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What is the general form of the valid polynomial equations over $mathbb Q$ relating $alpha,beta,gamma$ and $delta$?
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Let $alpha=i$, $beta =-i$, $gamma=sqrt2$ and $delta=-sqrt2$ and consider the permutations
$$R=beginpmatrixalpha&beta&gamma&delta\beta&alpha&gamma&deltaendpmatrix,~S=beginpmatrixalpha&beta&gamma&delta\alpha&beta&delta&gammaendpmatrix.$$
I need to prove that the permutations $R$ and $S$ preserve every valid polynomial equation over $mathbb Q$ relating $alpha,beta,gamma$ and $delta$.
Some of these valid polynomial equations are
$$alpha^2+1=0,~alpha+beta=0,~delta^2-2=0,~gamma+delta=0,~alphagamma-betadelta=0$$
We see that $R$ and $S$ preserve them, but I don't know what is the general form of these polynomials, in order to show that $R$ and $S$ preserve them. Could anyone help me, please?
abstract-algebra field-theory galois-theory
asked Aug 2 at 7:44
user547800
166210
add a comment |Â
up vote
0
down vote
favorite
Let $alpha=i$, $beta =-i$, $gamma=sqrt2$ and $delta=-sqrt2$ and consider the permutations
$$R=beginpmatrixalpha&beta&gamma&delta\beta&alpha&gamma&deltaendpmatrix,~S=beginpmatrixalpha&beta&gamma&delta\alpha&beta&delta&gammaendpmatrix.$$
I need to prove that the permutations $R$ and $S$ preserve every valid polynomial equation over $mathbb Q$ relating $alpha,beta,gamma$ and $delta$.
Some of these valid polynomial equations are
$$alpha^2+1=0,~alpha+beta=0,~delta^2-2=0,~gamma+delta=0,~alphagamma-betadelta=0$$
We see that $R$ and $S$ preserve them, but I don't know what is the general form of these polynomials, in order to show that $R$ and $S$ preserve them. Could anyone help me, please?
abstract-algebra field-theory galois-theory
asked Aug 2 at 7:44
user547800
166210
You may realise that $P_1(X)=X^2+1$ is the minimal polynomial of $alpha$ and $beta$ and so you would get $alpha^2+1=0$(because it is a minimal polynomial), $alpha+beta=0$ (sum of roots is equal to the coefficient of $x$) and the same goes for $P_2(X)=X^2-2$. You may also say that $alpha gamma=beta delta$ and $alpha delta=beta gamma$ satisfies the minimal polynomial $X^2+2=0$ which yields the last relationship
– daruma
Aug 2 at 8:01
@daruma But these are not all the valid polynomial equations. I want to show that $R$ and $S$ preserve all the valid polynomial equations.
– user547800
Aug 2 at 8:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $alpha=i$, $beta =-i$, $gamma=sqrt2$ and $delta=-sqrt2$ and consider the permutations
$$R=beginpmatrixalpha&beta&gamma&delta\beta&alpha&gamma&deltaendpmatrix,~S=beginpmatrixalpha&beta&gamma&delta\alpha&beta&delta&gammaendpmatrix.$$
I need to prove that the permutations $R$ and $S$ preserve every valid polynomial equation over $mathbb Q$ relating $alpha,beta,gamma$ and $delta$.
Some of these valid polynomial equations are
$$alpha^2+1=0,~alpha+beta=0,~delta^2-2=0,~gamma+delta=0,~alphagamma-betadelta=0$$
We see that $R$ and $S$ preserve them, but I don't know what is the general form of these polynomials, in order to show that $R$ and $S$ preserve them. Could anyone help me, please?
abstract-algebra field-theory galois-theory
asked Aug 2 at 7:44
user547800
166210
Let $alpha=i$, $beta =-i$, $gamma=sqrt2$ and $delta=-sqrt2$ and consider the permutations
$$R=beginpmatrixalpha&beta&gamma&delta\beta&alpha&gamma&deltaendpmatrix,~S=beginpmatrixalpha&beta&gamma&delta\alpha&beta&delta&gammaendpmatrix.$$
I need to prove that the permutations $R$ and $S$ preserve every valid polynomial equation over $mathbb Q$ relating $alpha,beta,gamma$ and $delta$.
Some of these valid polynomial equations are
$$alpha^2+1=0,~alpha+beta=0,~delta^2-2=0,~gamma+delta=0,~alphagamma-betadelta=0$$
We see that $R$ and $S$ preserve them, but I don't know what is the general form of these polynomials, in order to show that $R$ and $S$ preserve them. Could anyone help me, please?
abstract-algebra field-theory galois-theory
asked Aug 2 at 7:44
user547800
166210
edited Aug 3 at 5:58
asked Aug 2 at 7:44
user547800
166210
asked Aug 2 at 7:44
user547800
166210
asked Aug 2 at 7:44
user547800
166210
166210
You may realise that $P_1(X)=X^2+1$ is the minimal polynomial of $alpha$ and $beta$ and so you would get $alpha^2+1=0$(because it is a minimal polynomial), $alpha+beta=0$ (sum of roots is equal to the coefficient of $x$) and the same goes for $P_2(X)=X^2-2$. You may also say that $alpha gamma=beta delta$ and $alpha delta=beta gamma$ satisfies the minimal polynomial $X^2+2=0$ which yields the last relationship
– daruma
Aug 2 at 8:01
@daruma But these are not all the valid polynomial equations. I want to show that $R$ and $S$ preserve all the valid polynomial equations.
– user547800
Aug 2 at 8:03
add a comment |Â
You may realise that $P_1(X)=X^2+1$ is the minimal polynomial of $alpha$ and $beta$ and so you would get $alpha^2+1=0$(because it is a minimal polynomial), $alpha+beta=0$ (sum of roots is equal to the coefficient of $x$) and the same goes for $P_2(X)=X^2-2$. You may also say that $alpha gamma=beta delta$ and $alpha delta=beta gamma$ satisfies the minimal polynomial $X^2+2=0$ which yields the last relationship
– daruma
Aug 2 at 8:01
@daruma But these are not all the valid polynomial equations. I want to show that $R$ and $S$ preserve all the valid polynomial equations.
– user547800
Aug 2 at 8:03
You may realise that $P_1(X)=X^2+1$ is the minimal polynomial of $alpha$ and $beta$ and so you would get $alpha^2+1=0$(because it is a minimal polynomial), $alpha+beta=0$ (sum of roots is equal to the coefficient of $x$) and the same goes for $P_2(X)=X^2-2$. You may also say that $alpha gamma=beta delta$ and $alpha delta=beta gamma$ satisfies the minimal polynomial $X^2+2=0$ which yields the last relationship
– daruma
Aug 2 at 8:01
You may realise that $P_1(X)=X^2+1$ is the minimal polynomial of $alpha$ and $beta$ and so you would get $alpha^2+1=0$(because it is a minimal polynomial), $alpha+beta=0$ (sum of roots is equal to the coefficient of $x$) and the same goes for $P_2(X)=X^2-2$. You may also say that $alpha gamma=beta delta$ and $alpha delta=beta gamma$ satisfies the minimal polynomial $X^2+2=0$ which yields the last relationship
– daruma
Aug 2 at 8:01
@daruma But these are not all the valid polynomial equations. I want to show that $R$ and $S$ preserve all the valid polynomial equations.
– user547800
Aug 2 at 8:03
@daruma But these are not all the valid polynomial equations. I want to show that $R$ and $S$ preserve all the valid polynomial equations.
– user547800
Aug 2 at 8:03
add a comment |Â
1 Answer
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0
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For convenience, I am going to rewrite things a little bit:
Let $K=mathbbQ(alpha,delta)$
Let $R: Kto K$ defined by $alphamapsto -alpha$ and $gammamapsto gamma$
Let $S: Kto K$ defined by $alphamapsto alpha$ and $gammamapsto -gamma$
Now $G=langle R, Srangle $ is a group of order $4$.
You may try to prove that $[K:mathbbQ]=4$ (Not too difficult). Note that this is actually a normal extension and separable extension(characteristic of $K$ is clearly $0$ as it is an extension of $mathbbQ$) It is in fact a Galois extension.
Then, $G=Gal(K/mathbbQ)$. If we want to look at all the fixed elements under $G$, you may recall from FTGT that
$K^G=mathbbQ$(fixed field of $G$.) So in particular, $p(alpha,gamma)in mathbbQ$ where $p(X,Y)in mathbbQ[X]$ , it means $R$ and $S$ maps $p(alpha,gamma)$ to itself.
answered Aug 2 at 8:36
daruma
711512
Thanks. The problem is that I didn't study the Galois group yet, so that I'm looking for an elementary way.
– user547800
Aug 2 at 13:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For convenience, I am going to rewrite things a little bit:
Let $K=mathbbQ(alpha,delta)$
Let $R: Kto K$ defined by $alphamapsto -alpha$ and $gammamapsto gamma$
Let $S: Kto K$ defined by $alphamapsto alpha$ and $gammamapsto -gamma$
Now $G=langle R, Srangle $ is a group of order $4$.
You may try to prove that $[K:mathbbQ]=4$ (Not too difficult). Note that this is actually a normal extension and separable extension(characteristic of $K$ is clearly $0$ as it is an extension of $mathbbQ$) It is in fact a Galois extension.
Then, $G=Gal(K/mathbbQ)$. If we want to look at all the fixed elements under $G$, you may recall from FTGT that
$K^G=mathbbQ$(fixed field of $G$.) So in particular, $p(alpha,gamma)in mathbbQ$ where $p(X,Y)in mathbbQ[X]$ , it means $R$ and $S$ maps $p(alpha,gamma)$ to itself.
answered Aug 2 at 8:36
daruma
711512
Thanks. The problem is that I didn't study the Galois group yet, so that I'm looking for an elementary way.
– user547800
Aug 2 at 13:04
add a comment |Â
up vote
0
down vote
For convenience, I am going to rewrite things a little bit:
Let $K=mathbbQ(alpha,delta)$
Let $R: Kto K$ defined by $alphamapsto -alpha$ and $gammamapsto gamma$
Let $S: Kto K$ defined by $alphamapsto alpha$ and $gammamapsto -gamma$
Now $G=langle R, Srangle $ is a group of order $4$.
You may try to prove that $[K:mathbbQ]=4$ (Not too difficult). Note that this is actually a normal extension and separable extension(characteristic of $K$ is clearly $0$ as it is an extension of $mathbbQ$) It is in fact a Galois extension.
Then, $G=Gal(K/mathbbQ)$. If we want to look at all the fixed elements under $G$, you may recall from FTGT that
$K^G=mathbbQ$(fixed field of $G$.) So in particular, $p(alpha,gamma)in mathbbQ$ where $p(X,Y)in mathbbQ[X]$ , it means $R$ and $S$ maps $p(alpha,gamma)$ to itself.
answered Aug 2 at 8:36
daruma
711512
Thanks. The problem is that I didn't study the Galois group yet, so that I'm looking for an elementary way.
– user547800
Aug 2 at 13:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For convenience, I am going to rewrite things a little bit:
Let $K=mathbbQ(alpha,delta)$
Let $R: Kto K$ defined by $alphamapsto -alpha$ and $gammamapsto gamma$
Let $S: Kto K$ defined by $alphamapsto alpha$ and $gammamapsto -gamma$
Now $G=langle R, Srangle $ is a group of order $4$.
You may try to prove that $[K:mathbbQ]=4$ (Not too difficult). Note that this is actually a normal extension and separable extension(characteristic of $K$ is clearly $0$ as it is an extension of $mathbbQ$) It is in fact a Galois extension.
Then, $G=Gal(K/mathbbQ)$. If we want to look at all the fixed elements under $G$, you may recall from FTGT that
$K^G=mathbbQ$(fixed field of $G$.) So in particular, $p(alpha,gamma)in mathbbQ$ where $p(X,Y)in mathbbQ[X]$ , it means $R$ and $S$ maps $p(alpha,gamma)$ to itself.
answered Aug 2 at 8:36
daruma
711512
For convenience, I am going to rewrite things a little bit:
Let $K=mathbbQ(alpha,delta)$
Let $R: Kto K$ defined by $alphamapsto -alpha$ and $gammamapsto gamma$
Let $S: Kto K$ defined by $alphamapsto alpha$ and $gammamapsto -gamma$
Now $G=langle R, Srangle $ is a group of order $4$.
You may try to prove that $[K:mathbbQ]=4$ (Not too difficult). Note that this is actually a normal extension and separable extension(characteristic of $K$ is clearly $0$ as it is an extension of $mathbbQ$) It is in fact a Galois extension.
Then, $G=Gal(K/mathbbQ)$. If we want to look at all the fixed elements under $G$, you may recall from FTGT that
$K^G=mathbbQ$(fixed field of $G$.) So in particular, $p(alpha,gamma)in mathbbQ$ where $p(X,Y)in mathbbQ[X]$ , it means $R$ and $S$ maps $p(alpha,gamma)$ to itself.
answered Aug 2 at 8:36
daruma
711512
answered Aug 2 at 8:36
daruma
711512
answered Aug 2 at 8:36
daruma
711512
answered Aug 2 at 8:36
daruma
711512
711512
Thanks. The problem is that I didn't study the Galois group yet, so that I'm looking for an elementary way.
– user547800
Aug 2 at 13:04
add a comment |Â
Thanks. The problem is that I didn't study the Galois group yet, so that I'm looking for an elementary way.
– user547800
Aug 2 at 13:04
Thanks. The problem is that I didn't study the Galois group yet, so that I'm looking for an elementary way.
– user547800
Aug 2 at 13:04
Thanks. The problem is that I didn't study the Galois group yet, so that I'm looking for an elementary way.
– user547800
Aug 2 at 13:04
add a comment |Â
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You may realise that $P_1(X)=X^2+1$ is the minimal polynomial of $alpha$ and $beta$ and so you would get $alpha^2+1=0$(because it is a minimal polynomial), $alpha+beta=0$ (sum of roots is equal to the coefficient of $x$) and the same goes for $P_2(X)=X^2-2$. You may also say that $alpha gamma=beta delta$ and $alpha delta=beta gamma$ satisfies the minimal polynomial $X^2+2=0$ which yields the last relationship
– daruma
Aug 2 at 8:01
@daruma But these are not all the valid polynomial equations. I want to show that $R$ and $S$ preserve all the valid polynomial equations.
– user547800
Aug 2 at 8:03