Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 iff d= gcd(r,s)$
$begingroup$
Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 iff d= gcd(r,s)$.
I define $r=ad$ and $s=bd$, then we have $((x^{ad}-1),(x^{bd}-1)) = x^d-1$, but the annoying thing for me to proceed are those "$-1$".
From literature:
For non-negative integers a and b, where a and b are not both zero,
provable by considering the Euclidean algorithm in base n we know
$gcd(n^a − 1, n^b − 1) = n^{gcd(a,b)} − 1$.
Some hints for a proof that doesn't use the Euclidean Algorithm?
abstract-algebra field-theory
edited Dec 21 '18 at 17:42
Shaun
9,789113684
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
$endgroup$
|
show 1 more comment
$begingroup$
Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 iff d= gcd(r,s)$.
I define $r=ad$ and $s=bd$, then we have $((x^{ad}-1),(x^{bd}-1)) = x^d-1$, but the annoying thing for me to proceed are those "$-1$".
From literature:
For non-negative integers a and b, where a and b are not both zero,
provable by considering the Euclidean algorithm in base n we know
$gcd(n^a − 1, n^b − 1) = n^{gcd(a,b)} − 1$.
Some hints for a proof that doesn't use the Euclidean Algorithm?
abstract-algebra field-theory
edited Dec 21 '18 at 17:42
Shaun
9,789113684
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
$endgroup$
1
$begingroup$
math.stackexchange.com/questions/7473/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 9:39
$begingroup$
I can't understand this passage; if $n > m$, then $$gcd(a^n - 1, a^m - 1) =color{red}{ gcd(a^n - 1, a^n - a^{n-m}) = gcd(a^{n-m} - 1, a^m - 1)}$$
$endgroup$
– Alessar
Dec 18 '18 at 10:02
1
$begingroup$
If $d$ divide both, $d$ must divide $a^n-a^m=a^m(a^{n-m}-1)$ But $dnmid a^m$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:08
$begingroup$
In my case, your $d$ is my $x^d-1$, right? Sorry I want to understand clearly
$endgroup$
– Alessar
Dec 18 '18 at 10:23
1
$begingroup$
I've stated common divisor, ultimately this leads to the greatest common divisor
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:25
|
show 1 more comment
$begingroup$
Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 iff d= gcd(r,s)$.
I define $r=ad$ and $s=bd$, then we have $((x^{ad}-1),(x^{bd}-1)) = x^d-1$, but the annoying thing for me to proceed are those "$-1$".
From literature:
For non-negative integers a and b, where a and b are not both zero,
provable by considering the Euclidean algorithm in base n we know
$gcd(n^a − 1, n^b − 1) = n^{gcd(a,b)} − 1$.
Some hints for a proof that doesn't use the Euclidean Algorithm?
abstract-algebra field-theory
edited Dec 21 '18 at 17:42
Shaun
9,789113684
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
$endgroup$
Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 iff d= gcd(r,s)$.
I define $r=ad$ and $s=bd$, then we have $((x^{ad}-1),(x^{bd}-1)) = x^d-1$, but the annoying thing for me to proceed are those "$-1$".
From literature:
For non-negative integers a and b, where a and b are not both zero,
provable by considering the Euclidean algorithm in base n we know
$gcd(n^a − 1, n^b − 1) = n^{gcd(a,b)} − 1$.
Some hints for a proof that doesn't use the Euclidean Algorithm?
abstract-algebra field-theory
abstract-algebra field-theory
edited Dec 21 '18 at 17:42
Shaun
9,789113684
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
edited Dec 21 '18 at 17:42
Shaun
9,789113684
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
edited Dec 21 '18 at 17:42
Shaun
9,789113684
edited Dec 21 '18 at 17:42
Shaun
9,789113684
edited Dec 21 '18 at 17:42
Shaun
9,789113684
9,789113684
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
asked Dec 18 '18 at 9:33
AlessarAlessar
313115
313115
1
$begingroup$
math.stackexchange.com/questions/7473/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 9:39
$begingroup$
I can't understand this passage; if $n > m$, then $$gcd(a^n - 1, a^m - 1) =color{red}{ gcd(a^n - 1, a^n - a^{n-m}) = gcd(a^{n-m} - 1, a^m - 1)}$$
$endgroup$
– Alessar
Dec 18 '18 at 10:02
1
$begingroup$
If $d$ divide both, $d$ must divide $a^n-a^m=a^m(a^{n-m}-1)$ But $dnmid a^m$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:08
$begingroup$
In my case, your $d$ is my $x^d-1$, right? Sorry I want to understand clearly
$endgroup$
– Alessar
Dec 18 '18 at 10:23
1
$begingroup$
I've stated common divisor, ultimately this leads to the greatest common divisor
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:25
|
show 1 more comment
1
$begingroup$
math.stackexchange.com/questions/7473/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 9:39
$begingroup$
I can't understand this passage; if $n > m$, then $$gcd(a^n - 1, a^m - 1) =color{red}{ gcd(a^n - 1, a^n - a^{n-m}) = gcd(a^{n-m} - 1, a^m - 1)}$$
$endgroup$
– Alessar
Dec 18 '18 at 10:02
1
$begingroup$
If $d$ divide both, $d$ must divide $a^n-a^m=a^m(a^{n-m}-1)$ But $dnmid a^m$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:08
$begingroup$
In my case, your $d$ is my $x^d-1$, right? Sorry I want to understand clearly
$endgroup$
– Alessar
Dec 18 '18 at 10:23
1
$begingroup$
I've stated common divisor, ultimately this leads to the greatest common divisor
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:25
1
1
$begingroup$
math.stackexchange.com/questions/7473/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 9:39
$begingroup$
math.stackexchange.com/questions/7473/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 9:39
$begingroup$
I can't understand this passage; if $n > m$, then $$gcd(a^n - 1, a^m - 1) =color{red}{ gcd(a^n - 1, a^n - a^{n-m}) = gcd(a^{n-m} - 1, a^m - 1)}$$
$endgroup$
– Alessar
Dec 18 '18 at 10:02
$begingroup$
I can't understand this passage; if $n > m$, then $$gcd(a^n - 1, a^m - 1) =color{red}{ gcd(a^n - 1, a^n - a^{n-m}) = gcd(a^{n-m} - 1, a^m - 1)}$$
$endgroup$
– Alessar
Dec 18 '18 at 10:02
1
1
$begingroup$
If $d$ divide both, $d$ must divide $a^n-a^m=a^m(a^{n-m}-1)$ But $dnmid a^m$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:08
$begingroup$
If $d$ divide both, $d$ must divide $a^n-a^m=a^m(a^{n-m}-1)$ But $dnmid a^m$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:08
$begingroup$
In my case, your $d$ is my $x^d-1$, right? Sorry I want to understand clearly
$endgroup$
– Alessar
Dec 18 '18 at 10:23
$begingroup$
In my case, your $d$ is my $x^d-1$, right? Sorry I want to understand clearly
$endgroup$
– Alessar
Dec 18 '18 at 10:23
1
1
$begingroup$
I've stated common divisor, ultimately this leads to the greatest common divisor
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:25
$begingroup$
I've stated common divisor, ultimately this leads to the greatest common divisor
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:25
|
show 1 more comment
0
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1
$begingroup$
math.stackexchange.com/questions/7473/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 9:39
$begingroup$
I can't understand this passage; if $n > m$, then $$gcd(a^n - 1, a^m - 1) =color{red}{ gcd(a^n - 1, a^n - a^{n-m}) = gcd(a^{n-m} - 1, a^m - 1)}$$
$endgroup$
– Alessar
Dec 18 '18 at 10:02
1
$begingroup$
If $d$ divide both, $d$ must divide $a^n-a^m=a^m(a^{n-m}-1)$ But $dnmid a^m$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:08
$begingroup$
In my case, your $d$ is my $x^d-1$, right? Sorry I want to understand clearly
$endgroup$
– Alessar
Dec 18 '18 at 10:23
1
$begingroup$
I've stated common divisor, ultimately this leads to the greatest common divisor
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 10:25