Relatively prime polynomials in proof that there exists n-th root of unity in a field.
I was reading a proof to show that if we have a field of characteristic p and that p doesn't divide a natural number n, then there exists an n-th primitive root of unity. It starts by saying that if $p nmid n$ than $x^n-1$ and
$nx^{(n-1)}$ are relatively prime. This doesn't seem crazy to me, but what argument could I use to be sure about this?
abstract-algebra
asked Dec 1 at 1:10
roi_saumon
38517
add a comment |
I was reading a proof to show that if we have a field of characteristic p and that p doesn't divide a natural number n, then there exists an n-th primitive root of unity. It starts by saying that if $p nmid n$ than $x^n-1$ and
$nx^{(n-1)}$ are relatively prime. This doesn't seem crazy to me, but what argument could I use to be sure about this?
abstract-algebra
asked Dec 1 at 1:10
roi_saumon
38517
add a comment |
I was reading a proof to show that if we have a field of characteristic p and that p doesn't divide a natural number n, then there exists an n-th primitive root of unity. It starts by saying that if $p nmid n$ than $x^n-1$ and
$nx^{(n-1)}$ are relatively prime. This doesn't seem crazy to me, but what argument could I use to be sure about this?
abstract-algebra
asked Dec 1 at 1:10
roi_saumon
38517
I was reading a proof to show that if we have a field of characteristic p and that p doesn't divide a natural number n, then there exists an n-th primitive root of unity. It starts by saying that if $p nmid n$ than $x^n-1$ and
$nx^{(n-1)}$ are relatively prime. This doesn't seem crazy to me, but what argument could I use to be sure about this?
abstract-algebra
abstract-algebra
asked Dec 1 at 1:10
roi_saumon
38517
asked Dec 1 at 1:10
roi_saumon
38517
asked Dec 1 at 1:10
roi_saumon
38517
asked Dec 1 at 1:10
roi_saumon
38517
asked Dec 1 at 1:10
roi_saumon
38517
38517
add a comment |
add a comment |
2 Answers
2
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oldest
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Since $pnotmid n$, $ncdot 1_Fnot=0$. So there is an inverse, $t$, for $n$.
Now $(x^n-1)-tx(nx^{n-1})=-1$.
answered Dec 1 at 2:11
Chris Custer
10.7k3724
add a comment |
You are working on a UFD (actually a Euclidean domain). For two elements $f$ and $g$ to be relatively prime it suffices to show no prime divisor of $f$ is a divisor of $g$.
The second polynomial is a product of linear factors and a constant: multiply the linear polynomial $(x-0)$, with itself $n-1$ times along with constant $n$. As $(x-0)$ is a prime (irreducuble polynomial) and not a factor of $x^n-1$ (use remainder theorem) we are done.
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
add a comment |
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2 Answers
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2 Answers
2
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Since $pnotmid n$, $ncdot 1_Fnot=0$. So there is an inverse, $t$, for $n$.
Now $(x^n-1)-tx(nx^{n-1})=-1$.
answered Dec 1 at 2:11
Chris Custer
10.7k3724
add a comment |
Since $pnotmid n$, $ncdot 1_Fnot=0$. So there is an inverse, $t$, for $n$.
Now $(x^n-1)-tx(nx^{n-1})=-1$.
answered Dec 1 at 2:11
Chris Custer
10.7k3724
add a comment |
Since $pnotmid n$, $ncdot 1_Fnot=0$. So there is an inverse, $t$, for $n$.
Now $(x^n-1)-tx(nx^{n-1})=-1$.
answered Dec 1 at 2:11
Chris Custer
10.7k3724
Since $pnotmid n$, $ncdot 1_Fnot=0$. So there is an inverse, $t$, for $n$.
Now $(x^n-1)-tx(nx^{n-1})=-1$.
answered Dec 1 at 2:11
Chris Custer
10.7k3724
answered Dec 1 at 2:11
Chris Custer
10.7k3724
answered Dec 1 at 2:11
Chris Custer
10.7k3724
answered Dec 1 at 2:11
Chris Custer
10.7k3724
10.7k3724
add a comment |
add a comment |
You are working on a UFD (actually a Euclidean domain). For two elements $f$ and $g$ to be relatively prime it suffices to show no prime divisor of $f$ is a divisor of $g$.
The second polynomial is a product of linear factors and a constant: multiply the linear polynomial $(x-0)$, with itself $n-1$ times along with constant $n$. As $(x-0)$ is a prime (irreducuble polynomial) and not a factor of $x^n-1$ (use remainder theorem) we are done.
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
add a comment |
You are working on a UFD (actually a Euclidean domain). For two elements $f$ and $g$ to be relatively prime it suffices to show no prime divisor of $f$ is a divisor of $g$.
The second polynomial is a product of linear factors and a constant: multiply the linear polynomial $(x-0)$, with itself $n-1$ times along with constant $n$. As $(x-0)$ is a prime (irreducuble polynomial) and not a factor of $x^n-1$ (use remainder theorem) we are done.
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
add a comment |
You are working on a UFD (actually a Euclidean domain). For two elements $f$ and $g$ to be relatively prime it suffices to show no prime divisor of $f$ is a divisor of $g$.
The second polynomial is a product of linear factors and a constant: multiply the linear polynomial $(x-0)$, with itself $n-1$ times along with constant $n$. As $(x-0)$ is a prime (irreducuble polynomial) and not a factor of $x^n-1$ (use remainder theorem) we are done.
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
You are working on a UFD (actually a Euclidean domain). For two elements $f$ and $g$ to be relatively prime it suffices to show no prime divisor of $f$ is a divisor of $g$.
The second polynomial is a product of linear factors and a constant: multiply the linear polynomial $(x-0)$, with itself $n-1$ times along with constant $n$. As $(x-0)$ is a prime (irreducuble polynomial) and not a factor of $x^n-1$ (use remainder theorem) we are done.
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
answered Dec 1 at 2:14
P Vanchinathan
14.8k12136
14.8k12136
add a comment |
add a comment |
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