Mixing
![What are the Characteristics of the Ring $mathbf F_p[X]/g$ (mod $g$, $g$ is non-irreducible)?](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Relationship between cyclotomic polynomials
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let $n, m$ be two natural numbers and $Phi_n(q), Phi_m(q)$ the $n$-th and $m$-th cyclotomic polynomials respectively. Define a function $c_n,m colon mathbbN to left0,1rightcup left p mid p , textprimeright$ by
$$c_m,n =
begincases
0, , & textif $n = m$\
p, , & textif $fracnm = p^j$ where $p$ prime and $j in mathbbZsetminus 0$\
1, , & textif $fracnm$ is not an integer power of a prime
endcases$$
Show that $Phi_m(q) in sqrtleft( Phi_n(q)right) + (c_m,n)[q]$
This result should rely on properties of cyclotomic polynomials I am not aware of. Obviously if $c_n,m = 0$ the result is true since $n = m$ and $(c_n,m) = (0)$. If $c_n,m = 1$, then $(c_m,n) = (1) = mathbbZ$, and then it is true that $Phi_m(q) in sqrtleft( Phi_n(q)right) +mathbbZ[q]$, correct?
Any answer or any reference to literature where this results or proven would be appreciated. Thank you in advance.
abstract-algebra polynomials commutative-algebra ideals cyclotomic-polynomials
edited Jul 14 at 17:53
saulspatz
10.7k21323
asked Jul 14 at 17:24
user313212
622519
add a comment |Â
up vote
0
down vote
favorite
Let $n, m$ be two natural numbers and $Phi_n(q), Phi_m(q)$ the $n$-th and $m$-th cyclotomic polynomials respectively. Define a function $c_n,m colon mathbbN to left0,1rightcup left p mid p , textprimeright$ by
$$c_m,n =
begincases
0, , & textif $n = m$\
p, , & textif $fracnm = p^j$ where $p$ prime and $j in mathbbZsetminus 0$\
1, , & textif $fracnm$ is not an integer power of a prime
endcases$$
Show that $Phi_m(q) in sqrtleft( Phi_n(q)right) + (c_m,n)[q]$
This result should rely on properties of cyclotomic polynomials I am not aware of. Obviously if $c_n,m = 0$ the result is true since $n = m$ and $(c_n,m) = (0)$. If $c_n,m = 1$, then $(c_m,n) = (1) = mathbbZ$, and then it is true that $Phi_m(q) in sqrtleft( Phi_n(q)right) +mathbbZ[q]$, correct?
Any answer or any reference to literature where this results or proven would be appreciated. Thank you in advance.
abstract-algebra polynomials commutative-algebra ideals cyclotomic-polynomials
edited Jul 14 at 17:53
saulspatz
10.7k21323
asked Jul 14 at 17:24
user313212
622519
Your notation is really unclear. Using the $in$ symbol seems to indicate that you intend $sqrtleft( Phi_n(q)right) + (c_m,n)[q]$ to denote a set; what is the exact definition of this set?
– Greg Martin
Jul 14 at 18:03
@GregMartin The notation $sqrt(Phi_n(q))+ (c_m,n)[q]$ means the radical of $(Phi_n(q))+ (c_m,n)[q]$. I hope it is clear now.
– user313212
Jul 14 at 18:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $n, m$ be two natural numbers and $Phi_n(q), Phi_m(q)$ the $n$-th and $m$-th cyclotomic polynomials respectively. Define a function $c_n,m colon mathbbN to left0,1rightcup left p mid p , textprimeright$ by
$$c_m,n =
begincases
0, , & textif $n = m$\
p, , & textif $fracnm = p^j$ where $p$ prime and $j in mathbbZsetminus 0$\
1, , & textif $fracnm$ is not an integer power of a prime
endcases$$
Show that $Phi_m(q) in sqrtleft( Phi_n(q)right) + (c_m,n)[q]$
This result should rely on properties of cyclotomic polynomials I am not aware of. Obviously if $c_n,m = 0$ the result is true since $n = m$ and $(c_n,m) = (0)$. If $c_n,m = 1$, then $(c_m,n) = (1) = mathbbZ$, and then it is true that $Phi_m(q) in sqrtleft( Phi_n(q)right) +mathbbZ[q]$, correct?
Any answer or any reference to literature where this results or proven would be appreciated. Thank you in advance.
abstract-algebra polynomials commutative-algebra ideals cyclotomic-polynomials
edited Jul 14 at 17:53
saulspatz
10.7k21323
asked Jul 14 at 17:24
user313212
622519
Let $n, m$ be two natural numbers and $Phi_n(q), Phi_m(q)$ the $n$-th and $m$-th cyclotomic polynomials respectively. Define a function $c_n,m colon mathbbN to left0,1rightcup left p mid p , textprimeright$ by
$$c_m,n =
begincases
0, , & textif $n = m$\
p, , & textif $fracnm = p^j$ where $p$ prime and $j in mathbbZsetminus 0$\
1, , & textif $fracnm$ is not an integer power of a prime
endcases$$
Show that $Phi_m(q) in sqrtleft( Phi_n(q)right) + (c_m,n)[q]$
This result should rely on properties of cyclotomic polynomials I am not aware of. Obviously if $c_n,m = 0$ the result is true since $n = m$ and $(c_n,m) = (0)$. If $c_n,m = 1$, then $(c_m,n) = (1) = mathbbZ$, and then it is true that $Phi_m(q) in sqrtleft( Phi_n(q)right) +mathbbZ[q]$, correct?
Any answer or any reference to literature where this results or proven would be appreciated. Thank you in advance.
abstract-algebra polynomials commutative-algebra ideals cyclotomic-polynomials
edited Jul 14 at 17:53
saulspatz
10.7k21323
asked Jul 14 at 17:24
user313212
622519
edited Jul 14 at 17:53
saulspatz
10.7k21323
edited Jul 14 at 17:53
saulspatz
10.7k21323
edited Jul 14 at 17:53
saulspatz
10.7k21323
10.7k21323
asked Jul 14 at 17:24
user313212
622519
asked Jul 14 at 17:24
user313212
622519
asked Jul 14 at 17:24
user313212
622519
622519
Your notation is really unclear. Using the $in$ symbol seems to indicate that you intend $sqrtleft( Phi_n(q)right) + (c_m,n)[q]$ to denote a set; what is the exact definition of this set?
– Greg Martin
Jul 14 at 18:03
@GregMartin The notation $sqrt(Phi_n(q))+ (c_m,n)[q]$ means the radical of $(Phi_n(q))+ (c_m,n)[q]$. I hope it is clear now.
– user313212
Jul 14 at 18:39
add a comment |Â
Your notation is really unclear. Using the $in$ symbol seems to indicate that you intend $sqrtleft( Phi_n(q)right) + (c_m,n)[q]$ to denote a set; what is the exact definition of this set?
– Greg Martin
Jul 14 at 18:03
@GregMartin The notation $sqrt(Phi_n(q))+ (c_m,n)[q]$ means the radical of $(Phi_n(q))+ (c_m,n)[q]$. I hope it is clear now.
– user313212
Jul 14 at 18:39
Your notation is really unclear. Using the $in$ symbol seems to indicate that you intend $sqrtleft( Phi_n(q)right) + (c_m,n)[q]$ to denote a set; what is the exact definition of this set?
– Greg Martin
Jul 14 at 18:03
Your notation is really unclear. Using the $in$ symbol seems to indicate that you intend $sqrtleft( Phi_n(q)right) + (c_m,n)[q]$ to denote a set; what is the exact definition of this set?
– Greg Martin
Jul 14 at 18:03
@GregMartin The notation $sqrt(Phi_n(q))+ (c_m,n)[q]$ means the radical of $(Phi_n(q))+ (c_m,n)[q]$. I hope it is clear now.
– user313212
Jul 14 at 18:39
@GregMartin The notation $sqrt(Phi_n(q))+ (c_m,n)[q]$ means the radical of $(Phi_n(q))+ (c_m,n)[q]$. I hope it is clear now.
– user313212
Jul 14 at 18:39
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
As you write, the statement in case $c_m,n = 1$ is immediate because clearly $Phi_m(x) in mathbbZ[x] = mathbbZ[x] + (Phi_n(x)) = sqrtmathbbZ[x] + (Phi_n(x))$. Since this is true for all $m,n$, the statement in this case is also not meaningful.
If $n = m p^i$ so that $c_m,n = p$, then we have $Phi_n(x) = Phi_m(x^p^i)$ (elementary property of cyclotomic polynomials).
Therefore it suffices to prove that $$Phi_m(x)^p^i equiv Phi_m(x^p^i) mod p$$
But in fact this statement is more generally true of any polynomials over $mathbbZ/pmathbbZ$. It can be seen as an elaboration of the so-called "Freshman's Dream," that $(x+y)^p equiv x^p + y^p mod p$. I also found a reference here as lemma 4.3. The first property I mentioned is theorem 3.2 there.
answered Jul 14 at 22:19
Badam Baplan
3,356721
Thank you! The case where $c_m,n = 1$ is okay as I wrote on the question?
– user313212
Jul 15 at 11:01
yep, i included a sentence about that. Curious, does this exercise have some application?
– Badam Baplan
Jul 15 at 14:48
Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first...
– user313212
Jul 15 at 18:09
Yes if I came across this in a paper I also would've appreciated a quick sentence for justification.
– Badam Baplan
Jul 16 at 18:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As you write, the statement in case $c_m,n = 1$ is immediate because clearly $Phi_m(x) in mathbbZ[x] = mathbbZ[x] + (Phi_n(x)) = sqrtmathbbZ[x] + (Phi_n(x))$. Since this is true for all $m,n$, the statement in this case is also not meaningful.
If $n = m p^i$ so that $c_m,n = p$, then we have $Phi_n(x) = Phi_m(x^p^i)$ (elementary property of cyclotomic polynomials).
Therefore it suffices to prove that $$Phi_m(x)^p^i equiv Phi_m(x^p^i) mod p$$
But in fact this statement is more generally true of any polynomials over $mathbbZ/pmathbbZ$. It can be seen as an elaboration of the so-called "Freshman's Dream," that $(x+y)^p equiv x^p + y^p mod p$. I also found a reference here as lemma 4.3. The first property I mentioned is theorem 3.2 there.
answered Jul 14 at 22:19
Badam Baplan
3,356721
Thank you! The case where $c_m,n = 1$ is okay as I wrote on the question?
– user313212
Jul 15 at 11:01
yep, i included a sentence about that. Curious, does this exercise have some application?
– Badam Baplan
Jul 15 at 14:48
Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first...
– user313212
Jul 15 at 18:09
Yes if I came across this in a paper I also would've appreciated a quick sentence for justification.
– Badam Baplan
Jul 16 at 18:26
add a comment |Â
up vote
1
down vote
accepted
As you write, the statement in case $c_m,n = 1$ is immediate because clearly $Phi_m(x) in mathbbZ[x] = mathbbZ[x] + (Phi_n(x)) = sqrtmathbbZ[x] + (Phi_n(x))$. Since this is true for all $m,n$, the statement in this case is also not meaningful.
If $n = m p^i$ so that $c_m,n = p$, then we have $Phi_n(x) = Phi_m(x^p^i)$ (elementary property of cyclotomic polynomials).
Therefore it suffices to prove that $$Phi_m(x)^p^i equiv Phi_m(x^p^i) mod p$$
But in fact this statement is more generally true of any polynomials over $mathbbZ/pmathbbZ$. It can be seen as an elaboration of the so-called "Freshman's Dream," that $(x+y)^p equiv x^p + y^p mod p$. I also found a reference here as lemma 4.3. The first property I mentioned is theorem 3.2 there.
answered Jul 14 at 22:19
Badam Baplan
3,356721
Thank you! The case where $c_m,n = 1$ is okay as I wrote on the question?
– user313212
Jul 15 at 11:01
yep, i included a sentence about that. Curious, does this exercise have some application?
– Badam Baplan
Jul 15 at 14:48
Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first...
– user313212
Jul 15 at 18:09
Yes if I came across this in a paper I also would've appreciated a quick sentence for justification.
– Badam Baplan
Jul 16 at 18:26
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As you write, the statement in case $c_m,n = 1$ is immediate because clearly $Phi_m(x) in mathbbZ[x] = mathbbZ[x] + (Phi_n(x)) = sqrtmathbbZ[x] + (Phi_n(x))$. Since this is true for all $m,n$, the statement in this case is also not meaningful.
If $n = m p^i$ so that $c_m,n = p$, then we have $Phi_n(x) = Phi_m(x^p^i)$ (elementary property of cyclotomic polynomials).
Therefore it suffices to prove that $$Phi_m(x)^p^i equiv Phi_m(x^p^i) mod p$$
But in fact this statement is more generally true of any polynomials over $mathbbZ/pmathbbZ$. It can be seen as an elaboration of the so-called "Freshman's Dream," that $(x+y)^p equiv x^p + y^p mod p$. I also found a reference here as lemma 4.3. The first property I mentioned is theorem 3.2 there.
answered Jul 14 at 22:19
Badam Baplan
3,356721
As you write, the statement in case $c_m,n = 1$ is immediate because clearly $Phi_m(x) in mathbbZ[x] = mathbbZ[x] + (Phi_n(x)) = sqrtmathbbZ[x] + (Phi_n(x))$. Since this is true for all $m,n$, the statement in this case is also not meaningful.
If $n = m p^i$ so that $c_m,n = p$, then we have $Phi_n(x) = Phi_m(x^p^i)$ (elementary property of cyclotomic polynomials).
Therefore it suffices to prove that $$Phi_m(x)^p^i equiv Phi_m(x^p^i) mod p$$
But in fact this statement is more generally true of any polynomials over $mathbbZ/pmathbbZ$. It can be seen as an elaboration of the so-called "Freshman's Dream," that $(x+y)^p equiv x^p + y^p mod p$. I also found a reference here as lemma 4.3. The first property I mentioned is theorem 3.2 there.
answered Jul 14 at 22:19
Badam Baplan
3,356721
edited Jul 15 at 14:47
answered Jul 14 at 22:19
Badam Baplan
3,356721
answered Jul 14 at 22:19
Badam Baplan
3,356721
answered Jul 14 at 22:19
Badam Baplan
3,356721
3,356721
Thank you! The case where $c_m,n = 1$ is okay as I wrote on the question?
– user313212
Jul 15 at 11:01
yep, i included a sentence about that. Curious, does this exercise have some application?
– Badam Baplan
Jul 15 at 14:48
Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first...
– user313212
Jul 15 at 18:09
Yes if I came across this in a paper I also would've appreciated a quick sentence for justification.
– Badam Baplan
Jul 16 at 18:26
add a comment |Â
Thank you! The case where $c_m,n = 1$ is okay as I wrote on the question?
– user313212
Jul 15 at 11:01
yep, i included a sentence about that. Curious, does this exercise have some application?
– Badam Baplan
Jul 15 at 14:48
Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first...
– user313212
Jul 15 at 18:09
Yes if I came across this in a paper I also would've appreciated a quick sentence for justification.
– Badam Baplan
Jul 16 at 18:26
Thank you! The case where $c_m,n = 1$ is okay as I wrote on the question?
– user313212
Jul 15 at 11:01
Thank you! The case where $c_m,n = 1$ is okay as I wrote on the question?
– user313212
Jul 15 at 11:01
yep, i included a sentence about that. Curious, does this exercise have some application?
– Badam Baplan
Jul 15 at 14:48
yep, i included a sentence about that. Curious, does this exercise have some application?
– Badam Baplan
Jul 15 at 14:48
Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first...
– user313212
Jul 15 at 18:09
Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first...
– user313212
Jul 15 at 18:09
Yes if I came across this in a paper I also would've appreciated a quick sentence for justification.
– Badam Baplan
Jul 16 at 18:26
Yes if I came across this in a paper I also would've appreciated a quick sentence for justification.
– Badam Baplan
Jul 16 at 18:26
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2851797%2frelationship-between-cyclotomic-polynomials%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Your notation is really unclear. Using the $in$ symbol seems to indicate that you intend $sqrtleft( Phi_n(q)right) + (c_m,n)[q]$ to denote a set; what is the exact definition of this set?
– Greg Martin
Jul 14 at 18:03
@GregMartin The notation $sqrt(Phi_n(q))+ (c_m,n)[q]$ means the radical of $(Phi_n(q))+ (c_m,n)[q]$. I hope it is clear now.
– user313212
Jul 14 at 18:39