Cyclotomic cosets and quadratic residues
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Let $p$ be a prime and let $Q$ be the set of quadratic residues and $N$ the set of nonresidues. Assume $2 in Q$. When I look a the cyclotomic cosets mod $p$, I get ${{0}}, {{Q}}, {{N}}.$ For example, for $p = 7,17,23$. Is this true in all cases?
What's the theory behind this?
number-theory quadratic-residues cyclotomic-fields
asked Nov 26 at 20:55
the man
672715
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show 2 more comments
up vote
0
down vote
favorite
Let $p$ be a prime and let $Q$ be the set of quadratic residues and $N$ the set of nonresidues. Assume $2 in Q$. When I look a the cyclotomic cosets mod $p$, I get ${{0}}, {{Q}}, {{N}}.$ For example, for $p = 7,17,23$. Is this true in all cases?
What's the theory behind this?
number-theory quadratic-residues cyclotomic-fields
asked Nov 26 at 20:55
the man
672715
The cyclotomic coset of $n bmod p$ containing $i$ is ${ i n^l bmod p| l ge 0}$ ?
– reuns
Nov 26 at 21:05
Yes? I still don’t understand why the above is true, though?
– the man
Nov 26 at 23:13
.. which $n,i,p$ are you considering
– reuns
Nov 26 at 23:14
I’m looking at the cyclotomic cosets mod $p$, with $n=2$.
– the man
Nov 26 at 23:17
$Q$ is a subgroup of $(mathbb{Z}/pmathbb{Z})^times$ thus a cyclic group. What do you get for ${ i n^l bmod p| l ge 0}$ when $n$ a generator of $Q$ ? If $n in Q$ then ${ n^l bmod p| l ge 0} subset Q$, what do you get when $n$ isn't a generator of $Q$ ?
– reuns
Nov 26 at 23:20
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $p$ be a prime and let $Q$ be the set of quadratic residues and $N$ the set of nonresidues. Assume $2 in Q$. When I look a the cyclotomic cosets mod $p$, I get ${{0}}, {{Q}}, {{N}}.$ For example, for $p = 7,17,23$. Is this true in all cases?
What's the theory behind this?
number-theory quadratic-residues cyclotomic-fields
asked Nov 26 at 20:55
the man
672715
Let $p$ be a prime and let $Q$ be the set of quadratic residues and $N$ the set of nonresidues. Assume $2 in Q$. When I look a the cyclotomic cosets mod $p$, I get ${{0}}, {{Q}}, {{N}}.$ For example, for $p = 7,17,23$. Is this true in all cases?
What's the theory behind this?
number-theory quadratic-residues cyclotomic-fields
number-theory quadratic-residues cyclotomic-fields
asked Nov 26 at 20:55
the man
672715
asked Nov 26 at 20:55
the man
672715
asked Nov 26 at 20:55
the man
672715
asked Nov 26 at 20:55
the man
672715
asked Nov 26 at 20:55
the man
672715
672715
The cyclotomic coset of $n bmod p$ containing $i$ is ${ i n^l bmod p| l ge 0}$ ?
– reuns
Nov 26 at 21:05
Yes? I still don’t understand why the above is true, though?
– the man
Nov 26 at 23:13
.. which $n,i,p$ are you considering
– reuns
Nov 26 at 23:14
I’m looking at the cyclotomic cosets mod $p$, with $n=2$.
– the man
Nov 26 at 23:17
$Q$ is a subgroup of $(mathbb{Z}/pmathbb{Z})^times$ thus a cyclic group. What do you get for ${ i n^l bmod p| l ge 0}$ when $n$ a generator of $Q$ ? If $n in Q$ then ${ n^l bmod p| l ge 0} subset Q$, what do you get when $n$ isn't a generator of $Q$ ?
– reuns
Nov 26 at 23:20
|
show 2 more comments
The cyclotomic coset of $n bmod p$ containing $i$ is ${ i n^l bmod p| l ge 0}$ ?
– reuns
Nov 26 at 21:05
Yes? I still don’t understand why the above is true, though?
– the man
Nov 26 at 23:13
.. which $n,i,p$ are you considering
– reuns
Nov 26 at 23:14
I’m looking at the cyclotomic cosets mod $p$, with $n=2$.
– the man
Nov 26 at 23:17
$Q$ is a subgroup of $(mathbb{Z}/pmathbb{Z})^times$ thus a cyclic group. What do you get for ${ i n^l bmod p| l ge 0}$ when $n$ a generator of $Q$ ? If $n in Q$ then ${ n^l bmod p| l ge 0} subset Q$, what do you get when $n$ isn't a generator of $Q$ ?
– reuns
Nov 26 at 23:20
The cyclotomic coset of $n bmod p$ containing $i$ is ${ i n^l bmod p| l ge 0}$ ?
– reuns
Nov 26 at 21:05
The cyclotomic coset of $n bmod p$ containing $i$ is ${ i n^l bmod p| l ge 0}$ ?
– reuns
Nov 26 at 21:05
Yes? I still don’t understand why the above is true, though?
– the man
Nov 26 at 23:13
Yes? I still don’t understand why the above is true, though?
– the man
Nov 26 at 23:13
.. which $n,i,p$ are you considering
– reuns
Nov 26 at 23:14
.. which $n,i,p$ are you considering
– reuns
Nov 26 at 23:14
I’m looking at the cyclotomic cosets mod $p$, with $n=2$.
– the man
Nov 26 at 23:17
I’m looking at the cyclotomic cosets mod $p$, with $n=2$.
– the man
Nov 26 at 23:17
$Q$ is a subgroup of $(mathbb{Z}/pmathbb{Z})^times$ thus a cyclic group. What do you get for ${ i n^l bmod p| l ge 0}$ when $n$ a generator of $Q$ ? If $n in Q$ then ${ n^l bmod p| l ge 0} subset Q$, what do you get when $n$ isn't a generator of $Q$ ?
– reuns
Nov 26 at 23:20
$Q$ is a subgroup of $(mathbb{Z}/pmathbb{Z})^times$ thus a cyclic group. What do you get for ${ i n^l bmod p| l ge 0}$ when $n$ a generator of $Q$ ? If $n in Q$ then ${ n^l bmod p| l ge 0} subset Q$, what do you get when $n$ isn't a generator of $Q$ ?
– reuns
Nov 26 at 23:20
|
show 2 more comments
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The cyclotomic coset of $n bmod p$ containing $i$ is ${ i n^l bmod p| l ge 0}$ ?
– reuns
Nov 26 at 21:05
Yes? I still don’t understand why the above is true, though?
– the man
Nov 26 at 23:13
.. which $n,i,p$ are you considering
– reuns
Nov 26 at 23:14
I’m looking at the cyclotomic cosets mod $p$, with $n=2$.
– the man
Nov 26 at 23:17
$Q$ is a subgroup of $(mathbb{Z}/pmathbb{Z})^times$ thus a cyclic group. What do you get for ${ i n^l bmod p| l ge 0}$ when $n$ a generator of $Q$ ? If $n in Q$ then ${ n^l bmod p| l ge 0} subset Q$, what do you get when $n$ isn't a generator of $Q$ ?
– reuns
Nov 26 at 23:20