Let $F(x) in GF(p)[x]$ a prime polynomial deg $n>0$, if $gcd(F,x^{q^i}-x)=1$ for $i=1,2,…,[n/2]$ $iff$...
I want to prove this claim below as an exercise for exam:
Let $F(x) in GF(p)[x]$ a prime polynomial deg $n>0$, if
$gcd(F,x^{q^i}-x)=1$ for $i=1,2,...,[n/2]$ $iff$ $F$ is irreducible
Here a trace of my proof:
Suppose $F$ reducible.
Then $F(x)=G(x)H(x)$ product of at least two irreducible polynomial, and since $deg F=n Rightarrow degf geq degG times degH$. Suppose then $degG=degH=[n/2]$.
We know that for $q=p^k Rightarrow x^{q^i}-x=$ {product of all irreducible polynomial of $GF(q)$}. But from the product of thos polynomial, must emerge a polynomial with degree at least $[n/2]$, so $gcd(F,x^{q^i}-x) not =1$.
So in order to have $gcd(F,x^{q^i}-x) =1$, $F$ must be irreducible.
I have difficulties in translating my ideas in mathematical format so thanks for any kind of hints or corrections, thanks.
abstract-algebra finite-fields
asked Jan 4 at 13:42
AlessarAlessar
21113
|
show 1 more comment
I want to prove this claim below as an exercise for exam:
Let $F(x) in GF(p)[x]$ a prime polynomial deg $n>0$, if
$gcd(F,x^{q^i}-x)=1$ for $i=1,2,...,[n/2]$ $iff$ $F$ is irreducible
Here a trace of my proof:
Suppose $F$ reducible.
Then $F(x)=G(x)H(x)$ product of at least two irreducible polynomial, and since $deg F=n Rightarrow degf geq degG times degH$. Suppose then $degG=degH=[n/2]$.
We know that for $q=p^k Rightarrow x^{q^i}-x=$ {product of all irreducible polynomial of $GF(q)$}. But from the product of thos polynomial, must emerge a polynomial with degree at least $[n/2]$, so $gcd(F,x^{q^i}-x) not =1$.
So in order to have $gcd(F,x^{q^i}-x) =1$, $F$ must be irreducible.
I have difficulties in translating my ideas in mathematical format so thanks for any kind of hints or corrections, thanks.
abstract-algebra finite-fields
asked Jan 4 at 13:42
AlessarAlessar
21113
1
Contains some gaps.
– Wuestenfux
Jan 4 at 13:47
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
Jan 4 at 13:49
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
Jan 4 at 13:59
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
Jan 4 at 14:13
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
Jan 4 at 14:18
|
show 1 more comment
I want to prove this claim below as an exercise for exam:
Let $F(x) in GF(p)[x]$ a prime polynomial deg $n>0$, if
$gcd(F,x^{q^i}-x)=1$ for $i=1,2,...,[n/2]$ $iff$ $F$ is irreducible
Here a trace of my proof:
Suppose $F$ reducible.
Then $F(x)=G(x)H(x)$ product of at least two irreducible polynomial, and since $deg F=n Rightarrow degf geq degG times degH$. Suppose then $degG=degH=[n/2]$.
We know that for $q=p^k Rightarrow x^{q^i}-x=$ {product of all irreducible polynomial of $GF(q)$}. But from the product of thos polynomial, must emerge a polynomial with degree at least $[n/2]$, so $gcd(F,x^{q^i}-x) not =1$.
So in order to have $gcd(F,x^{q^i}-x) =1$, $F$ must be irreducible.
I have difficulties in translating my ideas in mathematical format so thanks for any kind of hints or corrections, thanks.
abstract-algebra finite-fields
asked Jan 4 at 13:42
AlessarAlessar
21113
I want to prove this claim below as an exercise for exam:
Let $F(x) in GF(p)[x]$ a prime polynomial deg $n>0$, if
$gcd(F,x^{q^i}-x)=1$ for $i=1,2,...,[n/2]$ $iff$ $F$ is irreducible
Here a trace of my proof:
Suppose $F$ reducible.
Then $F(x)=G(x)H(x)$ product of at least two irreducible polynomial, and since $deg F=n Rightarrow degf geq degG times degH$. Suppose then $degG=degH=[n/2]$.
We know that for $q=p^k Rightarrow x^{q^i}-x=$ {product of all irreducible polynomial of $GF(q)$}. But from the product of thos polynomial, must emerge a polynomial with degree at least $[n/2]$, so $gcd(F,x^{q^i}-x) not =1$.
So in order to have $gcd(F,x^{q^i}-x) =1$, $F$ must be irreducible.
I have difficulties in translating my ideas in mathematical format so thanks for any kind of hints or corrections, thanks.
abstract-algebra finite-fields
abstract-algebra finite-fields
asked Jan 4 at 13:42
AlessarAlessar
21113
asked Jan 4 at 13:42
AlessarAlessar
21113
asked Jan 4 at 13:42
AlessarAlessar
21113
asked Jan 4 at 13:42
AlessarAlessar
21113
asked Jan 4 at 13:42
AlessarAlessar
21113
21113
1
Contains some gaps.
– Wuestenfux
Jan 4 at 13:47
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
Jan 4 at 13:49
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
Jan 4 at 13:59
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
Jan 4 at 14:13
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
Jan 4 at 14:18
|
show 1 more comment
1
Contains some gaps.
– Wuestenfux
Jan 4 at 13:47
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
Jan 4 at 13:49
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
Jan 4 at 13:59
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
Jan 4 at 14:13
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
Jan 4 at 14:18
1
1
Contains some gaps.
– Wuestenfux
Jan 4 at 13:47
Contains some gaps.
– Wuestenfux
Jan 4 at 13:47
1
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
Jan 4 at 13:49
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
Jan 4 at 13:49
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
Jan 4 at 13:59
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
Jan 4 at 13:59
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
Jan 4 at 14:13
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
Jan 4 at 14:13
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
Jan 4 at 14:18
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
Jan 4 at 14:18
|
show 1 more comment
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1
Contains some gaps.
– Wuestenfux
Jan 4 at 13:47
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
Jan 4 at 13:49
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
Jan 4 at 13:59
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
Jan 4 at 14:13
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
Jan 4 at 14:18