Notation and interpretation of Polynomials in $mathbb{F}_{p}[x]$
i'm confused with some notation that involves reduction of polynomyals on $mathbb{Z}[x]$ to $mathbb{F}_p[x]$. It's part of the proof that Cyclotomic polynomials are irreducible over $mathbb{Q}[x]$.
I have the polynomial $f(x)=x^n-1$ that I assuming that is factorized in $f=gh$, with $g,h in mathbb{Z}[x]$ monic polynomials and such that $g$ is the minimal pol. of $zeta_{n}$. Let $p$ be an integer such that $gcd(n,p)=1$, and we assume that $zeta_{n}^p$ is a root of $h$, so $h(x^p)$ has $zeta_{n}$ as root and then $h=fz$, where by the Gauss Lema $z$ has intenger coefieficients.
Here is where I getting problem.
We reduce all the polynomial $mathbb{Z}[x]$ to $mathbb{F}_{p}[x]$ and the by the Little Teo. of Fermat:
$bar{h(x^p)}=bar{g}bar{z}$ and then $bar{h(zeta_n)}=bar{0}$.
So the text affirm that $x^n-bar{1}$ has multiple roots and did exctaly this:
"let $alpha in mathbb{Z}$ be the multiple root, then $alpha^n=1$ and $nalpha^{n-1}=0$ so $n=0$ and since p doest not divide n this is absurd."
This made me very confuse because we are still in $mathbb{F}_p[x]$ since the arguments of the absurd depends on the field characteristic.
In my way of view the sencence should be:
" So $zeta_n$ is a multiple root of $x^n-bar{1}$ and then $bar{zeta_n}^n=bar{1}$ and then $bar{n}bar{zeta_n}^{n-1}=bar{0}$. Since $p$ does not divide $n$ this is an absurd!"
My doubts are:
Is there a convention on how to write ordinary polynomials in $mathbb{F}_p[x]$?
Why did the text take $alpha in mathbb{Z}$ since we just prove that the multiple root is $zeta_n$?
group-theory polynomials irreducible-polynomials roots-of-unity cyclotomic-polynomials
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
add a comment |
i'm confused with some notation that involves reduction of polynomyals on $mathbb{Z}[x]$ to $mathbb{F}_p[x]$. It's part of the proof that Cyclotomic polynomials are irreducible over $mathbb{Q}[x]$.
I have the polynomial $f(x)=x^n-1$ that I assuming that is factorized in $f=gh$, with $g,h in mathbb{Z}[x]$ monic polynomials and such that $g$ is the minimal pol. of $zeta_{n}$. Let $p$ be an integer such that $gcd(n,p)=1$, and we assume that $zeta_{n}^p$ is a root of $h$, so $h(x^p)$ has $zeta_{n}$ as root and then $h=fz$, where by the Gauss Lema $z$ has intenger coefieficients.
Here is where I getting problem.
We reduce all the polynomial $mathbb{Z}[x]$ to $mathbb{F}_{p}[x]$ and the by the Little Teo. of Fermat:
$bar{h(x^p)}=bar{g}bar{z}$ and then $bar{h(zeta_n)}=bar{0}$.
So the text affirm that $x^n-bar{1}$ has multiple roots and did exctaly this:
"let $alpha in mathbb{Z}$ be the multiple root, then $alpha^n=1$ and $nalpha^{n-1}=0$ so $n=0$ and since p doest not divide n this is absurd."
This made me very confuse because we are still in $mathbb{F}_p[x]$ since the arguments of the absurd depends on the field characteristic.
In my way of view the sencence should be:
" So $zeta_n$ is a multiple root of $x^n-bar{1}$ and then $bar{zeta_n}^n=bar{1}$ and then $bar{n}bar{zeta_n}^{n-1}=bar{0}$. Since $p$ does not divide $n$ this is an absurd!"
My doubts are:
Is there a convention on how to write ordinary polynomials in $mathbb{F}_p[x]$?
Why did the text take $alpha in mathbb{Z}$ since we just prove that the multiple root is $zeta_n$?
group-theory polynomials irreducible-polynomials roots-of-unity cyclotomic-polynomials
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
What is mdc? I assume it is $gcd$ (greatest common divisor), but I wonder what language this is from..
– Hagen von Eitzen
Nov 21 '18 at 21:19
you right, thats was the usual notation on portuguese, I've edited.
– Eduardo Silva
Nov 21 '18 at 21:22
1
Error: when $h(zeta_n^p) = 0$, so $zeta_n$ is a root of $h(x^p)$, this means $h(x^p)$ is divisible in $mathbf Z[x]$ by $g(x)$, not $f(x) = x^n - 1$. For example, $i$ has minimal polynomial $x^2 + 1$ and $i$ is a root of $x^4 + 3x^2 + 2$, but this does not mean $x^4 + 3x^2 + 2$ is divisible by $x^4 - 1$; it is divisible by $x^2 + 1$. You can find a treatment of this proof in Theorem 2.5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf.
– KCd
Nov 21 '18 at 21:56
Oh, sorry, that should be $bar{h(x^p)}=bar{g}bar{z}$ instead
– Eduardo Silva
Nov 22 '18 at 2:02
add a comment |
i'm confused with some notation that involves reduction of polynomyals on $mathbb{Z}[x]$ to $mathbb{F}_p[x]$. It's part of the proof that Cyclotomic polynomials are irreducible over $mathbb{Q}[x]$.
I have the polynomial $f(x)=x^n-1$ that I assuming that is factorized in $f=gh$, with $g,h in mathbb{Z}[x]$ monic polynomials and such that $g$ is the minimal pol. of $zeta_{n}$. Let $p$ be an integer such that $gcd(n,p)=1$, and we assume that $zeta_{n}^p$ is a root of $h$, so $h(x^p)$ has $zeta_{n}$ as root and then $h=fz$, where by the Gauss Lema $z$ has intenger coefieficients.
Here is where I getting problem.
We reduce all the polynomial $mathbb{Z}[x]$ to $mathbb{F}_{p}[x]$ and the by the Little Teo. of Fermat:
$bar{h(x^p)}=bar{g}bar{z}$ and then $bar{h(zeta_n)}=bar{0}$.
So the text affirm that $x^n-bar{1}$ has multiple roots and did exctaly this:
"let $alpha in mathbb{Z}$ be the multiple root, then $alpha^n=1$ and $nalpha^{n-1}=0$ so $n=0$ and since p doest not divide n this is absurd."
This made me very confuse because we are still in $mathbb{F}_p[x]$ since the arguments of the absurd depends on the field characteristic.
In my way of view the sencence should be:
" So $zeta_n$ is a multiple root of $x^n-bar{1}$ and then $bar{zeta_n}^n=bar{1}$ and then $bar{n}bar{zeta_n}^{n-1}=bar{0}$. Since $p$ does not divide $n$ this is an absurd!"
My doubts are:
Is there a convention on how to write ordinary polynomials in $mathbb{F}_p[x]$?
Why did the text take $alpha in mathbb{Z}$ since we just prove that the multiple root is $zeta_n$?
group-theory polynomials irreducible-polynomials roots-of-unity cyclotomic-polynomials
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
i'm confused with some notation that involves reduction of polynomyals on $mathbb{Z}[x]$ to $mathbb{F}_p[x]$. It's part of the proof that Cyclotomic polynomials are irreducible over $mathbb{Q}[x]$.
I have the polynomial $f(x)=x^n-1$ that I assuming that is factorized in $f=gh$, with $g,h in mathbb{Z}[x]$ monic polynomials and such that $g$ is the minimal pol. of $zeta_{n}$. Let $p$ be an integer such that $gcd(n,p)=1$, and we assume that $zeta_{n}^p$ is a root of $h$, so $h(x^p)$ has $zeta_{n}$ as root and then $h=fz$, where by the Gauss Lema $z$ has intenger coefieficients.
Here is where I getting problem.
We reduce all the polynomial $mathbb{Z}[x]$ to $mathbb{F}_{p}[x]$ and the by the Little Teo. of Fermat:
$bar{h(x^p)}=bar{g}bar{z}$ and then $bar{h(zeta_n)}=bar{0}$.
So the text affirm that $x^n-bar{1}$ has multiple roots and did exctaly this:
"let $alpha in mathbb{Z}$ be the multiple root, then $alpha^n=1$ and $nalpha^{n-1}=0$ so $n=0$ and since p doest not divide n this is absurd."
This made me very confuse because we are still in $mathbb{F}_p[x]$ since the arguments of the absurd depends on the field characteristic.
In my way of view the sencence should be:
" So $zeta_n$ is a multiple root of $x^n-bar{1}$ and then $bar{zeta_n}^n=bar{1}$ and then $bar{n}bar{zeta_n}^{n-1}=bar{0}$. Since $p$ does not divide $n$ this is an absurd!"
My doubts are:
Is there a convention on how to write ordinary polynomials in $mathbb{F}_p[x]$?
Why did the text take $alpha in mathbb{Z}$ since we just prove that the multiple root is $zeta_n$?
group-theory polynomials irreducible-polynomials roots-of-unity cyclotomic-polynomials
group-theory polynomials irreducible-polynomials roots-of-unity cyclotomic-polynomials
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
edited Nov 22 '18 at 2:02
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
asked Nov 21 '18 at 21:14
Eduardo Silva
68239
68239
What is mdc? I assume it is $gcd$ (greatest common divisor), but I wonder what language this is from..
– Hagen von Eitzen
Nov 21 '18 at 21:19
you right, thats was the usual notation on portuguese, I've edited.
– Eduardo Silva
Nov 21 '18 at 21:22
1
Error: when $h(zeta_n^p) = 0$, so $zeta_n$ is a root of $h(x^p)$, this means $h(x^p)$ is divisible in $mathbf Z[x]$ by $g(x)$, not $f(x) = x^n - 1$. For example, $i$ has minimal polynomial $x^2 + 1$ and $i$ is a root of $x^4 + 3x^2 + 2$, but this does not mean $x^4 + 3x^2 + 2$ is divisible by $x^4 - 1$; it is divisible by $x^2 + 1$. You can find a treatment of this proof in Theorem 2.5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf.
– KCd
Nov 21 '18 at 21:56
Oh, sorry, that should be $bar{h(x^p)}=bar{g}bar{z}$ instead
– Eduardo Silva
Nov 22 '18 at 2:02
add a comment |
What is mdc? I assume it is $gcd$ (greatest common divisor), but I wonder what language this is from..
– Hagen von Eitzen
Nov 21 '18 at 21:19
you right, thats was the usual notation on portuguese, I've edited.
– Eduardo Silva
Nov 21 '18 at 21:22
1
Error: when $h(zeta_n^p) = 0$, so $zeta_n$ is a root of $h(x^p)$, this means $h(x^p)$ is divisible in $mathbf Z[x]$ by $g(x)$, not $f(x) = x^n - 1$. For example, $i$ has minimal polynomial $x^2 + 1$ and $i$ is a root of $x^4 + 3x^2 + 2$, but this does not mean $x^4 + 3x^2 + 2$ is divisible by $x^4 - 1$; it is divisible by $x^2 + 1$. You can find a treatment of this proof in Theorem 2.5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf.
– KCd
Nov 21 '18 at 21:56
Oh, sorry, that should be $bar{h(x^p)}=bar{g}bar{z}$ instead
– Eduardo Silva
Nov 22 '18 at 2:02
What is mdc? I assume it is $gcd$ (greatest common divisor), but I wonder what language this is from..
– Hagen von Eitzen
Nov 21 '18 at 21:19
What is mdc? I assume it is $gcd$ (greatest common divisor), but I wonder what language this is from..
– Hagen von Eitzen
Nov 21 '18 at 21:19
you right, thats was the usual notation on portuguese, I've edited.
– Eduardo Silva
Nov 21 '18 at 21:22
you right, thats was the usual notation on portuguese, I've edited.
– Eduardo Silva
Nov 21 '18 at 21:22
1
1
Error: when $h(zeta_n^p) = 0$, so $zeta_n$ is a root of $h(x^p)$, this means $h(x^p)$ is divisible in $mathbf Z[x]$ by $g(x)$, not $f(x) = x^n - 1$. For example, $i$ has minimal polynomial $x^2 + 1$ and $i$ is a root of $x^4 + 3x^2 + 2$, but this does not mean $x^4 + 3x^2 + 2$ is divisible by $x^4 - 1$; it is divisible by $x^2 + 1$. You can find a treatment of this proof in Theorem 2.5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf.
– KCd
Nov 21 '18 at 21:56
Error: when $h(zeta_n^p) = 0$, so $zeta_n$ is a root of $h(x^p)$, this means $h(x^p)$ is divisible in $mathbf Z[x]$ by $g(x)$, not $f(x) = x^n - 1$. For example, $i$ has minimal polynomial $x^2 + 1$ and $i$ is a root of $x^4 + 3x^2 + 2$, but this does not mean $x^4 + 3x^2 + 2$ is divisible by $x^4 - 1$; it is divisible by $x^2 + 1$. You can find a treatment of this proof in Theorem 2.5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf.
– KCd
Nov 21 '18 at 21:56
Oh, sorry, that should be $bar{h(x^p)}=bar{g}bar{z}$ instead
– Eduardo Silva
Nov 22 '18 at 2:02
Oh, sorry, that should be $bar{h(x^p)}=bar{g}bar{z}$ instead
– Eduardo Silva
Nov 22 '18 at 2:02
add a comment |
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What is mdc? I assume it is $gcd$ (greatest common divisor), but I wonder what language this is from..
– Hagen von Eitzen
Nov 21 '18 at 21:19
you right, thats was the usual notation on portuguese, I've edited.
– Eduardo Silva
Nov 21 '18 at 21:22
1
Error: when $h(zeta_n^p) = 0$, so $zeta_n$ is a root of $h(x^p)$, this means $h(x^p)$ is divisible in $mathbf Z[x]$ by $g(x)$, not $f(x) = x^n - 1$. For example, $i$ has minimal polynomial $x^2 + 1$ and $i$ is a root of $x^4 + 3x^2 + 2$, but this does not mean $x^4 + 3x^2 + 2$ is divisible by $x^4 - 1$; it is divisible by $x^2 + 1$. You can find a treatment of this proof in Theorem 2.5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf.
– KCd
Nov 21 '18 at 21:56
Oh, sorry, that should be $bar{h(x^p)}=bar{g}bar{z}$ instead
– Eduardo Silva
Nov 22 '18 at 2:02