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 18 hours ago
Alessar
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 18 hours ago
Alessar
21113
1
Contains some gaps.
– Wuestenfux
18 hours ago
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
18 hours ago
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
17 hours ago
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
17 hours ago
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
17 hours ago
|
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 18 hours ago
Alessar
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 18 hours ago
Alessar
21113
asked 18 hours ago
Alessar
21113
asked 18 hours ago
Alessar
21113
asked 18 hours ago
Alessar
21113
asked 18 hours ago
Alessar
21113
21113
1
Contains some gaps.
– Wuestenfux
18 hours ago
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
18 hours ago
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
17 hours ago
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
17 hours ago
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
17 hours ago
|
show 1 more comment
1
Contains some gaps.
– Wuestenfux
18 hours ago
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
18 hours ago
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
17 hours ago
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
17 hours ago
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
17 hours ago
1
1
Contains some gaps.
– Wuestenfux
18 hours ago
Contains some gaps.
– Wuestenfux
18 hours ago
1
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
18 hours ago
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
18 hours ago
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
17 hours ago
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
17 hours ago
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
17 hours ago
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
17 hours ago
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
17 hours ago
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
17 hours ago
|
show 1 more comment
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1
Contains some gaps.
– Wuestenfux
18 hours ago
1
Over a field, what's the difference between a prime and an irreducible polynomial?
– Bernard
18 hours ago
Over a field, prime polynomial and irreducible polynomial are the same, polynomials that cannot be factored into polynomials of lower degree
– Alessar
17 hours ago
@Wuestenfux how can I complete my proof? Can you give some hints of those gaps?
– Alessar
17 hours ago
You can assume that one of the polynomials $G, H$ has degree $leq n/2$.
– Wuestenfux
17 hours ago