Mixing
algebraic integer in $mathbb{Q}$($zeta_n$)
$begingroup$
Let $chi(cdot)$ be an irreducible (over $mathbb{C}$) character of an representation in a finite Group G:
Show that $chi(g)$ is an algebraic integer in the cyclotomic
field $mathbb{Q}$($zeta_n$), where $zeta_n$ := $e^{2pi i/n}$.
I already got an answer here: my old question
But I struggle with the part, where he follows the similarity of M(g) to some diagonal matrix.
So any help or even a new approach?
abstract-algebra group-theory representation-theory cyclotomic-fields
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
$endgroup$
add a comment |
$begingroup$
Let $chi(cdot)$ be an irreducible (over $mathbb{C}$) character of an representation in a finite Group G:
Show that $chi(g)$ is an algebraic integer in the cyclotomic
field $mathbb{Q}$($zeta_n$), where $zeta_n$ := $e^{2pi i/n}$.
I already got an answer here: my old question
But I struggle with the part, where he follows the similarity of M(g) to some diagonal matrix.
So any help or even a new approach?
abstract-algebra group-theory representation-theory cyclotomic-fields
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
$endgroup$
1
$begingroup$
Use Jordan canonical form.
$endgroup$
– jgon
Jan 6 at 14:33
add a comment |
$begingroup$
Let $chi(cdot)$ be an irreducible (over $mathbb{C}$) character of an representation in a finite Group G:
Show that $chi(g)$ is an algebraic integer in the cyclotomic
field $mathbb{Q}$($zeta_n$), where $zeta_n$ := $e^{2pi i/n}$.
I already got an answer here: my old question
But I struggle with the part, where he follows the similarity of M(g) to some diagonal matrix.
So any help or even a new approach?
abstract-algebra group-theory representation-theory cyclotomic-fields
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
$endgroup$
Let $chi(cdot)$ be an irreducible (over $mathbb{C}$) character of an representation in a finite Group G:
Show that $chi(g)$ is an algebraic integer in the cyclotomic
field $mathbb{Q}$($zeta_n$), where $zeta_n$ := $e^{2pi i/n}$.
I already got an answer here: my old question
But I struggle with the part, where he follows the similarity of M(g) to some diagonal matrix.
So any help or even a new approach?
abstract-algebra group-theory representation-theory cyclotomic-fields
abstract-algebra group-theory representation-theory cyclotomic-fields
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
asked Jan 6 at 14:28
FabianSchneiderFabianSchneider
556
556
1
$begingroup$
Use Jordan canonical form.
$endgroup$
– jgon
Jan 6 at 14:33
add a comment |
1
$begingroup$
Use Jordan canonical form.
$endgroup$
– jgon
Jan 6 at 14:33
1
1
$begingroup$
Use Jordan canonical form.
$endgroup$
– jgon
Jan 6 at 14:33
$begingroup$
Use Jordan canonical form.
$endgroup$
– jgon
Jan 6 at 14:33
add a comment |
1 Answer
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$begingroup$
A fact from linear algebra: If the minimal polynomial of a matrix $Min mathrm{M}_n(mathbb{C})$ has no repeated roots then $M$ is diagonalizable. You can prove this as follows. If $p,q$ are coprime polynomials then using Bézout's identity we have $ker(pq(M))=ker(p(M))oplus ker(q(M))$. So in particular if the minimal polynomial of $M$ is say $(z-lambda_1)timescdotstimes (z-lambda_k)$ with the $lambda_i's$ distinct, then $mathbb{C}^n = ker(M-lambda_1 I_n)oplus cdots oplus (M-lambda_k I_n)$, so $M$ is diagonalizable.
Now if $rho:Grightarrow mathrm{GL}_n(mathbb{C})$ is a representation with character $chi(g)=mathrm{tr}left(rho(g)right)$ then $I_n=rholeft(g^{|G|} right)= rholeft(g right)^{|G|}$. Hence the minimal polynomial of $rholeft(g right)$, which is a divisor of $X^{|G|}-1$, has distinct roots. Hence $rholeft(g right)$ is diagonalizable with eigenvalues (which are some of the roots of $X^{|G|}-1$) in $mathbb{Q}(zeta_{|G|})$ and so is their sum, which is $chi(g)$.
answered Jan 6 at 17:02
moutheticsmouthetics
50137
$endgroup$
add a comment |
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$begingroup$
A fact from linear algebra: If the minimal polynomial of a matrix $Min mathrm{M}_n(mathbb{C})$ has no repeated roots then $M$ is diagonalizable. You can prove this as follows. If $p,q$ are coprime polynomials then using Bézout's identity we have $ker(pq(M))=ker(p(M))oplus ker(q(M))$. So in particular if the minimal polynomial of $M$ is say $(z-lambda_1)timescdotstimes (z-lambda_k)$ with the $lambda_i's$ distinct, then $mathbb{C}^n = ker(M-lambda_1 I_n)oplus cdots oplus (M-lambda_k I_n)$, so $M$ is diagonalizable.
Now if $rho:Grightarrow mathrm{GL}_n(mathbb{C})$ is a representation with character $chi(g)=mathrm{tr}left(rho(g)right)$ then $I_n=rholeft(g^{|G|} right)= rholeft(g right)^{|G|}$. Hence the minimal polynomial of $rholeft(g right)$, which is a divisor of $X^{|G|}-1$, has distinct roots. Hence $rholeft(g right)$ is diagonalizable with eigenvalues (which are some of the roots of $X^{|G|}-1$) in $mathbb{Q}(zeta_{|G|})$ and so is their sum, which is $chi(g)$.
answered Jan 6 at 17:02
moutheticsmouthetics
50137
$endgroup$
add a comment |
$begingroup$
A fact from linear algebra: If the minimal polynomial of a matrix $Min mathrm{M}_n(mathbb{C})$ has no repeated roots then $M$ is diagonalizable. You can prove this as follows. If $p,q$ are coprime polynomials then using Bézout's identity we have $ker(pq(M))=ker(p(M))oplus ker(q(M))$. So in particular if the minimal polynomial of $M$ is say $(z-lambda_1)timescdotstimes (z-lambda_k)$ with the $lambda_i's$ distinct, then $mathbb{C}^n = ker(M-lambda_1 I_n)oplus cdots oplus (M-lambda_k I_n)$, so $M$ is diagonalizable.
Now if $rho:Grightarrow mathrm{GL}_n(mathbb{C})$ is a representation with character $chi(g)=mathrm{tr}left(rho(g)right)$ then $I_n=rholeft(g^{|G|} right)= rholeft(g right)^{|G|}$. Hence the minimal polynomial of $rholeft(g right)$, which is a divisor of $X^{|G|}-1$, has distinct roots. Hence $rholeft(g right)$ is diagonalizable with eigenvalues (which are some of the roots of $X^{|G|}-1$) in $mathbb{Q}(zeta_{|G|})$ and so is their sum, which is $chi(g)$.
answered Jan 6 at 17:02
moutheticsmouthetics
50137
$endgroup$
add a comment |
$begingroup$
A fact from linear algebra: If the minimal polynomial of a matrix $Min mathrm{M}_n(mathbb{C})$ has no repeated roots then $M$ is diagonalizable. You can prove this as follows. If $p,q$ are coprime polynomials then using Bézout's identity we have $ker(pq(M))=ker(p(M))oplus ker(q(M))$. So in particular if the minimal polynomial of $M$ is say $(z-lambda_1)timescdotstimes (z-lambda_k)$ with the $lambda_i's$ distinct, then $mathbb{C}^n = ker(M-lambda_1 I_n)oplus cdots oplus (M-lambda_k I_n)$, so $M$ is diagonalizable.
Now if $rho:Grightarrow mathrm{GL}_n(mathbb{C})$ is a representation with character $chi(g)=mathrm{tr}left(rho(g)right)$ then $I_n=rholeft(g^{|G|} right)= rholeft(g right)^{|G|}$. Hence the minimal polynomial of $rholeft(g right)$, which is a divisor of $X^{|G|}-1$, has distinct roots. Hence $rholeft(g right)$ is diagonalizable with eigenvalues (which are some of the roots of $X^{|G|}-1$) in $mathbb{Q}(zeta_{|G|})$ and so is their sum, which is $chi(g)$.
answered Jan 6 at 17:02
moutheticsmouthetics
50137
$endgroup$
A fact from linear algebra: If the minimal polynomial of a matrix $Min mathrm{M}_n(mathbb{C})$ has no repeated roots then $M$ is diagonalizable. You can prove this as follows. If $p,q$ are coprime polynomials then using Bézout's identity we have $ker(pq(M))=ker(p(M))oplus ker(q(M))$. So in particular if the minimal polynomial of $M$ is say $(z-lambda_1)timescdotstimes (z-lambda_k)$ with the $lambda_i's$ distinct, then $mathbb{C}^n = ker(M-lambda_1 I_n)oplus cdots oplus (M-lambda_k I_n)$, so $M$ is diagonalizable.
Now if $rho:Grightarrow mathrm{GL}_n(mathbb{C})$ is a representation with character $chi(g)=mathrm{tr}left(rho(g)right)$ then $I_n=rholeft(g^{|G|} right)= rholeft(g right)^{|G|}$. Hence the minimal polynomial of $rholeft(g right)$, which is a divisor of $X^{|G|}-1$, has distinct roots. Hence $rholeft(g right)$ is diagonalizable with eigenvalues (which are some of the roots of $X^{|G|}-1$) in $mathbb{Q}(zeta_{|G|})$ and so is their sum, which is $chi(g)$.
answered Jan 6 at 17:02
moutheticsmouthetics
50137
answered Jan 6 at 17:02
moutheticsmouthetics
50137
answered Jan 6 at 17:02
moutheticsmouthetics
50137
answered Jan 6 at 17:02
moutheticsmouthetics
50137
50137
add a comment |
add a comment |
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Use Jordan canonical form.
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– jgon
Jan 6 at 14:33