Mixing
Generator of $mathbb{F}_{q}^*$ where $q$ is not prime.
0
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I'm trying to find the generator of the multiplicative group $mathbb{F}_{8}^*$, where $mathbb{F}_8$ is a field. If the order of the field is prime, then it is easy since in that case the field would be isomorphic to the integers modulo, say p. However, for cases where the order of the field is not prime, I'm stuck. Can anyone help me out?
Thanks in advance.
abstract-algebra field-theory
edited Jan 1 at 12:25
toric_actions
898
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
$endgroup$
add a comment |
0
$begingroup$
I'm trying to find the generator of the multiplicative group $mathbb{F}_{8}^*$, where $mathbb{F}_8$ is a field. If the order of the field is prime, then it is easy since in that case the field would be isomorphic to the integers modulo, say p. However, for cases where the order of the field is not prime, I'm stuck. Can anyone help me out?
Thanks in advance.
abstract-algebra field-theory
edited Jan 1 at 12:25
toric_actions
898
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
$endgroup$
3
$begingroup$
By $Bbb F_q^x$ do you mean the multiplicative group of nonzero elements of $Bbb F_q$? This is usually denoted as $Bbb F_q^times$ or $Bbb F_q^*$. When $q=8$, it has seven elements, and seven is a prime number, so any element other than the identity is a generator.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:56
$begingroup$
@Xenidia Can you please explain your multiplicative group $mathbb{F}^x_q$? Is it the same as $mathbb{F}^*_q$
$endgroup$
– toric_actions
Jan 1 at 12:04
$begingroup$
See also the answers to this question on $Bbb{F}_q^{times}$. In general, see also this duplicate.
$endgroup$
– Dietrich Burde
Jan 1 at 13:56
$begingroup$
So $t + (t^3+t+1)$ is a generator of $mathbb{F}_2[t]/(t^3+t+1)^times cong mathbb{F}_8^times$
$endgroup$
– reuns
Jan 1 at 17:02
add a comment |
0
0
0
$begingroup$
I'm trying to find the generator of the multiplicative group $mathbb{F}_{8}^*$, where $mathbb{F}_8$ is a field. If the order of the field is prime, then it is easy since in that case the field would be isomorphic to the integers modulo, say p. However, for cases where the order of the field is not prime, I'm stuck. Can anyone help me out?
Thanks in advance.
abstract-algebra field-theory
edited Jan 1 at 12:25
toric_actions
898
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
$endgroup$
I'm trying to find the generator of the multiplicative group $mathbb{F}_{8}^*$, where $mathbb{F}_8$ is a field. If the order of the field is prime, then it is easy since in that case the field would be isomorphic to the integers modulo, say p. However, for cases where the order of the field is not prime, I'm stuck. Can anyone help me out?
Thanks in advance.
abstract-algebra field-theory
abstract-algebra field-theory
edited Jan 1 at 12:25
toric_actions
898
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
edited Jan 1 at 12:25
toric_actions
898
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
edited Jan 1 at 12:25
toric_actions
898
edited Jan 1 at 12:25
toric_actions
898
edited Jan 1 at 12:25
toric_actions
898
898
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
asked Jan 1 at 11:54
XenidiaXenidia
1,275729
1,275729
3
$begingroup$
By $Bbb F_q^x$ do you mean the multiplicative group of nonzero elements of $Bbb F_q$? This is usually denoted as $Bbb F_q^times$ or $Bbb F_q^*$. When $q=8$, it has seven elements, and seven is a prime number, so any element other than the identity is a generator.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:56
$begingroup$
@Xenidia Can you please explain your multiplicative group $mathbb{F}^x_q$? Is it the same as $mathbb{F}^*_q$
$endgroup$
– toric_actions
Jan 1 at 12:04
$begingroup$
See also the answers to this question on $Bbb{F}_q^{times}$. In general, see also this duplicate.
$endgroup$
– Dietrich Burde
Jan 1 at 13:56
$begingroup$
So $t + (t^3+t+1)$ is a generator of $mathbb{F}_2[t]/(t^3+t+1)^times cong mathbb{F}_8^times$
$endgroup$
– reuns
Jan 1 at 17:02
add a comment |
3
$begingroup$
By $Bbb F_q^x$ do you mean the multiplicative group of nonzero elements of $Bbb F_q$? This is usually denoted as $Bbb F_q^times$ or $Bbb F_q^*$. When $q=8$, it has seven elements, and seven is a prime number, so any element other than the identity is a generator.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:56
$begingroup$
@Xenidia Can you please explain your multiplicative group $mathbb{F}^x_q$? Is it the same as $mathbb{F}^*_q$
$endgroup$
– toric_actions
Jan 1 at 12:04
$begingroup$
See also the answers to this question on $Bbb{F}_q^{times}$. In general, see also this duplicate.
$endgroup$
– Dietrich Burde
Jan 1 at 13:56
$begingroup$
So $t + (t^3+t+1)$ is a generator of $mathbb{F}_2[t]/(t^3+t+1)^times cong mathbb{F}_8^times$
$endgroup$
– reuns
Jan 1 at 17:02
3
3
$begingroup$
By $Bbb F_q^x$ do you mean the multiplicative group of nonzero elements of $Bbb F_q$? This is usually denoted as $Bbb F_q^times$ or $Bbb F_q^*$. When $q=8$, it has seven elements, and seven is a prime number, so any element other than the identity is a generator.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:56
$begingroup$
By $Bbb F_q^x$ do you mean the multiplicative group of nonzero elements of $Bbb F_q$? This is usually denoted as $Bbb F_q^times$ or $Bbb F_q^*$. When $q=8$, it has seven elements, and seven is a prime number, so any element other than the identity is a generator.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:56
$begingroup$
@Xenidia Can you please explain your multiplicative group $mathbb{F}^x_q$? Is it the same as $mathbb{F}^*_q$
$endgroup$
– toric_actions
Jan 1 at 12:04
$begingroup$
@Xenidia Can you please explain your multiplicative group $mathbb{F}^x_q$? Is it the same as $mathbb{F}^*_q$
$endgroup$
– toric_actions
Jan 1 at 12:04
$begingroup$
See also the answers to this question on $Bbb{F}_q^{times}$. In general, see also this duplicate.
$endgroup$
– Dietrich Burde
Jan 1 at 13:56
$begingroup$
See also the answers to this question on $Bbb{F}_q^{times}$. In general, see also this duplicate.
$endgroup$
– Dietrich Burde
Jan 1 at 13:56
$begingroup$
So $t + (t^3+t+1)$ is a generator of $mathbb{F}_2[t]/(t^3+t+1)^times cong mathbb{F}_8^times$
$endgroup$
– reuns
Jan 1 at 17:02
$begingroup$
So $t + (t^3+t+1)$ is a generator of $mathbb{F}_2[t]/(t^3+t+1)^times cong mathbb{F}_8^times$
$endgroup$
– reuns
Jan 1 at 17:02
add a comment |
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3
$begingroup$
By $Bbb F_q^x$ do you mean the multiplicative group of nonzero elements of $Bbb F_q$? This is usually denoted as $Bbb F_q^times$ or $Bbb F_q^*$. When $q=8$, it has seven elements, and seven is a prime number, so any element other than the identity is a generator.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:56
$begingroup$
@Xenidia Can you please explain your multiplicative group $mathbb{F}^x_q$? Is it the same as $mathbb{F}^*_q$
$endgroup$
– toric_actions
Jan 1 at 12:04
$begingroup$
See also the answers to this question on $Bbb{F}_q^{times}$. In general, see also this duplicate.
$endgroup$
– Dietrich Burde
Jan 1 at 13:56
$begingroup$
So $t + (t^3+t+1)$ is a generator of $mathbb{F}_2[t]/(t^3+t+1)^times cong mathbb{F}_8^times$
$endgroup$
– reuns
Jan 1 at 17:02