Show that $|A|$ divides $q-1$ for $Ain SL(2,mathbbF_q)$ if $x^2-texttr(A)x+1=0$ has distinct solutions in $mathbbF_q$
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This problem has been giving me a lot of trouble. Let $mathbbF_q$ be the finite field of order $q$. Let $Ain SL(2,mathbbF_q)$, the multiplicative group of $2times 2$ matrices with determinant $1$. Note that the characteristic equation of $A$ is $x^2-texttr(A)x+1=0$. Show that if the characteristic equation for $A$ has distinct solutions in $mathbbF_q$, then $|A|$ divides $q-1$, and if the characteristic equation of $A$ does not have any solutions in $mathbbF_q$, then $|A|$ divides $q+1$.
A piece of useful information is the fact that $SL(2,mathbbF_q)$ has an order of $q^3-q$. Showing that $|A|$ divides $qpm 1$ is equivalent to showing that $A^qpm 1=I_2$, the $2times 2$ identity matrix. I'm not sure where to go from here though. I get the impression this problem may involve field extensions as well.
abstract-algebra matrices finite-fields
asked Aug 18 at 22:20
Julian Benali
23713
This question has an open bounty worth +50
reputation from Julian Benali ending ending at 2018-09-04 23:34:36Z">in 4 days.
The current answers do not contain enough detail.
I would like an explanation or proof with enough information to follow along to the conclusion.
add a comment |Â
up vote
2
down vote
favorite
This problem has been giving me a lot of trouble. Let $mathbbF_q$ be the finite field of order $q$. Let $Ain SL(2,mathbbF_q)$, the multiplicative group of $2times 2$ matrices with determinant $1$. Note that the characteristic equation of $A$ is $x^2-texttr(A)x+1=0$. Show that if the characteristic equation for $A$ has distinct solutions in $mathbbF_q$, then $|A|$ divides $q-1$, and if the characteristic equation of $A$ does not have any solutions in $mathbbF_q$, then $|A|$ divides $q+1$.
A piece of useful information is the fact that $SL(2,mathbbF_q)$ has an order of $q^3-q$. Showing that $|A|$ divides $qpm 1$ is equivalent to showing that $A^qpm 1=I_2$, the $2times 2$ identity matrix. I'm not sure where to go from here though. I get the impression this problem may involve field extensions as well.
abstract-algebra matrices finite-fields
asked Aug 18 at 22:20
Julian Benali
23713
This question has an open bounty worth +50
reputation from Julian Benali ending ending at 2018-09-04 23:34:36Z">in 4 days.
The current answers do not contain enough detail.
I would like an explanation or proof with enough information to follow along to the conclusion.
1
what does $|A|$ mean ?
– Will Jagy
Aug 18 at 22:25
1
Perhaps the order of $A$ as an element of $SL(2,mathbbF_q)$?
– Joshua Mundinger
Aug 18 at 22:48
Yes, the order of $A$. Sorry, I see how that may look like cardinality.
– Julian Benali
Aug 19 at 3:13
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This problem has been giving me a lot of trouble. Let $mathbbF_q$ be the finite field of order $q$. Let $Ain SL(2,mathbbF_q)$, the multiplicative group of $2times 2$ matrices with determinant $1$. Note that the characteristic equation of $A$ is $x^2-texttr(A)x+1=0$. Show that if the characteristic equation for $A$ has distinct solutions in $mathbbF_q$, then $|A|$ divides $q-1$, and if the characteristic equation of $A$ does not have any solutions in $mathbbF_q$, then $|A|$ divides $q+1$.
A piece of useful information is the fact that $SL(2,mathbbF_q)$ has an order of $q^3-q$. Showing that $|A|$ divides $qpm 1$ is equivalent to showing that $A^qpm 1=I_2$, the $2times 2$ identity matrix. I'm not sure where to go from here though. I get the impression this problem may involve field extensions as well.
abstract-algebra matrices finite-fields
asked Aug 18 at 22:20
Julian Benali
23713
This problem has been giving me a lot of trouble. Let $mathbbF_q$ be the finite field of order $q$. Let $Ain SL(2,mathbbF_q)$, the multiplicative group of $2times 2$ matrices with determinant $1$. Note that the characteristic equation of $A$ is $x^2-texttr(A)x+1=0$. Show that if the characteristic equation for $A$ has distinct solutions in $mathbbF_q$, then $|A|$ divides $q-1$, and if the characteristic equation of $A$ does not have any solutions in $mathbbF_q$, then $|A|$ divides $q+1$.
A piece of useful information is the fact that $SL(2,mathbbF_q)$ has an order of $q^3-q$. Showing that $|A|$ divides $qpm 1$ is equivalent to showing that $A^qpm 1=I_2$, the $2times 2$ identity matrix. I'm not sure where to go from here though. I get the impression this problem may involve field extensions as well.
abstract-algebra matrices finite-fields
asked Aug 18 at 22:20
Julian Benali
23713
edited Aug 19 at 3:37
asked Aug 18 at 22:20
Julian Benali
23713
asked Aug 18 at 22:20
Julian Benali
23713
asked Aug 18 at 22:20
Julian Benali
23713
23713
This question has an open bounty worth +50
reputation from Julian Benali ending ending at 2018-09-04 23:34:36Z">in 4 days.
The current answers do not contain enough detail.
I would like an explanation or proof with enough information to follow along to the conclusion.
This question has an open bounty worth +50
reputation from Julian Benali ending ending at 2018-09-04 23:34:36Z">in 4 days.
The current answers do not contain enough detail.
I would like an explanation or proof with enough information to follow along to the conclusion.
1
what does $|A|$ mean ?
– Will Jagy
Aug 18 at 22:25
1
Perhaps the order of $A$ as an element of $SL(2,mathbbF_q)$?
– Joshua Mundinger
Aug 18 at 22:48
Yes, the order of $A$. Sorry, I see how that may look like cardinality.
– Julian Benali
Aug 19 at 3:13
add a comment |Â
1
what does $|A|$ mean ?
– Will Jagy
Aug 18 at 22:25
1
Perhaps the order of $A$ as an element of $SL(2,mathbbF_q)$?
– Joshua Mundinger
Aug 18 at 22:48
Yes, the order of $A$. Sorry, I see how that may look like cardinality.
– Julian Benali
Aug 19 at 3:13
1
1
what does $|A|$ mean ?
– Will Jagy
Aug 18 at 22:25
what does $|A|$ mean ?
– Will Jagy
Aug 18 at 22:25
1
1
Perhaps the order of $A$ as an element of $SL(2,mathbbF_q)$?
– Joshua Mundinger
Aug 18 at 22:48
Perhaps the order of $A$ as an element of $SL(2,mathbbF_q)$?
– Joshua Mundinger
Aug 18 at 22:48
Yes, the order of $A$. Sorry, I see how that may look like cardinality.
– Julian Benali
Aug 19 at 3:13
Yes, the order of $A$. Sorry, I see how that may look like cardinality.
– Julian Benali
Aug 19 at 3:13
add a comment |Â
1 Answer
1
active
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votes
up vote
5
down vote
If the (distinct) eigenvalues $alpha$ and $beta$ lie in $Bbb F_q$ then
$alpha^q-1=beta^q-1=1$.
If $alpha$ and $beta$ are outside $Bbb F_q$ then $beta=alpha^q$
(Frobenius automorphism) and $alphabeta=1$ (determinant).
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
2
I would like to add that, in the case where $alpha$ and $beta$ are in $mathbbF_q$, the assumption that they be distinct is quite important. For example, let $g$ be a generator of the multiplicative group $mathbbF_p^times$ and $$A:=beginbmatrix -1&1\0&-1endbmatrix,,$$ then the order of $A$ is precisely $2p$ if $q=p^r$, where $p$ is an odd prime natural number and $rinmathbbZ_>0$, and clearly, $2p$ divides neither $q-1$ nor $q+1$.
– Batominovski
Aug 18 at 22:38
Forgive me, but I don't quite follow everything. Clearly $alpha^q-1=beta^q-1=1$ when $alpha,betainmathbbF_q$ since $mathbbF_qsetminus0$ is a multiplicative group of order $q-1$, but I don't see why this implies $A^q-1$ is the identity matrix. I also fail to see how you got $beta=alpha^q$ from the Frobenius automorphism when $alpha,betainmathbbF_q^algsetminusmathbbF_q$ where $mathbbF_q^alg$ denotes the algebraic closure of $mathbbF_q$. Thank you!
– Julian Benali
Aug 19 at 3:36
1
$alpha^q-1$ and $beta^q-1$ are the eigenvalues of $A^q-1$. If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$.
– Lord Shark the Unknown
Aug 19 at 7:15
I see now that $alpha^q-1$ and $beta^q-1$ are eigenvalues of $A^q-1$. I had forgotten that the eigenvalues of the power of a matrix are the powers of its eigenvalues. However, just because the eigenvalues of $A^q-1$ are both $1$ does not mean $A^q-1$ is the identity matrix. For example, $$beginbmatrix2&1\ -1&0 endbmatrix$$ has eigenvalues of $1$. Finally, I don't see how you determined "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". Thank you for the help!
– Julian Benali
Aug 19 at 20:15
Actually, I have figured out your claim that "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". If $alpha$ is a root of the quadratic polynomial which does not lie in $mathbbF_q$, then the field extension $mathbbF_q(alpha)$ has degree $2$ over $mathbbF_q$. Furthermore, because $mathbbF_q$ and $mathbbF_q(alpha)$ have the same characteristic. Using the Frobenius automorphism $varphi$ you previously mentioned and the fact that it is the identity function over $mathbbF_q$, we have
– Julian Benali
Aug 19 at 20:34
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
If the (distinct) eigenvalues $alpha$ and $beta$ lie in $Bbb F_q$ then
$alpha^q-1=beta^q-1=1$.
If $alpha$ and $beta$ are outside $Bbb F_q$ then $beta=alpha^q$
(Frobenius automorphism) and $alphabeta=1$ (determinant).
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
2
I would like to add that, in the case where $alpha$ and $beta$ are in $mathbbF_q$, the assumption that they be distinct is quite important. For example, let $g$ be a generator of the multiplicative group $mathbbF_p^times$ and $$A:=beginbmatrix -1&1\0&-1endbmatrix,,$$ then the order of $A$ is precisely $2p$ if $q=p^r$, where $p$ is an odd prime natural number and $rinmathbbZ_>0$, and clearly, $2p$ divides neither $q-1$ nor $q+1$.
– Batominovski
Aug 18 at 22:38
Forgive me, but I don't quite follow everything. Clearly $alpha^q-1=beta^q-1=1$ when $alpha,betainmathbbF_q$ since $mathbbF_qsetminus0$ is a multiplicative group of order $q-1$, but I don't see why this implies $A^q-1$ is the identity matrix. I also fail to see how you got $beta=alpha^q$ from the Frobenius automorphism when $alpha,betainmathbbF_q^algsetminusmathbbF_q$ where $mathbbF_q^alg$ denotes the algebraic closure of $mathbbF_q$. Thank you!
– Julian Benali
Aug 19 at 3:36
1
$alpha^q-1$ and $beta^q-1$ are the eigenvalues of $A^q-1$. If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$.
– Lord Shark the Unknown
Aug 19 at 7:15
I see now that $alpha^q-1$ and $beta^q-1$ are eigenvalues of $A^q-1$. I had forgotten that the eigenvalues of the power of a matrix are the powers of its eigenvalues. However, just because the eigenvalues of $A^q-1$ are both $1$ does not mean $A^q-1$ is the identity matrix. For example, $$beginbmatrix2&1\ -1&0 endbmatrix$$ has eigenvalues of $1$. Finally, I don't see how you determined "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". Thank you for the help!
– Julian Benali
Aug 19 at 20:15
Actually, I have figured out your claim that "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". If $alpha$ is a root of the quadratic polynomial which does not lie in $mathbbF_q$, then the field extension $mathbbF_q(alpha)$ has degree $2$ over $mathbbF_q$. Furthermore, because $mathbbF_q$ and $mathbbF_q(alpha)$ have the same characteristic. Using the Frobenius automorphism $varphi$ you previously mentioned and the fact that it is the identity function over $mathbbF_q$, we have
– Julian Benali
Aug 19 at 20:34
 |Â
show 3 more comments
up vote
5
down vote
If the (distinct) eigenvalues $alpha$ and $beta$ lie in $Bbb F_q$ then
$alpha^q-1=beta^q-1=1$.
If $alpha$ and $beta$ are outside $Bbb F_q$ then $beta=alpha^q$
(Frobenius automorphism) and $alphabeta=1$ (determinant).
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
2
I would like to add that, in the case where $alpha$ and $beta$ are in $mathbbF_q$, the assumption that they be distinct is quite important. For example, let $g$ be a generator of the multiplicative group $mathbbF_p^times$ and $$A:=beginbmatrix -1&1\0&-1endbmatrix,,$$ then the order of $A$ is precisely $2p$ if $q=p^r$, where $p$ is an odd prime natural number and $rinmathbbZ_>0$, and clearly, $2p$ divides neither $q-1$ nor $q+1$.
– Batominovski
Aug 18 at 22:38
Forgive me, but I don't quite follow everything. Clearly $alpha^q-1=beta^q-1=1$ when $alpha,betainmathbbF_q$ since $mathbbF_qsetminus0$ is a multiplicative group of order $q-1$, but I don't see why this implies $A^q-1$ is the identity matrix. I also fail to see how you got $beta=alpha^q$ from the Frobenius automorphism when $alpha,betainmathbbF_q^algsetminusmathbbF_q$ where $mathbbF_q^alg$ denotes the algebraic closure of $mathbbF_q$. Thank you!
– Julian Benali
Aug 19 at 3:36
1
$alpha^q-1$ and $beta^q-1$ are the eigenvalues of $A^q-1$. If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$.
– Lord Shark the Unknown
Aug 19 at 7:15
I see now that $alpha^q-1$ and $beta^q-1$ are eigenvalues of $A^q-1$. I had forgotten that the eigenvalues of the power of a matrix are the powers of its eigenvalues. However, just because the eigenvalues of $A^q-1$ are both $1$ does not mean $A^q-1$ is the identity matrix. For example, $$beginbmatrix2&1\ -1&0 endbmatrix$$ has eigenvalues of $1$. Finally, I don't see how you determined "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". Thank you for the help!
– Julian Benali
Aug 19 at 20:15
Actually, I have figured out your claim that "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". If $alpha$ is a root of the quadratic polynomial which does not lie in $mathbbF_q$, then the field extension $mathbbF_q(alpha)$ has degree $2$ over $mathbbF_q$. Furthermore, because $mathbbF_q$ and $mathbbF_q(alpha)$ have the same characteristic. Using the Frobenius automorphism $varphi$ you previously mentioned and the fact that it is the identity function over $mathbbF_q$, we have
– Julian Benali
Aug 19 at 20:34
 |Â
show 3 more comments
up vote
5
down vote
up vote
5
down vote
If the (distinct) eigenvalues $alpha$ and $beta$ lie in $Bbb F_q$ then
$alpha^q-1=beta^q-1=1$.
If $alpha$ and $beta$ are outside $Bbb F_q$ then $beta=alpha^q$
(Frobenius automorphism) and $alphabeta=1$ (determinant).
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
If the (distinct) eigenvalues $alpha$ and $beta$ lie in $Bbb F_q$ then
$alpha^q-1=beta^q-1=1$.
If $alpha$ and $beta$ are outside $Bbb F_q$ then $beta=alpha^q$
(Frobenius automorphism) and $alphabeta=1$ (determinant).
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
answered Aug 18 at 22:25
Lord Shark the Unknown
87.5k953114
87.5k953114
2
I would like to add that, in the case where $alpha$ and $beta$ are in $mathbbF_q$, the assumption that they be distinct is quite important. For example, let $g$ be a generator of the multiplicative group $mathbbF_p^times$ and $$A:=beginbmatrix -1&1\0&-1endbmatrix,,$$ then the order of $A$ is precisely $2p$ if $q=p^r$, where $p$ is an odd prime natural number and $rinmathbbZ_>0$, and clearly, $2p$ divides neither $q-1$ nor $q+1$.
– Batominovski
Aug 18 at 22:38
Forgive me, but I don't quite follow everything. Clearly $alpha^q-1=beta^q-1=1$ when $alpha,betainmathbbF_q$ since $mathbbF_qsetminus0$ is a multiplicative group of order $q-1$, but I don't see why this implies $A^q-1$ is the identity matrix. I also fail to see how you got $beta=alpha^q$ from the Frobenius automorphism when $alpha,betainmathbbF_q^algsetminusmathbbF_q$ where $mathbbF_q^alg$ denotes the algebraic closure of $mathbbF_q$. Thank you!
– Julian Benali
Aug 19 at 3:36
1
$alpha^q-1$ and $beta^q-1$ are the eigenvalues of $A^q-1$. If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$.
– Lord Shark the Unknown
Aug 19 at 7:15
I see now that $alpha^q-1$ and $beta^q-1$ are eigenvalues of $A^q-1$. I had forgotten that the eigenvalues of the power of a matrix are the powers of its eigenvalues. However, just because the eigenvalues of $A^q-1$ are both $1$ does not mean $A^q-1$ is the identity matrix. For example, $$beginbmatrix2&1\ -1&0 endbmatrix$$ has eigenvalues of $1$. Finally, I don't see how you determined "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". Thank you for the help!
– Julian Benali
Aug 19 at 20:15
Actually, I have figured out your claim that "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". If $alpha$ is a root of the quadratic polynomial which does not lie in $mathbbF_q$, then the field extension $mathbbF_q(alpha)$ has degree $2$ over $mathbbF_q$. Furthermore, because $mathbbF_q$ and $mathbbF_q(alpha)$ have the same characteristic. Using the Frobenius automorphism $varphi$ you previously mentioned and the fact that it is the identity function over $mathbbF_q$, we have
– Julian Benali
Aug 19 at 20:34
 |Â
show 3 more comments
2
I would like to add that, in the case where $alpha$ and $beta$ are in $mathbbF_q$, the assumption that they be distinct is quite important. For example, let $g$ be a generator of the multiplicative group $mathbbF_p^times$ and $$A:=beginbmatrix -1&1\0&-1endbmatrix,,$$ then the order of $A$ is precisely $2p$ if $q=p^r$, where $p$ is an odd prime natural number and $rinmathbbZ_>0$, and clearly, $2p$ divides neither $q-1$ nor $q+1$.
– Batominovski
Aug 18 at 22:38
Forgive me, but I don't quite follow everything. Clearly $alpha^q-1=beta^q-1=1$ when $alpha,betainmathbbF_q$ since $mathbbF_qsetminus0$ is a multiplicative group of order $q-1$, but I don't see why this implies $A^q-1$ is the identity matrix. I also fail to see how you got $beta=alpha^q$ from the Frobenius automorphism when $alpha,betainmathbbF_q^algsetminusmathbbF_q$ where $mathbbF_q^alg$ denotes the algebraic closure of $mathbbF_q$. Thank you!
– Julian Benali
Aug 19 at 3:36
1
$alpha^q-1$ and $beta^q-1$ are the eigenvalues of $A^q-1$. If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$.
– Lord Shark the Unknown
Aug 19 at 7:15
I see now that $alpha^q-1$ and $beta^q-1$ are eigenvalues of $A^q-1$. I had forgotten that the eigenvalues of the power of a matrix are the powers of its eigenvalues. However, just because the eigenvalues of $A^q-1$ are both $1$ does not mean $A^q-1$ is the identity matrix. For example, $$beginbmatrix2&1\ -1&0 endbmatrix$$ has eigenvalues of $1$. Finally, I don't see how you determined "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". Thank you for the help!
– Julian Benali
Aug 19 at 20:15
Actually, I have figured out your claim that "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". If $alpha$ is a root of the quadratic polynomial which does not lie in $mathbbF_q$, then the field extension $mathbbF_q(alpha)$ has degree $2$ over $mathbbF_q$. Furthermore, because $mathbbF_q$ and $mathbbF_q(alpha)$ have the same characteristic. Using the Frobenius automorphism $varphi$ you previously mentioned and the fact that it is the identity function over $mathbbF_q$, we have
– Julian Benali
Aug 19 at 20:34
2
2
I would like to add that, in the case where $alpha$ and $beta$ are in $mathbbF_q$, the assumption that they be distinct is quite important. For example, let $g$ be a generator of the multiplicative group $mathbbF_p^times$ and $$A:=beginbmatrix -1&1\0&-1endbmatrix,,$$ then the order of $A$ is precisely $2p$ if $q=p^r$, where $p$ is an odd prime natural number and $rinmathbbZ_>0$, and clearly, $2p$ divides neither $q-1$ nor $q+1$.
– Batominovski
Aug 18 at 22:38
I would like to add that, in the case where $alpha$ and $beta$ are in $mathbbF_q$, the assumption that they be distinct is quite important. For example, let $g$ be a generator of the multiplicative group $mathbbF_p^times$ and $$A:=beginbmatrix -1&1\0&-1endbmatrix,,$$ then the order of $A$ is precisely $2p$ if $q=p^r$, where $p$ is an odd prime natural number and $rinmathbbZ_>0$, and clearly, $2p$ divides neither $q-1$ nor $q+1$.
– Batominovski
Aug 18 at 22:38
Forgive me, but I don't quite follow everything. Clearly $alpha^q-1=beta^q-1=1$ when $alpha,betainmathbbF_q$ since $mathbbF_qsetminus0$ is a multiplicative group of order $q-1$, but I don't see why this implies $A^q-1$ is the identity matrix. I also fail to see how you got $beta=alpha^q$ from the Frobenius automorphism when $alpha,betainmathbbF_q^algsetminusmathbbF_q$ where $mathbbF_q^alg$ denotes the algebraic closure of $mathbbF_q$. Thank you!
– Julian Benali
Aug 19 at 3:36
Forgive me, but I don't quite follow everything. Clearly $alpha^q-1=beta^q-1=1$ when $alpha,betainmathbbF_q$ since $mathbbF_qsetminus0$ is a multiplicative group of order $q-1$, but I don't see why this implies $A^q-1$ is the identity matrix. I also fail to see how you got $beta=alpha^q$ from the Frobenius automorphism when $alpha,betainmathbbF_q^algsetminusmathbbF_q$ where $mathbbF_q^alg$ denotes the algebraic closure of $mathbbF_q$. Thank you!
– Julian Benali
Aug 19 at 3:36
1
1
$alpha^q-1$ and $beta^q-1$ are the eigenvalues of $A^q-1$. If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$.
– Lord Shark the Unknown
Aug 19 at 7:15
$alpha^q-1$ and $beta^q-1$ are the eigenvalues of $A^q-1$. If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$.
– Lord Shark the Unknown
Aug 19 at 7:15
I see now that $alpha^q-1$ and $beta^q-1$ are eigenvalues of $A^q-1$. I had forgotten that the eigenvalues of the power of a matrix are the powers of its eigenvalues. However, just because the eigenvalues of $A^q-1$ are both $1$ does not mean $A^q-1$ is the identity matrix. For example, $$beginbmatrix2&1\ -1&0 endbmatrix$$ has eigenvalues of $1$. Finally, I don't see how you determined "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". Thank you for the help!
– Julian Benali
Aug 19 at 20:15
I see now that $alpha^q-1$ and $beta^q-1$ are eigenvalues of $A^q-1$. I had forgotten that the eigenvalues of the power of a matrix are the powers of its eigenvalues. However, just because the eigenvalues of $A^q-1$ are both $1$ does not mean $A^q-1$ is the identity matrix. For example, $$beginbmatrix2&1\ -1&0 endbmatrix$$ has eigenvalues of $1$. Finally, I don't see how you determined "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". Thank you for the help!
– Julian Benali
Aug 19 at 20:15
Actually, I have figured out your claim that "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". If $alpha$ is a root of the quadratic polynomial which does not lie in $mathbbF_q$, then the field extension $mathbbF_q(alpha)$ has degree $2$ over $mathbbF_q$. Furthermore, because $mathbbF_q$ and $mathbbF_q(alpha)$ have the same characteristic. Using the Frobenius automorphism $varphi$ you previously mentioned and the fact that it is the identity function over $mathbbF_q$, we have
– Julian Benali
Aug 19 at 20:34
Actually, I have figured out your claim that "If your quadratic polynomial is irreducible, then if $alpha$ is a zero, then so is $alpha^q$". If $alpha$ is a root of the quadratic polynomial which does not lie in $mathbbF_q$, then the field extension $mathbbF_q(alpha)$ has degree $2$ over $mathbbF_q$. Furthermore, because $mathbbF_q$ and $mathbbF_q(alpha)$ have the same characteristic. Using the Frobenius automorphism $varphi$ you previously mentioned and the fact that it is the identity function over $mathbbF_q$, we have
– Julian Benali
Aug 19 at 20:34
 |Â
show 3 more comments
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1
what does $|A|$ mean ?
– Will Jagy
Aug 18 at 22:25
1
Perhaps the order of $A$ as an element of $SL(2,mathbbF_q)$?
– Joshua Mundinger
Aug 18 at 22:48
Yes, the order of $A$. Sorry, I see how that may look like cardinality.
– Julian Benali
Aug 19 at 3:13