Mixing
If $degleft(fright) = minleft(left{d in mathbb{N}^times;; fvert X^{q^d} - Xright}right),$, then $f$ is...
Let for some prime power $q$ $mathbb{F}_q$ be a finite field and consider $f in mathbb{F}_qleft[Xright]$.
I want to show the implication mentioned in the title, i.e.
$$
degleft(,fright) =
minleft(left{d in mathbb{N}^times;; f,vert, X^{q^d} - Xright}right) implies f ;textit{is irreducible}
$$
I could really use some help proving this.
If you want to see my current approach (which I couldn't yet turn into a proof), read on.
I already know that for $d in mathbb{N}^times$, $P_d:=X^{q^d} - X$ is the squarefree product of all irreducible polynomials $r in mathbb{F}_q[X]$ such that $degleft(rright),mid,d$.
Another fact that might be useful is that if $r$ is an irreducible factor of $P_d$, then $degleft(rright) ,mid, d$.
My current proof idea is as follows:
Assume that $degleft(fright)$ has the minimality property mentioned above and that $f$ is not irreducible. We hence have, for $2 leq m in mathbb{N}^times$ and prime factors $f_i in mathbb{F}_qleft[Xright]$
$$
f = prod_{i=1}^m f_i
$$
I'm not sure if this helps, but on a side note, all the $f_i$ are coprime ($f$ is squarefree, since $P_d$ is).
Anyways, consider $mu:=text{lcm}left(left{degleft(f_iright);; 1 leq i leq mright}right)$.
Using one of the facts mentioned above, note that for all $i$, since $f_i$ is an irreducible factor of $f$ which divides $P_d$, we have $degleft(f_iright),mid,degleft(fright)$. By definition of $mu$, this implies $mu ,mid, degleft(fright)$.
It thus would suffice to show that $mu lt degleft(fright)$ to arrive at a contradiction to the minimality property of $degleft(fright)$.
Can somebody help me out here?
UPDATE: The answer here leads me to believe my approach might be wrong. Not sure though, just found this.
abstract-algebra polynomials finite-fields irreducible-polynomials
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
asked Nov 20 '18 at 17:28
polynomial_donut
597217
add a comment |
Let for some prime power $q$ $mathbb{F}_q$ be a finite field and consider $f in mathbb{F}_qleft[Xright]$.
I want to show the implication mentioned in the title, i.e.
$$
degleft(,fright) =
minleft(left{d in mathbb{N}^times;; f,vert, X^{q^d} - Xright}right) implies f ;textit{is irreducible}
$$
I could really use some help proving this.
If you want to see my current approach (which I couldn't yet turn into a proof), read on.
I already know that for $d in mathbb{N}^times$, $P_d:=X^{q^d} - X$ is the squarefree product of all irreducible polynomials $r in mathbb{F}_q[X]$ such that $degleft(rright),mid,d$.
Another fact that might be useful is that if $r$ is an irreducible factor of $P_d$, then $degleft(rright) ,mid, d$.
My current proof idea is as follows:
Assume that $degleft(fright)$ has the minimality property mentioned above and that $f$ is not irreducible. We hence have, for $2 leq m in mathbb{N}^times$ and prime factors $f_i in mathbb{F}_qleft[Xright]$
$$
f = prod_{i=1}^m f_i
$$
I'm not sure if this helps, but on a side note, all the $f_i$ are coprime ($f$ is squarefree, since $P_d$ is).
Anyways, consider $mu:=text{lcm}left(left{degleft(f_iright);; 1 leq i leq mright}right)$.
Using one of the facts mentioned above, note that for all $i$, since $f_i$ is an irreducible factor of $f$ which divides $P_d$, we have $degleft(f_iright),mid,degleft(fright)$. By definition of $mu$, this implies $mu ,mid, degleft(fright)$.
It thus would suffice to show that $mu lt degleft(fright)$ to arrive at a contradiction to the minimality property of $degleft(fright)$.
Can somebody help me out here?
UPDATE: The answer here leads me to believe my approach might be wrong. Not sure though, just found this.
abstract-algebra polynomials finite-fields irreducible-polynomials
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
asked Nov 20 '18 at 17:28
polynomial_donut
597217
1
The other question you linked to is different from yours because it has the extra assumption that $d$ is prime.
– Eric Wofsey
Nov 20 '18 at 23:48
Oh, good point. Only quickly scanned the answer and thought this was something to be concluded.
– polynomial_donut
Nov 21 '18 at 15:02
add a comment |
Let for some prime power $q$ $mathbb{F}_q$ be a finite field and consider $f in mathbb{F}_qleft[Xright]$.
I want to show the implication mentioned in the title, i.e.
$$
degleft(,fright) =
minleft(left{d in mathbb{N}^times;; f,vert, X^{q^d} - Xright}right) implies f ;textit{is irreducible}
$$
I could really use some help proving this.
If you want to see my current approach (which I couldn't yet turn into a proof), read on.
I already know that for $d in mathbb{N}^times$, $P_d:=X^{q^d} - X$ is the squarefree product of all irreducible polynomials $r in mathbb{F}_q[X]$ such that $degleft(rright),mid,d$.
Another fact that might be useful is that if $r$ is an irreducible factor of $P_d$, then $degleft(rright) ,mid, d$.
My current proof idea is as follows:
Assume that $degleft(fright)$ has the minimality property mentioned above and that $f$ is not irreducible. We hence have, for $2 leq m in mathbb{N}^times$ and prime factors $f_i in mathbb{F}_qleft[Xright]$
$$
f = prod_{i=1}^m f_i
$$
I'm not sure if this helps, but on a side note, all the $f_i$ are coprime ($f$ is squarefree, since $P_d$ is).
Anyways, consider $mu:=text{lcm}left(left{degleft(f_iright);; 1 leq i leq mright}right)$.
Using one of the facts mentioned above, note that for all $i$, since $f_i$ is an irreducible factor of $f$ which divides $P_d$, we have $degleft(f_iright),mid,degleft(fright)$. By definition of $mu$, this implies $mu ,mid, degleft(fright)$.
It thus would suffice to show that $mu lt degleft(fright)$ to arrive at a contradiction to the minimality property of $degleft(fright)$.
Can somebody help me out here?
UPDATE: The answer here leads me to believe my approach might be wrong. Not sure though, just found this.
abstract-algebra polynomials finite-fields irreducible-polynomials
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
asked Nov 20 '18 at 17:28
polynomial_donut
597217
Let for some prime power $q$ $mathbb{F}_q$ be a finite field and consider $f in mathbb{F}_qleft[Xright]$.
I want to show the implication mentioned in the title, i.e.
$$
degleft(,fright) =
minleft(left{d in mathbb{N}^times;; f,vert, X^{q^d} - Xright}right) implies f ;textit{is irreducible}
$$
I could really use some help proving this.
If you want to see my current approach (which I couldn't yet turn into a proof), read on.
I already know that for $d in mathbb{N}^times$, $P_d:=X^{q^d} - X$ is the squarefree product of all irreducible polynomials $r in mathbb{F}_q[X]$ such that $degleft(rright),mid,d$.
Another fact that might be useful is that if $r$ is an irreducible factor of $P_d$, then $degleft(rright) ,mid, d$.
My current proof idea is as follows:
Assume that $degleft(fright)$ has the minimality property mentioned above and that $f$ is not irreducible. We hence have, for $2 leq m in mathbb{N}^times$ and prime factors $f_i in mathbb{F}_qleft[Xright]$
$$
f = prod_{i=1}^m f_i
$$
I'm not sure if this helps, but on a side note, all the $f_i$ are coprime ($f$ is squarefree, since $P_d$ is).
Anyways, consider $mu:=text{lcm}left(left{degleft(f_iright);; 1 leq i leq mright}right)$.
Using one of the facts mentioned above, note that for all $i$, since $f_i$ is an irreducible factor of $f$ which divides $P_d$, we have $degleft(f_iright),mid,degleft(fright)$. By definition of $mu$, this implies $mu ,mid, degleft(fright)$.
It thus would suffice to show that $mu lt degleft(fright)$ to arrive at a contradiction to the minimality property of $degleft(fright)$.
Can somebody help me out here?
UPDATE: The answer here leads me to believe my approach might be wrong. Not sure though, just found this.
abstract-algebra polynomials finite-fields irreducible-polynomials
abstract-algebra polynomials finite-fields irreducible-polynomials
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
asked Nov 20 '18 at 17:28
polynomial_donut
597217
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
asked Nov 20 '18 at 17:28
polynomial_donut
597217
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
edited Nov 21 '18 at 1:29
Batominovski
33.8k33292
33.8k33292
asked Nov 20 '18 at 17:28
polynomial_donut
597217
asked Nov 20 '18 at 17:28
polynomial_donut
597217
asked Nov 20 '18 at 17:28
polynomial_donut
597217
597217
1
The other question you linked to is different from yours because it has the extra assumption that $d$ is prime.
– Eric Wofsey
Nov 20 '18 at 23:48
Oh, good point. Only quickly scanned the answer and thought this was something to be concluded.
– polynomial_donut
Nov 21 '18 at 15:02
add a comment |
1
The other question you linked to is different from yours because it has the extra assumption that $d$ is prime.
– Eric Wofsey
Nov 20 '18 at 23:48
Oh, good point. Only quickly scanned the answer and thought this was something to be concluded.
– polynomial_donut
Nov 21 '18 at 15:02
1
1
The other question you linked to is different from yours because it has the extra assumption that $d$ is prime.
– Eric Wofsey
Nov 20 '18 at 23:48
The other question you linked to is different from yours because it has the extra assumption that $d$ is prime.
– Eric Wofsey
Nov 20 '18 at 23:48
Oh, good point. Only quickly scanned the answer and thought this was something to be concluded.
– polynomial_donut
Nov 21 '18 at 15:02
Oh, good point. Only quickly scanned the answer and thought this was something to be concluded.
– polynomial_donut
Nov 21 '18 at 15:02
add a comment |
1 Answer
1
active
oldest
votes
This is false. For instance, if $q=2$ and $f(x)=x(x^2+x+1)(x^3+x+1)$ then $f$ is not irreducible but the smallest $d>0$ such that $f$ divides $x^{2^d}-x$ is $6$, the degree of $f$. Indeed, the irreducible factors of $f$ have degree $1,2,$ and $3$, and so $f$ divides $x^{2^d}-x$ iff $d$ is divisible by $1,2,$ and $3$. The least such $d$ is $6$.
In general, by the same reasoning, if $f$ is squarefree then the least such $d$ will be the least common multiple of the degrees of the irreducible factors of $f$.
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
Thank you. This now led me to another question, maybe you're interested...
– polynomial_donut
Nov 21 '18 at 14:48
add a comment |
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This is false. For instance, if $q=2$ and $f(x)=x(x^2+x+1)(x^3+x+1)$ then $f$ is not irreducible but the smallest $d>0$ such that $f$ divides $x^{2^d}-x$ is $6$, the degree of $f$. Indeed, the irreducible factors of $f$ have degree $1,2,$ and $3$, and so $f$ divides $x^{2^d}-x$ iff $d$ is divisible by $1,2,$ and $3$. The least such $d$ is $6$.
In general, by the same reasoning, if $f$ is squarefree then the least such $d$ will be the least common multiple of the degrees of the irreducible factors of $f$.
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
Thank you. This now led me to another question, maybe you're interested...
– polynomial_donut
Nov 21 '18 at 14:48
add a comment |
This is false. For instance, if $q=2$ and $f(x)=x(x^2+x+1)(x^3+x+1)$ then $f$ is not irreducible but the smallest $d>0$ such that $f$ divides $x^{2^d}-x$ is $6$, the degree of $f$. Indeed, the irreducible factors of $f$ have degree $1,2,$ and $3$, and so $f$ divides $x^{2^d}-x$ iff $d$ is divisible by $1,2,$ and $3$. The least such $d$ is $6$.
In general, by the same reasoning, if $f$ is squarefree then the least such $d$ will be the least common multiple of the degrees of the irreducible factors of $f$.
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
Thank you. This now led me to another question, maybe you're interested...
– polynomial_donut
Nov 21 '18 at 14:48
add a comment |
This is false. For instance, if $q=2$ and $f(x)=x(x^2+x+1)(x^3+x+1)$ then $f$ is not irreducible but the smallest $d>0$ such that $f$ divides $x^{2^d}-x$ is $6$, the degree of $f$. Indeed, the irreducible factors of $f$ have degree $1,2,$ and $3$, and so $f$ divides $x^{2^d}-x$ iff $d$ is divisible by $1,2,$ and $3$. The least such $d$ is $6$.
In general, by the same reasoning, if $f$ is squarefree then the least such $d$ will be the least common multiple of the degrees of the irreducible factors of $f$.
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
This is false. For instance, if $q=2$ and $f(x)=x(x^2+x+1)(x^3+x+1)$ then $f$ is not irreducible but the smallest $d>0$ such that $f$ divides $x^{2^d}-x$ is $6$, the degree of $f$. Indeed, the irreducible factors of $f$ have degree $1,2,$ and $3$, and so $f$ divides $x^{2^d}-x$ iff $d$ is divisible by $1,2,$ and $3$. The least such $d$ is $6$.
In general, by the same reasoning, if $f$ is squarefree then the least such $d$ will be the least common multiple of the degrees of the irreducible factors of $f$.
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
answered Nov 20 '18 at 23:46
Eric Wofsey
180k12207335
180k12207335
Thank you. This now led me to another question, maybe you're interested...
– polynomial_donut
Nov 21 '18 at 14:48
add a comment |
Thank you. This now led me to another question, maybe you're interested...
– polynomial_donut
Nov 21 '18 at 14:48
Thank you. This now led me to another question, maybe you're interested...
– polynomial_donut
Nov 21 '18 at 14:48
Thank you. This now led me to another question, maybe you're interested...
– polynomial_donut
Nov 21 '18 at 14:48
add a comment |
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The other question you linked to is different from yours because it has the extra assumption that $d$ is prime.
– Eric Wofsey
Nov 20 '18 at 23:48
Oh, good point. Only quickly scanned the answer and thought this was something to be concluded.
– polynomial_donut
Nov 21 '18 at 15:02