Cyclotomic scheme is a Association schemeIrreducible cyclotomic polynomialCyclotomic polynomials, irreducibilityIs this proof that $mathbb F_q^*$ is cyclic correct?elementary properties of cyclotomic polynomialsFinding units in cyclotomic fieldsExample of non-commutative association schemereferences of discrete association schemeirreducibility of cyclotomic polynomialscyclotomic field automorphismCyclotomic cosets and minimal polynomials
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Cyclotomic scheme is a Association scheme
Irreducible cyclotomic polynomialCyclotomic polynomials, irreducibilityIs this proof that $mathbb F_q^*$ is cyclic correct?elementary properties of cyclotomic polynomialsFinding units in cyclotomic fieldsExample of non-commutative association schemereferences of discrete association schemeirreducibility of cyclotomic polynomialscyclotomic field automorphismCyclotomic cosets and minimal polynomials
$begingroup$
I try to show that the following defines an association scheme:
Let $mathbbF_q$ be a field, $omega$ a primitive element of $mathbbF_q^times$ and $s$ divides $q-1$. Define $r=fracq-1s$, $C_0=0$, $C_1=langle omega^srangle=omega^sk:k=0,1,ldots,r-1$ and$$
C_i=omega^i-1C_1quadtext for i=2,ldots,s.
$$I have to show that the $C_i$ partition the set $mathbbF_q$. I tried to construct an group operations which has the $C_i$ as its orbits, but so far I did not succeed. Does anyone have an idea?
Sincerely,
Hypertrooper
combinatorics finite-fields algebraic-combinatorics association-schemes
edited Mar 29 at 11:31
azimut
16.5k1052101
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
$endgroup$
add a comment |
$begingroup$
I try to show that the following defines an association scheme:
Let $mathbbF_q$ be a field, $omega$ a primitive element of $mathbbF_q^times$ and $s$ divides $q-1$. Define $r=fracq-1s$, $C_0=0$, $C_1=langle omega^srangle=omega^sk:k=0,1,ldots,r-1$ and$$
C_i=omega^i-1C_1quadtext for i=2,ldots,s.
$$I have to show that the $C_i$ partition the set $mathbbF_q$. I tried to construct an group operations which has the $C_i$ as its orbits, but so far I did not succeed. Does anyone have an idea?
Sincerely,
Hypertrooper
combinatorics finite-fields algebraic-combinatorics association-schemes
edited Mar 29 at 11:31
azimut
16.5k1052101
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
$endgroup$
$begingroup$
See the answer below.
$endgroup$
– Hypertrooper
Jun 11 '18 at 16:10
add a comment |
$begingroup$
I try to show that the following defines an association scheme:
Let $mathbbF_q$ be a field, $omega$ a primitive element of $mathbbF_q^times$ and $s$ divides $q-1$. Define $r=fracq-1s$, $C_0=0$, $C_1=langle omega^srangle=omega^sk:k=0,1,ldots,r-1$ and$$
C_i=omega^i-1C_1quadtext for i=2,ldots,s.
$$I have to show that the $C_i$ partition the set $mathbbF_q$. I tried to construct an group operations which has the $C_i$ as its orbits, but so far I did not succeed. Does anyone have an idea?
Sincerely,
Hypertrooper
combinatorics finite-fields algebraic-combinatorics association-schemes
edited Mar 29 at 11:31
azimut
16.5k1052101
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
$endgroup$
I try to show that the following defines an association scheme:
Let $mathbbF_q$ be a field, $omega$ a primitive element of $mathbbF_q^times$ and $s$ divides $q-1$. Define $r=fracq-1s$, $C_0=0$, $C_1=langle omega^srangle=omega^sk:k=0,1,ldots,r-1$ and$$
C_i=omega^i-1C_1quadtext for i=2,ldots,s.
$$I have to show that the $C_i$ partition the set $mathbbF_q$. I tried to construct an group operations which has the $C_i$ as its orbits, but so far I did not succeed. Does anyone have an idea?
Sincerely,
Hypertrooper
combinatorics finite-fields algebraic-combinatorics association-schemes
combinatorics finite-fields algebraic-combinatorics association-schemes
edited Mar 29 at 11:31
azimut
16.5k1052101
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
edited Mar 29 at 11:31
azimut
16.5k1052101
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
edited Mar 29 at 11:31
azimut
16.5k1052101
edited Mar 29 at 11:31
azimut
16.5k1052101
edited Mar 29 at 11:31
azimut
16.5k1052101
16.5k1052101
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
asked Jun 11 '18 at 9:38
HypertrooperHypertrooper
797
797
$begingroup$
See the answer below.
$endgroup$
– Hypertrooper
Jun 11 '18 at 16:10
add a comment |
$begingroup$
See the answer below.
$endgroup$
– Hypertrooper
Jun 11 '18 at 16:10
$begingroup$
See the answer below.
$endgroup$
– Hypertrooper
Jun 11 '18 at 16:10
$begingroup$
See the answer below.
$endgroup$
– Hypertrooper
Jun 11 '18 at 16:10
add a comment |
1 Answer
1
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oldest
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$begingroup$
Sometimes, it is to difficult to see the easiest solution. The $C_i$ are the equivalence classes of the quotient $mathbbF_q^*/C_1$.
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
$endgroup$
add a comment |
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$begingroup$
Sometimes, it is to difficult to see the easiest solution. The $C_i$ are the equivalence classes of the quotient $mathbbF_q^*/C_1$.
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
$endgroup$
add a comment |
$begingroup$
Sometimes, it is to difficult to see the easiest solution. The $C_i$ are the equivalence classes of the quotient $mathbbF_q^*/C_1$.
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
$endgroup$
add a comment |
$begingroup$
Sometimes, it is to difficult to see the easiest solution. The $C_i$ are the equivalence classes of the quotient $mathbbF_q^*/C_1$.
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
$endgroup$
Sometimes, it is to difficult to see the easiest solution. The $C_i$ are the equivalence classes of the quotient $mathbbF_q^*/C_1$.
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
answered Jun 11 '18 at 16:12
HypertrooperHypertrooper
797
797
add a comment |
add a comment |
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– Hypertrooper
Jun 11 '18 at 16:10