I need to clean my answer a little. It is essentially what I am doing, but I need to recheck my answer a little, in light of your answer.
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– Malkoun
Jan 9 at 8:20
add a comment |
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I will try to answer some of the motivation part of your post. Thanks to reuns's comment below, I realize now that I am making the assumption that $q$ is an odd prime power. I will first show how one may embed $mathbb{F}_{q^2}$ in the matrix space $M_2(mathbb{F}_q)$. We start with a finite field $mathbb{F}_q$. We would like to build $mathbb{F}_{q^2}$ from it. There is a standard construction for that. First one finds an irreducible (let us say monic) quadratic polynomial, say
$ f(x) = x^2 + ax + b in mathbb{F}_q[x] $
and then defines $mathbb{F}_{q^2}$ to be $mathbb{F}_q[x]/(f(x))$. Let
$delta = a^2 - 4 b$
be the discriminant of $f$. Then we could alternatively have defined $mathbb{F}_{q^2}$ from $mathbb{F}_q$ by adjoining $sqrt{delta}$ to it. In other words,
$mathbb{F}_q[y]/(y^2-delta) simeq mathbb{F}_{q^2}$.
So an arbitrary element $u$ of $mathbb{F}_{q^2}$ can be written as:
$u = c + dsqrt{delta}$,
where $c$ and $d$ are elements of $mathbb{F}_q$. We would like to map $sqrt{delta}$ to the element
$ left( begin{array}{cc} 0 & delta \
1 & 0 end{array} right) in GL(2;mathbb{F}_q)$.
Note that an element $u$ as above, gets then mapped to
$ left( begin{array}{cc} c & delta d \
d & c end{array} right)$.
I am sure that using group actions may make the above construction cleaner (indeed, see Daniel Schepler's argument in the comments below), but at least it is explicit. We have thus found an embedding $f: mathbb{F}_{q^2} to M_2(mathbb{F}_q)$.
The rest of this answer comes directly from Jyrki Lahtonen's answer below, as well as his comment. I include it here for completeness (you can give the points to Jyrki Lahtonen for that part!).
Define
$J = left( begin{array}{cc} 1 & 0 \ 0 & -1 end{array} right)$
Note that $J^2 = I$, with $I$ being the identity. Then, for all $u in mathbb{F}_{q^2}$, we have $J f(u) = f(bar{u}) J$.
One can now check that the map $F$ which maps $u_1 + j u_2$ to $f(u_1) + J f(u_2)$ is indeed an isomorphism, as required. Thank you Jyrki Lahtonen once more for this part of the answer.
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Just to say $mathbb{F}_{q^2} = mathbb{F}_q(sqrt{delta}) $ works iff $q$ is an odd prime power. Otherwise $a mapsto a^2$ is an automorphism of $mathbb{F}_{2^n}^*$ so $x^2-delta in mathbb{F}_{2^n}[x]$ is never irreducible
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– reuns
Jan 8 at 8:23
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true. I will edit. Thank you.
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– Malkoun
Jan 8 at 8:48
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Thank you very much. It is amazing to see the $Delta=b^2-4ac$ known from school is also working for finite fields ! I have tested in GAP that for $q=3$ matrix $pmatrix{1&1\1&0}$ is generator of $F_9$ and matrix $j=pmatrix{1&0\-1&-1}$ will work for Cayley-Dickson. The one I mentioned in the question does not work.
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– Marek Mitros
Jan 8 at 12:40
2
$begingroup$
Another way to describe an embedding would be: $mathbb{F}_{q^2}$ is a vector space of dimension two over $mathbb{F}_q$; and also, for any $alpha in mathbb{F}_{q^2}$, multiplication by $alpha$ is a linear operator on this vector space. So, if you take a basis of $mathbb{F}_{q^2}$ - for example, ${ 1, beta }$ will work whenever $beta in mathbb{F}_{q^2} setminus mathbb{F}_q$ - then the matrix representations of $alpha cdot$ with respect to this basis will form the desired embedding.
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– Daniel Schepler
Jan 8 at 22:42
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@DanielSchepler, yes, this is the cleaner argument I was referring to. Thank you. I wrote it this way, because it is a little more explicit. If I was writing an article, I would have done it the cleaner way, using the argument you gave.
$endgroup$
– Malkoun
Jan 9 at 8:06
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $x^2+ax+binmathbb{F}_q[x]$ be an irreducible polynomial one of whose roots is a generator of the cyclic group $mathbb{F}_{q^2}^times$ of order $q^2-1$. Define
$$A:=begin{bmatrix}0&-b\1&-aend{bmatrix},.$$
Show that the order of $A$ is precisely $q^2-1$.
$endgroup$
$begingroup$
I would prove by considering $A$ as matrix over $F_{q^2}$. We can diagonalize it and the roots of this polynomial are on the diagonal. It is remaining to show that both roots of polynomial are the same order and they are conjugate. If $t$ is root of the polynomial then $pmatrix{-b\t}$ is eigenvector with eigen value $t$.
$endgroup$
– Marek Mitros
Jan 8 at 14:31
2
$begingroup$
You only need to know that one root has exactly order $q^2-1$. That is why I placed the condition that one of the roots of $x^2+ax+b$ generates $mathbb{F}_{q^2}^times$.
$endgroup$
– Batominovski
Jan 8 at 22:14
add a comment |
$begingroup$
Let $x^2+ax+binmathbb{F}_q[x]$ be an irreducible polynomial one of whose roots is a generator of the cyclic group $mathbb{F}_{q^2}^times$ of order $q^2-1$. Define
$$A:=begin{bmatrix}0&-b\1&-aend{bmatrix},.$$
Show that the order of $A$ is precisely $q^2-1$.
$endgroup$
$begingroup$
I would prove by considering $A$ as matrix over $F_{q^2}$. We can diagonalize it and the roots of this polynomial are on the diagonal. It is remaining to show that both roots of polynomial are the same order and they are conjugate. If $t$ is root of the polynomial then $pmatrix{-b\t}$ is eigenvector with eigen value $t$.
$endgroup$
– Marek Mitros
Jan 8 at 14:31
2
$begingroup$
You only need to know that one root has exactly order $q^2-1$. That is why I placed the condition that one of the roots of $x^2+ax+b$ generates $mathbb{F}_{q^2}^times$.
$endgroup$
– Batominovski
Jan 8 at 22:14
add a comment |
$begingroup$
Let $x^2+ax+binmathbb{F}_q[x]$ be an irreducible polynomial one of whose roots is a generator of the cyclic group $mathbb{F}_{q^2}^times$ of order $q^2-1$. Define
$$A:=begin{bmatrix}0&-b\1&-aend{bmatrix},.$$
Show that the order of $A$ is precisely $q^2-1$.
$endgroup$
Let $x^2+ax+binmathbb{F}_q[x]$ be an irreducible polynomial one of whose roots is a generator of the cyclic group $mathbb{F}_{q^2}^times$ of order $q^2-1$. Define
$$A:=begin{bmatrix}0&-b\1&-aend{bmatrix},.$$
Show that the order of $A$ is precisely $q^2-1$.
answered Jan 8 at 6:30
BatominovskiBatominovski
1
1
$begingroup$
I would prove by considering $A$ as matrix over $F_{q^2}$. We can diagonalize it and the roots of this polynomial are on the diagonal. It is remaining to show that both roots of polynomial are the same order and they are conjugate. If $t$ is root of the polynomial then $pmatrix{-b\t}$ is eigenvector with eigen value $t$.
$endgroup$
– Marek Mitros
Jan 8 at 14:31
2
$begingroup$
You only need to know that one root has exactly order $q^2-1$. That is why I placed the condition that one of the roots of $x^2+ax+b$ generates $mathbb{F}_{q^2}^times$.
$endgroup$
– Batominovski
Jan 8 at 22:14
add a comment |
$begingroup$
I would prove by considering $A$ as matrix over $F_{q^2}$. We can diagonalize it and the roots of this polynomial are on the diagonal. It is remaining to show that both roots of polynomial are the same order and they are conjugate. If $t$ is root of the polynomial then $pmatrix{-b\t}$ is eigenvector with eigen value $t$.
$endgroup$
– Marek Mitros
Jan 8 at 14:31
2
$begingroup$
You only need to know that one root has exactly order $q^2-1$. That is why I placed the condition that one of the roots of $x^2+ax+b$ generates $mathbb{F}_{q^2}^times$.
$endgroup$
– Batominovski
Jan 8 at 22:14
$begingroup$
I would prove by considering $A$ as matrix over $F_{q^2}$. We can diagonalize it and the roots of this polynomial are on the diagonal. It is remaining to show that both roots of polynomial are the same order and they are conjugate. If $t$ is root of the polynomial then $pmatrix{-b\t}$ is eigenvector with eigen value $t$.
$endgroup$
– Marek Mitros
Jan 8 at 14:31
$begingroup$
I would prove by considering $A$ as matrix over $F_{q^2}$. We can diagonalize it and the roots of this polynomial are on the diagonal. It is remaining to show that both roots of polynomial are the same order and they are conjugate. If $t$ is root of the polynomial then $pmatrix{-b\t}$ is eigenvector with eigen value $t$.
$endgroup$
– Marek Mitros
Jan 8 at 14:31
2
2
$begingroup$
You only need to know that one root has exactly order $q^2-1$. That is why I placed the condition that one of the roots of $x^2+ax+b$ generates $mathbb{F}_{q^2}^times$.
$endgroup$
– Batominovski
Jan 8 at 22:14
$begingroup$
You only need to know that one root has exactly order $q^2-1$. That is why I placed the condition that one of the roots of $x^2+ax+b$ generates $mathbb{F}_{q^2}^times$.
$endgroup$
– Batominovski
Jan 8 at 22:14
add a comment |
$begingroup$
The main question (the part with Cayley-Dickson) follows from the Skolem-Noether theorem.
Let's denote $K=Bbb{F}_q, L=Bbb{F}_{q^2}$, and for all $zin L$ let $z^q=overline{z}$ be the Frobenius automorphism. As the other answers have explained there exist ways of embedding $L$ into $M_2(K)$. Let's fix one such embedding $phi:Lto M_2(K)$. Then $zmapsto phi(overline{z})$ is another embedding of $L$, so by Skolem-Noether there exists an invertible matrix $uin M_2(K)$ such that $uphi(z)=phi(overline{z})u$ for all $zin L$. Identifying $L$ with $phi(L)$ allows us to rewrite this as
$$uz=overline{z}uquadtext{for all $zin L$.}qquad(*)$$
A priori there is no reason to think that $u^2$ would be equal to the identity. However, we can always arrange that to be the case. Namely, if $cin L$ is arbitrary, it follows that $cz=zc$ for all $zin L$ implying that $u'=uc$ can take the role of $u$ in equation $(*)$. Actually, the double centralizer theorem tells us that any two matrices satisfying $(*)$ are gotten from each other by multiplication by an element of $L$.
Anyway, applying $(*)$ twice tells us that
$$u^2z=u(uz)=u(overline{z}u)=(uoverline{z})u=(zu)u=zu^2$$
for all $zin L$. In other words $u^2$ commutes with all the matrices in $L$. Trivially $u^2$ commutes with $u$. The sum of subspaces $L+uL$ must be direct, and hence all of $M_2(K)$. We have just shown that $u^2$ is in the center of $M_2(K)$. Therefore $u^2$ is a scalar matrix, and we can think of it as an element $u^2=alphain K$.
But, for all $cin L$
$$(uc)^2=u(cu)c=u^2overline{c}c=u^2N(c),$$
where $N:Lto K, zmapsto zoverline{z}$ is the norm map. In the case of finite fields the norm map is surjective, so we can find an element $c_0$ such that $N(c_0)=1/alpha$.
Therefore
$$
j=uc_0
$$
satisfies the relation $j^2=1$ as well as the relation $jz=overline{z}j$ for all $zin L$
as prescribed.
$endgroup$
$begingroup$
It should not be too difficult to find the matrix $j$ given an embedding $phi$. Different choices of $phi$ lead to different matrices $j$. They seem to form a conjugacy class in $GL_2(K)$.
$endgroup$
– Jyrki Lahtonen
Jan 8 at 18:52
$begingroup$
nice answer. I need to clean my answer a little. It is essentially what I am doing, but I need to recheck my answer a little, in light of your answer.
$endgroup$
– Malkoun
Jan 9 at 8:20
add a comment |
$begingroup$
The main question (the part with Cayley-Dickson) follows from the Skolem-Noether theorem.
Let's denote $K=Bbb{F}_q, L=Bbb{F}_{q^2}$, and for all $zin L$ let $z^q=overline{z}$ be the Frobenius automorphism. As the other answers have explained there exist ways of embedding $L$ into $M_2(K)$. Let's fix one such embedding $phi:Lto M_2(K)$. Then $zmapsto phi(overline{z})$ is another embedding of $L$, so by Skolem-Noether there exists an invertible matrix $uin M_2(K)$ such that $uphi(z)=phi(overline{z})u$ for all $zin L$. Identifying $L$ with $phi(L)$ allows us to rewrite this as
$$uz=overline{z}uquadtext{for all $zin L$.}qquad(*)$$
A priori there is no reason to think that $u^2$ would be equal to the identity. However, we can always arrange that to be the case. Namely, if $cin L$ is arbitrary, it follows that $cz=zc$ for all $zin L$ implying that $u'=uc$ can take the role of $u$ in equation $(*)$. Actually, the double centralizer theorem tells us that any two matrices satisfying $(*)$ are gotten from each other by multiplication by an element of $L$.
Anyway, applying $(*)$ twice tells us that
$$u^2z=u(uz)=u(overline{z}u)=(uoverline{z})u=(zu)u=zu^2$$
for all $zin L$. In other words $u^2$ commutes with all the matrices in $L$. Trivially $u^2$ commutes with $u$. The sum of subspaces $L+uL$ must be direct, and hence all of $M_2(K)$. We have just shown that $u^2$ is in the center of $M_2(K)$. Therefore $u^2$ is a scalar matrix, and we can think of it as an element $u^2=alphain K$.
But, for all $cin L$
$$(uc)^2=u(cu)c=u^2overline{c}c=u^2N(c),$$
where $N:Lto K, zmapsto zoverline{z}$ is the norm map. In the case of finite fields the norm map is surjective, so we can find an element $c_0$ such that $N(c_0)=1/alpha$.
Therefore
$$
j=uc_0
$$
satisfies the relation $j^2=1$ as well as the relation $jz=overline{z}j$ for all $zin L$
as prescribed.
$endgroup$
$begingroup$
It should not be too difficult to find the matrix $j$ given an embedding $phi$. Different choices of $phi$ lead to different matrices $j$. They seem to form a conjugacy class in $GL_2(K)$.
$endgroup$
– Jyrki Lahtonen
Jan 8 at 18:52
$begingroup$
nice answer. I need to clean my answer a little. It is essentially what I am doing, but I need to recheck my answer a little, in light of your answer.
$endgroup$
– Malkoun
Jan 9 at 8:20
add a comment |
$begingroup$
The main question (the part with Cayley-Dickson) follows from the Skolem-Noether theorem.
Let's denote $K=Bbb{F}_q, L=Bbb{F}_{q^2}$, and for all $zin L$ let $z^q=overline{z}$ be the Frobenius automorphism. As the other answers have explained there exist ways of embedding $L$ into $M_2(K)$. Let's fix one such embedding $phi:Lto M_2(K)$. Then $zmapsto phi(overline{z})$ is another embedding of $L$, so by Skolem-Noether there exists an invertible matrix $uin M_2(K)$ such that $uphi(z)=phi(overline{z})u$ for all $zin L$. Identifying $L$ with $phi(L)$ allows us to rewrite this as
$$uz=overline{z}uquadtext{for all $zin L$.}qquad(*)$$
A priori there is no reason to think that $u^2$ would be equal to the identity. However, we can always arrange that to be the case. Namely, if $cin L$ is arbitrary, it follows that $cz=zc$ for all $zin L$ implying that $u'=uc$ can take the role of $u$ in equation $(*)$. Actually, the double centralizer theorem tells us that any two matrices satisfying $(*)$ are gotten from each other by multiplication by an element of $L$.
Anyway, applying $(*)$ twice tells us that
$$u^2z=u(uz)=u(overline{z}u)=(uoverline{z})u=(zu)u=zu^2$$
for all $zin L$. In other words $u^2$ commutes with all the matrices in $L$. Trivially $u^2$ commutes with $u$. The sum of subspaces $L+uL$ must be direct, and hence all of $M_2(K)$. We have just shown that $u^2$ is in the center of $M_2(K)$. Therefore $u^2$ is a scalar matrix, and we can think of it as an element $u^2=alphain K$.
But, for all $cin L$
$$(uc)^2=u(cu)c=u^2overline{c}c=u^2N(c),$$
where $N:Lto K, zmapsto zoverline{z}$ is the norm map. In the case of finite fields the norm map is surjective, so we can find an element $c_0$ such that $N(c_0)=1/alpha$.
Therefore
$$
j=uc_0
$$
satisfies the relation $j^2=1$ as well as the relation $jz=overline{z}j$ for all $zin L$
as prescribed.
$endgroup$
The main question (the part with Cayley-Dickson) follows from the Skolem-Noether theorem.
Let's denote $K=Bbb{F}_q, L=Bbb{F}_{q^2}$, and for all $zin L$ let $z^q=overline{z}$ be the Frobenius automorphism. As the other answers have explained there exist ways of embedding $L$ into $M_2(K)$. Let's fix one such embedding $phi:Lto M_2(K)$. Then $zmapsto phi(overline{z})$ is another embedding of $L$, so by Skolem-Noether there exists an invertible matrix $uin M_2(K)$ such that $uphi(z)=phi(overline{z})u$ for all $zin L$. Identifying $L$ with $phi(L)$ allows us to rewrite this as
$$uz=overline{z}uquadtext{for all $zin L$.}qquad(*)$$
A priori there is no reason to think that $u^2$ would be equal to the identity. However, we can always arrange that to be the case. Namely, if $cin L$ is arbitrary, it follows that $cz=zc$ for all $zin L$ implying that $u'=uc$ can take the role of $u$ in equation $(*)$. Actually, the double centralizer theorem tells us that any two matrices satisfying $(*)$ are gotten from each other by multiplication by an element of $L$.
Anyway, applying $(*)$ twice tells us that
$$u^2z=u(uz)=u(overline{z}u)=(uoverline{z})u=(zu)u=zu^2$$
for all $zin L$. In other words $u^2$ commutes with all the matrices in $L$. Trivially $u^2$ commutes with $u$. The sum of subspaces $L+uL$ must be direct, and hence all of $M_2(K)$. We have just shown that $u^2$ is in the center of $M_2(K)$. Therefore $u^2$ is a scalar matrix, and we can think of it as an element $u^2=alphain K$.
But, for all $cin L$
$$(uc)^2=u(cu)c=u^2overline{c}c=u^2N(c),$$
where $N:Lto K, zmapsto zoverline{z}$ is the norm map. In the case of finite fields the norm map is surjective, so we can find an element $c_0$ such that $N(c_0)=1/alpha$.
Therefore
$$
j=uc_0
$$
satisfies the relation $j^2=1$ as well as the relation $jz=overline{z}j$ for all $zin L$
as prescribed.
edited Jan 8 at 18:45
answered Jan 8 at 18:39
Jyrki LahtonenJyrki Lahtonen
108k12166369
108k12166369
$begingroup$
It should not be too difficult to find the matrix $j$ given an embedding $phi$. Different choices of $phi$ lead to different matrices $j$. They seem to form a conjugacy class in $GL_2(K)$.
$endgroup$
– Jyrki Lahtonen
Jan 8 at 18:52
$begingroup$
nice answer. I need to clean my answer a little. It is essentially what I am doing, but I need to recheck my answer a little, in light of your answer.
$endgroup$
– Malkoun
Jan 9 at 8:20
add a comment |
$begingroup$
It should not be too difficult to find the matrix $j$ given an embedding $phi$. Different choices of $phi$ lead to different matrices $j$. They seem to form a conjugacy class in $GL_2(K)$.
$endgroup$
– Jyrki Lahtonen
Jan 8 at 18:52
$begingroup$
nice answer. I need to clean my answer a little. It is essentially what I am doing, but I need to recheck my answer a little, in light of your answer.
$endgroup$
– Malkoun
Jan 9 at 8:20
$begingroup$
It should not be too difficult to find the matrix $j$ given an embedding $phi$. Different choices of $phi$ lead to different matrices $j$. They seem to form a conjugacy class in $GL_2(K)$.
$endgroup$
– Jyrki Lahtonen
Jan 8 at 18:52
$begingroup$
It should not be too difficult to find the matrix $j$ given an embedding $phi$. Different choices of $phi$ lead to different matrices $j$. They seem to form a conjugacy class in $GL_2(K)$.
$endgroup$
– Jyrki Lahtonen
Jan 8 at 18:52
$begingroup$
nice answer. I need to clean my answer a little. It is essentially what I am doing, but I need to recheck my answer a little, in light of your answer.
$endgroup$
– Malkoun
Jan 9 at 8:20
$begingroup$
nice answer. I need to clean my answer a little. It is essentially what I am doing, but I need to recheck my answer a little, in light of your answer.
$endgroup$
– Malkoun
Jan 9 at 8:20
add a comment |
$begingroup$
I will try to answer some of the motivation part of your post. Thanks to reuns's comment below, I realize now that I am making the assumption that $q$ is an odd prime power. I will first show how one may embed $mathbb{F}_{q^2}$ in the matrix space $M_2(mathbb{F}_q)$. We start with a finite field $mathbb{F}_q$. We would like to build $mathbb{F}_{q^2}$ from it. There is a standard construction for that. First one finds an irreducible (let us say monic) quadratic polynomial, say
$ f(x) = x^2 + ax + b in mathbb{F}_q[x] $
and then defines $mathbb{F}_{q^2}$ to be $mathbb{F}_q[x]/(f(x))$. Let
$delta = a^2 - 4 b$
be the discriminant of $f$. Then we could alternatively have defined $mathbb{F}_{q^2}$ from $mathbb{F}_q$ by adjoining $sqrt{delta}$ to it. In other words,
$mathbb{F}_q[y]/(y^2-delta) simeq mathbb{F}_{q^2}$.
So an arbitrary element $u$ of $mathbb{F}_{q^2}$ can be written as:
$u = c + dsqrt{delta}$,
where $c$ and $d$ are elements of $mathbb{F}_q$. We would like to map $sqrt{delta}$ to the element
$ left( begin{array}{cc} 0 & delta \
1 & 0 end{array} right) in GL(2;mathbb{F}_q)$.
Note that an element $u$ as above, gets then mapped to
$ left( begin{array}{cc} c & delta d \
d & c end{array} right)$.
I am sure that using group actions may make the above construction cleaner (indeed, see Daniel Schepler's argument in the comments below), but at least it is explicit. We have thus found an embedding $f: mathbb{F}_{q^2} to M_2(mathbb{F}_q)$.
The rest of this answer comes directly from Jyrki Lahtonen's answer below, as well as his comment. I include it here for completeness (you can give the points to Jyrki Lahtonen for that part!).
Define
$J = left( begin{array}{cc} 1 & 0 \ 0 & -1 end{array} right)$
Note that $J^2 = I$, with $I$ being the identity. Then, for all $u in mathbb{F}_{q^2}$, we have $J f(u) = f(bar{u}) J$.
One can now check that the map $F$ which maps $u_1 + j u_2$ to $f(u_1) + J f(u_2)$ is indeed an isomorphism, as required. Thank you Jyrki Lahtonen once more for this part of the answer.
$endgroup$
$begingroup$
Just to say $mathbb{F}_{q^2} = mathbb{F}_q(sqrt{delta}) $ works iff $q$ is an odd prime power. Otherwise $a mapsto a^2$ is an automorphism of $mathbb{F}_{2^n}^*$ so $x^2-delta in mathbb{F}_{2^n}[x]$ is never irreducible
$endgroup$
– reuns
Jan 8 at 8:23
$begingroup$
true. I will edit. Thank you.
$endgroup$
– Malkoun
Jan 8 at 8:48
$begingroup$
Thank you very much. It is amazing to see the $Delta=b^2-4ac$ known from school is also working for finite fields ! I have tested in GAP that for $q=3$ matrix $pmatrix{1&1\1&0}$ is generator of $F_9$ and matrix $j=pmatrix{1&0\-1&-1}$ will work for Cayley-Dickson. The one I mentioned in the question does not work.
$endgroup$
– Marek Mitros
Jan 8 at 12:40
2
$begingroup$
Another way to describe an embedding would be: $mathbb{F}_{q^2}$ is a vector space of dimension two over $mathbb{F}_q$; and also, for any $alpha in mathbb{F}_{q^2}$, multiplication by $alpha$ is a linear operator on this vector space. So, if you take a basis of $mathbb{F}_{q^2}$ - for example, ${ 1, beta }$ will work whenever $beta in mathbb{F}_{q^2} setminus mathbb{F}_q$ - then the matrix representations of $alpha cdot$ with respect to this basis will form the desired embedding.
$endgroup$
– Daniel Schepler
Jan 8 at 22:42
$begingroup$
@DanielSchepler, yes, this is the cleaner argument I was referring to. Thank you. I wrote it this way, because it is a little more explicit. If I was writing an article, I would have done it the cleaner way, using the argument you gave.
$endgroup$
– Malkoun
Jan 9 at 8:06
|
show 3 more comments
$begingroup$
I will try to answer some of the motivation part of your post. Thanks to reuns's comment below, I realize now that I am making the assumption that $q$ is an odd prime power. I will first show how one may embed $mathbb{F}_{q^2}$ in the matrix space $M_2(mathbb{F}_q)$. We start with a finite field $mathbb{F}_q$. We would like to build $mathbb{F}_{q^2}$ from it. There is a standard construction for that. First one finds an irreducible (let us say monic) quadratic polynomial, say
$ f(x) = x^2 + ax + b in mathbb{F}_q[x] $
and then defines $mathbb{F}_{q^2}$ to be $mathbb{F}_q[x]/(f(x))$. Let
$delta = a^2 - 4 b$
be the discriminant of $f$. Then we could alternatively have defined $mathbb{F}_{q^2}$ from $mathbb{F}_q$ by adjoining $sqrt{delta}$ to it. In other words,
$mathbb{F}_q[y]/(y^2-delta) simeq mathbb{F}_{q^2}$.
So an arbitrary element $u$ of $mathbb{F}_{q^2}$ can be written as:
$u = c + dsqrt{delta}$,
where $c$ and $d$ are elements of $mathbb{F}_q$. We would like to map $sqrt{delta}$ to the element
$ left( begin{array}{cc} 0 & delta \
1 & 0 end{array} right) in GL(2;mathbb{F}_q)$.
Note that an element $u$ as above, gets then mapped to
$ left( begin{array}{cc} c & delta d \
d & c end{array} right)$.
I am sure that using group actions may make the above construction cleaner (indeed, see Daniel Schepler's argument in the comments below), but at least it is explicit. We have thus found an embedding $f: mathbb{F}_{q^2} to M_2(mathbb{F}_q)$.
The rest of this answer comes directly from Jyrki Lahtonen's answer below, as well as his comment. I include it here for completeness (you can give the points to Jyrki Lahtonen for that part!).
Define
$J = left( begin{array}{cc} 1 & 0 \ 0 & -1 end{array} right)$
Note that $J^2 = I$, with $I$ being the identity. Then, for all $u in mathbb{F}_{q^2}$, we have $J f(u) = f(bar{u}) J$.
One can now check that the map $F$ which maps $u_1 + j u_2$ to $f(u_1) + J f(u_2)$ is indeed an isomorphism, as required. Thank you Jyrki Lahtonen once more for this part of the answer.
$endgroup$
$begingroup$
Just to say $mathbb{F}_{q^2} = mathbb{F}_q(sqrt{delta}) $ works iff $q$ is an odd prime power. Otherwise $a mapsto a^2$ is an automorphism of $mathbb{F}_{2^n}^*$ so $x^2-delta in mathbb{F}_{2^n}[x]$ is never irreducible
$endgroup$
– reuns
Jan 8 at 8:23
$begingroup$
true. I will edit. Thank you.
$endgroup$
– Malkoun
Jan 8 at 8:48
$begingroup$
Thank you very much. It is amazing to see the $Delta=b^2-4ac$ known from school is also working for finite fields ! I have tested in GAP that for $q=3$ matrix $pmatrix{1&1\1&0}$ is generator of $F_9$ and matrix $j=pmatrix{1&0\-1&-1}$ will work for Cayley-Dickson. The one I mentioned in the question does not work.
$endgroup$
– Marek Mitros
Jan 8 at 12:40
2
$begingroup$
Another way to describe an embedding would be: $mathbb{F}_{q^2}$ is a vector space of dimension two over $mathbb{F}_q$; and also, for any $alpha in mathbb{F}_{q^2}$, multiplication by $alpha$ is a linear operator on this vector space. So, if you take a basis of $mathbb{F}_{q^2}$ - for example, ${ 1, beta }$ will work whenever $beta in mathbb{F}_{q^2} setminus mathbb{F}_q$ - then the matrix representations of $alpha cdot$ with respect to this basis will form the desired embedding.
$endgroup$
– Daniel Schepler
Jan 8 at 22:42
$begingroup$
@DanielSchepler, yes, this is the cleaner argument I was referring to. Thank you. I wrote it this way, because it is a little more explicit. If I was writing an article, I would have done it the cleaner way, using the argument you gave.
$endgroup$
– Malkoun
Jan 9 at 8:06
|
show 3 more comments
$begingroup$
I will try to answer some of the motivation part of your post. Thanks to reuns's comment below, I realize now that I am making the assumption that $q$ is an odd prime power. I will first show how one may embed $mathbb{F}_{q^2}$ in the matrix space $M_2(mathbb{F}_q)$. We start with a finite field $mathbb{F}_q$. We would like to build $mathbb{F}_{q^2}$ from it. There is a standard construction for that. First one finds an irreducible (let us say monic) quadratic polynomial, say
$ f(x) = x^2 + ax + b in mathbb{F}_q[x] $
and then defines $mathbb{F}_{q^2}$ to be $mathbb{F}_q[x]/(f(x))$. Let
$delta = a^2 - 4 b$
be the discriminant of $f$. Then we could alternatively have defined $mathbb{F}_{q^2}$ from $mathbb{F}_q$ by adjoining $sqrt{delta}$ to it. In other words,
$mathbb{F}_q[y]/(y^2-delta) simeq mathbb{F}_{q^2}$.
So an arbitrary element $u$ of $mathbb{F}_{q^2}$ can be written as:
$u = c + dsqrt{delta}$,
where $c$ and $d$ are elements of $mathbb{F}_q$. We would like to map $sqrt{delta}$ to the element
$ left( begin{array}{cc} 0 & delta \
1 & 0 end{array} right) in GL(2;mathbb{F}_q)$.
Note that an element $u$ as above, gets then mapped to
$ left( begin{array}{cc} c & delta d \
d & c end{array} right)$.
I am sure that using group actions may make the above construction cleaner (indeed, see Daniel Schepler's argument in the comments below), but at least it is explicit. We have thus found an embedding $f: mathbb{F}_{q^2} to M_2(mathbb{F}_q)$.
The rest of this answer comes directly from Jyrki Lahtonen's answer below, as well as his comment. I include it here for completeness (you can give the points to Jyrki Lahtonen for that part!).
Define
$J = left( begin{array}{cc} 1 & 0 \ 0 & -1 end{array} right)$
Note that $J^2 = I$, with $I$ being the identity. Then, for all $u in mathbb{F}_{q^2}$, we have $J f(u) = f(bar{u}) J$.
One can now check that the map $F$ which maps $u_1 + j u_2$ to $f(u_1) + J f(u_2)$ is indeed an isomorphism, as required. Thank you Jyrki Lahtonen once more for this part of the answer.
$endgroup$
I will try to answer some of the motivation part of your post. Thanks to reuns's comment below, I realize now that I am making the assumption that $q$ is an odd prime power. I will first show how one may embed $mathbb{F}_{q^2}$ in the matrix space $M_2(mathbb{F}_q)$. We start with a finite field $mathbb{F}_q$. We would like to build $mathbb{F}_{q^2}$ from it. There is a standard construction for that. First one finds an irreducible (let us say monic) quadratic polynomial, say
$ f(x) = x^2 + ax + b in mathbb{F}_q[x] $
and then defines $mathbb{F}_{q^2}$ to be $mathbb{F}_q[x]/(f(x))$. Let
$delta = a^2 - 4 b$
be the discriminant of $f$. Then we could alternatively have defined $mathbb{F}_{q^2}$ from $mathbb{F}_q$ by adjoining $sqrt{delta}$ to it. In other words,
$mathbb{F}_q[y]/(y^2-delta) simeq mathbb{F}_{q^2}$.
So an arbitrary element $u$ of $mathbb{F}_{q^2}$ can be written as:
$u = c + dsqrt{delta}$,
where $c$ and $d$ are elements of $mathbb{F}_q$. We would like to map $sqrt{delta}$ to the element
$ left( begin{array}{cc} 0 & delta \
1 & 0 end{array} right) in GL(2;mathbb{F}_q)$.
Note that an element $u$ as above, gets then mapped to
$ left( begin{array}{cc} c & delta d \
d & c end{array} right)$.
I am sure that using group actions may make the above construction cleaner (indeed, see Daniel Schepler's argument in the comments below), but at least it is explicit. We have thus found an embedding $f: mathbb{F}_{q^2} to M_2(mathbb{F}_q)$.
The rest of this answer comes directly from Jyrki Lahtonen's answer below, as well as his comment. I include it here for completeness (you can give the points to Jyrki Lahtonen for that part!).
Define
$J = left( begin{array}{cc} 1 & 0 \ 0 & -1 end{array} right)$
Note that $J^2 = I$, with $I$ being the identity. Then, for all $u in mathbb{F}_{q^2}$, we have $J f(u) = f(bar{u}) J$.
One can now check that the map $F$ which maps $u_1 + j u_2$ to $f(u_1) + J f(u_2)$ is indeed an isomorphism, as required. Thank you Jyrki Lahtonen once more for this part of the answer.
edited Jan 9 at 8:53
answered Jan 8 at 7:46
MalkounMalkoun
1,8811612
1,8811612
$begingroup$
Just to say $mathbb{F}_{q^2} = mathbb{F}_q(sqrt{delta}) $ works iff $q$ is an odd prime power. Otherwise $a mapsto a^2$ is an automorphism of $mathbb{F}_{2^n}^*$ so $x^2-delta in mathbb{F}_{2^n}[x]$ is never irreducible
$endgroup$
– reuns
Jan 8 at 8:23
$begingroup$
true. I will edit. Thank you.
$endgroup$
– Malkoun
Jan 8 at 8:48
$begingroup$
Thank you very much. It is amazing to see the $Delta=b^2-4ac$ known from school is also working for finite fields ! I have tested in GAP that for $q=3$ matrix $pmatrix{1&1\1&0}$ is generator of $F_9$ and matrix $j=pmatrix{1&0\-1&-1}$ will work for Cayley-Dickson. The one I mentioned in the question does not work.
$endgroup$
– Marek Mitros
Jan 8 at 12:40
2
$begingroup$
Another way to describe an embedding would be: $mathbb{F}_{q^2}$ is a vector space of dimension two over $mathbb{F}_q$; and also, for any $alpha in mathbb{F}_{q^2}$, multiplication by $alpha$ is a linear operator on this vector space. So, if you take a basis of $mathbb{F}_{q^2}$ - for example, ${ 1, beta }$ will work whenever $beta in mathbb{F}_{q^2} setminus mathbb{F}_q$ - then the matrix representations of $alpha cdot$ with respect to this basis will form the desired embedding.
$endgroup$
– Daniel Schepler
Jan 8 at 22:42
$begingroup$
@DanielSchepler, yes, this is the cleaner argument I was referring to. Thank you. I wrote it this way, because it is a little more explicit. If I was writing an article, I would have done it the cleaner way, using the argument you gave.
$endgroup$
– Malkoun
Jan 9 at 8:06
|
show 3 more comments
$begingroup$
Just to say $mathbb{F}_{q^2} = mathbb{F}_q(sqrt{delta}) $ works iff $q$ is an odd prime power. Otherwise $a mapsto a^2$ is an automorphism of $mathbb{F}_{2^n}^*$ so $x^2-delta in mathbb{F}_{2^n}[x]$ is never irreducible
$endgroup$
– reuns
Jan 8 at 8:23
$begingroup$
true. I will edit. Thank you.
$endgroup$
– Malkoun
Jan 8 at 8:48
$begingroup$
Thank you very much. It is amazing to see the $Delta=b^2-4ac$ known from school is also working for finite fields ! I have tested in GAP that for $q=3$ matrix $pmatrix{1&1\1&0}$ is generator of $F_9$ and matrix $j=pmatrix{1&0\-1&-1}$ will work for Cayley-Dickson. The one I mentioned in the question does not work.
$endgroup$
– Marek Mitros
Jan 8 at 12:40
2
$begingroup$
Another way to describe an embedding would be: $mathbb{F}_{q^2}$ is a vector space of dimension two over $mathbb{F}_q$; and also, for any $alpha in mathbb{F}_{q^2}$, multiplication by $alpha$ is a linear operator on this vector space. So, if you take a basis of $mathbb{F}_{q^2}$ - for example, ${ 1, beta }$ will work whenever $beta in mathbb{F}_{q^2} setminus mathbb{F}_q$ - then the matrix representations of $alpha cdot$ with respect to this basis will form the desired embedding.
$endgroup$
– Daniel Schepler
Jan 8 at 22:42
$begingroup$
@DanielSchepler, yes, this is the cleaner argument I was referring to. Thank you. I wrote it this way, because it is a little more explicit. If I was writing an article, I would have done it the cleaner way, using the argument you gave.
$endgroup$
– Malkoun
Jan 9 at 8:06
$begingroup$
Just to say $mathbb{F}_{q^2} = mathbb{F}_q(sqrt{delta}) $ works iff $q$ is an odd prime power. Otherwise $a mapsto a^2$ is an automorphism of $mathbb{F}_{2^n}^*$ so $x^2-delta in mathbb{F}_{2^n}[x]$ is never irreducible
$endgroup$
– reuns
Jan 8 at 8:23
$begingroup$
Just to say $mathbb{F}_{q^2} = mathbb{F}_q(sqrt{delta}) $ works iff $q$ is an odd prime power. Otherwise $a mapsto a^2$ is an automorphism of $mathbb{F}_{2^n}^*$ so $x^2-delta in mathbb{F}_{2^n}[x]$ is never irreducible
$endgroup$
– reuns
Jan 8 at 8:23
$begingroup$
true. I will edit. Thank you.
$endgroup$
– Malkoun
Jan 8 at 8:48
$begingroup$
true. I will edit. Thank you.
$endgroup$
– Malkoun
Jan 8 at 8:48
$begingroup$
Thank you very much. It is amazing to see the $Delta=b^2-4ac$ known from school is also working for finite fields ! I have tested in GAP that for $q=3$ matrix $pmatrix{1&1\1&0}$ is generator of $F_9$ and matrix $j=pmatrix{1&0\-1&-1}$ will work for Cayley-Dickson. The one I mentioned in the question does not work.
$endgroup$
– Marek Mitros
Jan 8 at 12:40
$begingroup$
Thank you very much. It is amazing to see the $Delta=b^2-4ac$ known from school is also working for finite fields ! I have tested in GAP that for $q=3$ matrix $pmatrix{1&1\1&0}$ is generator of $F_9$ and matrix $j=pmatrix{1&0\-1&-1}$ will work for Cayley-Dickson. The one I mentioned in the question does not work.
$endgroup$
– Marek Mitros
Jan 8 at 12:40
2
2
$begingroup$
Another way to describe an embedding would be: $mathbb{F}_{q^2}$ is a vector space of dimension two over $mathbb{F}_q$; and also, for any $alpha in mathbb{F}_{q^2}$, multiplication by $alpha$ is a linear operator on this vector space. So, if you take a basis of $mathbb{F}_{q^2}$ - for example, ${ 1, beta }$ will work whenever $beta in mathbb{F}_{q^2} setminus mathbb{F}_q$ - then the matrix representations of $alpha cdot$ with respect to this basis will form the desired embedding.
$endgroup$
– Daniel Schepler
Jan 8 at 22:42
$begingroup$
Another way to describe an embedding would be: $mathbb{F}_{q^2}$ is a vector space of dimension two over $mathbb{F}_q$; and also, for any $alpha in mathbb{F}_{q^2}$, multiplication by $alpha$ is a linear operator on this vector space. So, if you take a basis of $mathbb{F}_{q^2}$ - for example, ${ 1, beta }$ will work whenever $beta in mathbb{F}_{q^2} setminus mathbb{F}_q$ - then the matrix representations of $alpha cdot$ with respect to this basis will form the desired embedding.
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– Daniel Schepler
Jan 8 at 22:42
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@DanielSchepler, yes, this is the cleaner argument I was referring to. Thank you. I wrote it this way, because it is a little more explicit. If I was writing an article, I would have done it the cleaner way, using the argument you gave.
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– Malkoun
Jan 9 at 8:06
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@DanielSchepler, yes, this is the cleaner argument I was referring to. Thank you. I wrote it this way, because it is a little more explicit. If I was writing an article, I would have done it the cleaner way, using the argument you gave.
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– Malkoun
Jan 9 at 8:06
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– Aloizio Macedo♦
Jan 9 at 1:33