Abelian Galois group of even order












2














I'm stuck at the following problem.




Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$. Let $K= mathbb{Q}(a+bi)$ be a simple extension of $mathbb{Q}$. Suppose that $K$ is a Galois extention over $mathbb{Q}$ and the the Galois group $G(K/mathbb{Q})$ is abelian. Moreover, assume that $a^2+b^2 in mathbb{Q}$ and $b neq 0$.
(1) Prove that the order of $G(K/mathbb{Q})$ is even.
(2) Let the order of $G(K/mathbb{Q})$ is $2m$. Prove that for each positive divisor $d$ of $m$, there exists an irreducible polynomial over $mathbb{Q}$ whose zeros are all real and degree $d$.




This is an attempt to solve this problem.




  1. Let $rho$ be the complex conjugate transformation and $f(x)$ be the minimal polynomial of $a+bi$. Since $f(x)$ is invariant under $rho$, $rho$ restricted on $K$ is a member of the Galois group.

  2. Let $E$ be the fixed field of $rho$. We have an intermediate field $K-E-mathbb{Q}$ where $K=E[i]$.

  3. Degree of $K$ over $E$ is $2$. Hence the order of the Galois group is even. Moreover the Galois group of $E$ over $mathbb{Q}$ is abelian. Therefore from the converse of the Lagrange's theorem, for each divisor $d$ of $m$, there corresponds a subgroup of the group and an irreducible polynomial.


But I have not used the condition $a^2+b^2 in mathbb{Q}$. This should be wrong.






abstract-algebra field-theory galois-theory irreducible-polynomials







share|cite|improve this question
























  • Maybe you might use it to show that $K$ is (totally) complex?
    – Lubin
    Nov 26 at 20:54










  • Thanks reuns. But I still wonder where the condition $a^2+b^2 in mathbb{Q}$ enters. Do we need it to show that $K-E-mathbb{Q}$ is an intermediate field so that the complex conjugation fixes $E$?
    – seoneo
    Nov 27 at 1:24










  • But, @reuns, what is complex conjugation of $a+bi$ if $a$ and $b$ are not real? OP has just said that they are algebraic numbers...
    – Lubin
    Nov 27 at 2:07










  • @Lubin. I made a serious errata. $a$ and $b$ are real algebraic. I'm sorry. Now it is fixed.
    – seoneo
    Nov 27 at 2:09








  • 1




    There is no reason for $i$ to be an element of $K$. Consider the case of $a+bi=1+sqrt3 i$. What you can use is that complex conjugation is a non-trivial automorphism of $K$. What is the order of complex conjugation? What does that tell you about the order of $G$.
    – Jyrki Lahtonen
    Nov 27 at 5:09
















2














I'm stuck at the following problem.




Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$. Let $K= mathbb{Q}(a+bi)$ be a simple extension of $mathbb{Q}$. Suppose that $K$ is a Galois extention over $mathbb{Q}$ and the the Galois group $G(K/mathbb{Q})$ is abelian. Moreover, assume that $a^2+b^2 in mathbb{Q}$ and $b neq 0$.
(1) Prove that the order of $G(K/mathbb{Q})$ is even.
(2) Let the order of $G(K/mathbb{Q})$ is $2m$. Prove that for each positive divisor $d$ of $m$, there exists an irreducible polynomial over $mathbb{Q}$ whose zeros are all real and degree $d$.




This is an attempt to solve this problem.




  1. Let $rho$ be the complex conjugate transformation and $f(x)$ be the minimal polynomial of $a+bi$. Since $f(x)$ is invariant under $rho$, $rho$ restricted on $K$ is a member of the Galois group.

  2. Let $E$ be the fixed field of $rho$. We have an intermediate field $K-E-mathbb{Q}$ where $K=E[i]$.

  3. Degree of $K$ over $E$ is $2$. Hence the order of the Galois group is even. Moreover the Galois group of $E$ over $mathbb{Q}$ is abelian. Therefore from the converse of the Lagrange's theorem, for each divisor $d$ of $m$, there corresponds a subgroup of the group and an irreducible polynomial.


But I have not used the condition $a^2+b^2 in mathbb{Q}$. This should be wrong.






abstract-algebra field-theory galois-theory irreducible-polynomials







share|cite|improve this question
























  • Maybe you might use it to show that $K$ is (totally) complex?
    – Lubin
    Nov 26 at 20:54










  • Thanks reuns. But I still wonder where the condition $a^2+b^2 in mathbb{Q}$ enters. Do we need it to show that $K-E-mathbb{Q}$ is an intermediate field so that the complex conjugation fixes $E$?
    – seoneo
    Nov 27 at 1:24










  • But, @reuns, what is complex conjugation of $a+bi$ if $a$ and $b$ are not real? OP has just said that they are algebraic numbers...
    – Lubin
    Nov 27 at 2:07










  • @Lubin. I made a serious errata. $a$ and $b$ are real algebraic. I'm sorry. Now it is fixed.
    – seoneo
    Nov 27 at 2:09








  • 1




    There is no reason for $i$ to be an element of $K$. Consider the case of $a+bi=1+sqrt3 i$. What you can use is that complex conjugation is a non-trivial automorphism of $K$. What is the order of complex conjugation? What does that tell you about the order of $G$.
    – Jyrki Lahtonen
    Nov 27 at 5:09














2












2








2







I'm stuck at the following problem.




Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$. Let $K= mathbb{Q}(a+bi)$ be a simple extension of $mathbb{Q}$. Suppose that $K$ is a Galois extention over $mathbb{Q}$ and the the Galois group $G(K/mathbb{Q})$ is abelian. Moreover, assume that $a^2+b^2 in mathbb{Q}$ and $b neq 0$.
(1) Prove that the order of $G(K/mathbb{Q})$ is even.
(2) Let the order of $G(K/mathbb{Q})$ is $2m$. Prove that for each positive divisor $d$ of $m$, there exists an irreducible polynomial over $mathbb{Q}$ whose zeros are all real and degree $d$.




This is an attempt to solve this problem.




  1. Let $rho$ be the complex conjugate transformation and $f(x)$ be the minimal polynomial of $a+bi$. Since $f(x)$ is invariant under $rho$, $rho$ restricted on $K$ is a member of the Galois group.

  2. Let $E$ be the fixed field of $rho$. We have an intermediate field $K-E-mathbb{Q}$ where $K=E[i]$.

  3. Degree of $K$ over $E$ is $2$. Hence the order of the Galois group is even. Moreover the Galois group of $E$ over $mathbb{Q}$ is abelian. Therefore from the converse of the Lagrange's theorem, for each divisor $d$ of $m$, there corresponds a subgroup of the group and an irreducible polynomial.


But I have not used the condition $a^2+b^2 in mathbb{Q}$. This should be wrong.






abstract-algebra field-theory galois-theory irreducible-polynomials







share|cite|improve this question















I'm stuck at the following problem.




Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$. Let $K= mathbb{Q}(a+bi)$ be a simple extension of $mathbb{Q}$. Suppose that $K$ is a Galois extention over $mathbb{Q}$ and the the Galois group $G(K/mathbb{Q})$ is abelian. Moreover, assume that $a^2+b^2 in mathbb{Q}$ and $b neq 0$.
(1) Prove that the order of $G(K/mathbb{Q})$ is even.
(2) Let the order of $G(K/mathbb{Q})$ is $2m$. Prove that for each positive divisor $d$ of $m$, there exists an irreducible polynomial over $mathbb{Q}$ whose zeros are all real and degree $d$.




This is an attempt to solve this problem.




  1. Let $rho$ be the complex conjugate transformation and $f(x)$ be the minimal polynomial of $a+bi$. Since $f(x)$ is invariant under $rho$, $rho$ restricted on $K$ is a member of the Galois group.

  2. Let $E$ be the fixed field of $rho$. We have an intermediate field $K-E-mathbb{Q}$ where $K=E[i]$.

  3. Degree of $K$ over $E$ is $2$. Hence the order of the Galois group is even. Moreover the Galois group of $E$ over $mathbb{Q}$ is abelian. Therefore from the converse of the Lagrange's theorem, for each divisor $d$ of $m$, there corresponds a subgroup of the group and an irreducible polynomial.


But I have not used the condition $a^2+b^2 in mathbb{Q}$. This should be wrong.






abstract-algebra field-theory galois-theory irreducible-polynomials




abstract-algebra field-theory galois-theory irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 2:08

























asked Nov 26 at 20:28









seoneo

492213




492213












  • Maybe you might use it to show that $K$ is (totally) complex?
    – Lubin
    Nov 26 at 20:54










  • Thanks reuns. But I still wonder where the condition $a^2+b^2 in mathbb{Q}$ enters. Do we need it to show that $K-E-mathbb{Q}$ is an intermediate field so that the complex conjugation fixes $E$?
    – seoneo
    Nov 27 at 1:24










  • But, @reuns, what is complex conjugation of $a+bi$ if $a$ and $b$ are not real? OP has just said that they are algebraic numbers...
    – Lubin
    Nov 27 at 2:07










  • @Lubin. I made a serious errata. $a$ and $b$ are real algebraic. I'm sorry. Now it is fixed.
    – seoneo
    Nov 27 at 2:09








  • 1




    There is no reason for $i$ to be an element of $K$. Consider the case of $a+bi=1+sqrt3 i$. What you can use is that complex conjugation is a non-trivial automorphism of $K$. What is the order of complex conjugation? What does that tell you about the order of $G$.
    – Jyrki Lahtonen
    Nov 27 at 5:09


















  • Maybe you might use it to show that $K$ is (totally) complex?
    – Lubin
    Nov 26 at 20:54










  • Thanks reuns. But I still wonder where the condition $a^2+b^2 in mathbb{Q}$ enters. Do we need it to show that $K-E-mathbb{Q}$ is an intermediate field so that the complex conjugation fixes $E$?
    – seoneo
    Nov 27 at 1:24










  • But, @reuns, what is complex conjugation of $a+bi$ if $a$ and $b$ are not real? OP has just said that they are algebraic numbers...
    – Lubin
    Nov 27 at 2:07










  • @Lubin. I made a serious errata. $a$ and $b$ are real algebraic. I'm sorry. Now it is fixed.
    – seoneo
    Nov 27 at 2:09








  • 1




    There is no reason for $i$ to be an element of $K$. Consider the case of $a+bi=1+sqrt3 i$. What you can use is that complex conjugation is a non-trivial automorphism of $K$. What is the order of complex conjugation? What does that tell you about the order of $G$.
    – Jyrki Lahtonen
    Nov 27 at 5:09
















Maybe you might use it to show that $K$ is (totally) complex?
– Lubin
Nov 26 at 20:54




Maybe you might use it to show that $K$ is (totally) complex?
– Lubin
Nov 26 at 20:54












Thanks reuns. But I still wonder where the condition $a^2+b^2 in mathbb{Q}$ enters. Do we need it to show that $K-E-mathbb{Q}$ is an intermediate field so that the complex conjugation fixes $E$?
– seoneo
Nov 27 at 1:24




Thanks reuns. But I still wonder where the condition $a^2+b^2 in mathbb{Q}$ enters. Do we need it to show that $K-E-mathbb{Q}$ is an intermediate field so that the complex conjugation fixes $E$?
– seoneo
Nov 27 at 1:24












But, @reuns, what is complex conjugation of $a+bi$ if $a$ and $b$ are not real? OP has just said that they are algebraic numbers...
– Lubin
Nov 27 at 2:07




But, @reuns, what is complex conjugation of $a+bi$ if $a$ and $b$ are not real? OP has just said that they are algebraic numbers...
– Lubin
Nov 27 at 2:07












@Lubin. I made a serious errata. $a$ and $b$ are real algebraic. I'm sorry. Now it is fixed.
– seoneo
Nov 27 at 2:09






@Lubin. I made a serious errata. $a$ and $b$ are real algebraic. I'm sorry. Now it is fixed.
– seoneo
Nov 27 at 2:09






1




1




There is no reason for $i$ to be an element of $K$. Consider the case of $a+bi=1+sqrt3 i$. What you can use is that complex conjugation is a non-trivial automorphism of $K$. What is the order of complex conjugation? What does that tell you about the order of $G$.
– Jyrki Lahtonen
Nov 27 at 5:09




There is no reason for $i$ to be an element of $K$. Consider the case of $a+bi=1+sqrt3 i$. What you can use is that complex conjugation is a non-trivial automorphism of $K$. What is the order of complex conjugation? What does that tell you about the order of $G$.
– Jyrki Lahtonen
Nov 27 at 5:09










1 Answer
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Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$ and let $K = mathbb{Q}(a+bi)$. Assume that $K$ over $mathbb{Q}$ is a Galois extension whose Galois group is abelian.



Since $K-mathbb{Q}$ is Galois extension, we have $a-bi in K$ and so $a in K$. Consider an intermediate field $K-E-mathbb{Q}$ where $E = mathbb{Q}(a)$. Consider $x^2-2ax+a^2+b^2 in E[x]$. This is irreducible since $b neq 0$ and $E subseteq mathbb{R}$. This is the minimal polynomial of $a+bi$ over $E$. Therefore $K-E$ is degree $2$. This tells us that the order of the Galois group is even, say $2m$ where $m = [E:mathbb{Q}]$.



Since $G=G(K/mathbb{Q})$ is abelian, it follows that $G(E/mathbb{Q})$ is also abelian. Hence by the inverse of the Lagrange's theorem, for each divisor $d$ of $m$, we have a subgroup $H$ of $G$ of order $m/d$. Let $F$ be the fixed field in $E$ by $H$. Then $[F:mathbb{Q}] = d$. Since characteristic is zero, $F = mathbb{Q}(alpha)$, we have that $hbox{irr}(alpha, mathbb{Q})$ has exactly $d$ real solutions degree $d$ polynomial. One more thing we can require is that the polynomial has zeros on $K cap mathbb{R}$.






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    Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$ and let $K = mathbb{Q}(a+bi)$. Assume that $K$ over $mathbb{Q}$ is a Galois extension whose Galois group is abelian.



    Since $K-mathbb{Q}$ is Galois extension, we have $a-bi in K$ and so $a in K$. Consider an intermediate field $K-E-mathbb{Q}$ where $E = mathbb{Q}(a)$. Consider $x^2-2ax+a^2+b^2 in E[x]$. This is irreducible since $b neq 0$ and $E subseteq mathbb{R}$. This is the minimal polynomial of $a+bi$ over $E$. Therefore $K-E$ is degree $2$. This tells us that the order of the Galois group is even, say $2m$ where $m = [E:mathbb{Q}]$.



    Since $G=G(K/mathbb{Q})$ is abelian, it follows that $G(E/mathbb{Q})$ is also abelian. Hence by the inverse of the Lagrange's theorem, for each divisor $d$ of $m$, we have a subgroup $H$ of $G$ of order $m/d$. Let $F$ be the fixed field in $E$ by $H$. Then $[F:mathbb{Q}] = d$. Since characteristic is zero, $F = mathbb{Q}(alpha)$, we have that $hbox{irr}(alpha, mathbb{Q})$ has exactly $d$ real solutions degree $d$ polynomial. One more thing we can require is that the polynomial has zeros on $K cap mathbb{R}$.






    share|cite|improve this answer


























      1














      Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$ and let $K = mathbb{Q}(a+bi)$. Assume that $K$ over $mathbb{Q}$ is a Galois extension whose Galois group is abelian.



      Since $K-mathbb{Q}$ is Galois extension, we have $a-bi in K$ and so $a in K$. Consider an intermediate field $K-E-mathbb{Q}$ where $E = mathbb{Q}(a)$. Consider $x^2-2ax+a^2+b^2 in E[x]$. This is irreducible since $b neq 0$ and $E subseteq mathbb{R}$. This is the minimal polynomial of $a+bi$ over $E$. Therefore $K-E$ is degree $2$. This tells us that the order of the Galois group is even, say $2m$ where $m = [E:mathbb{Q}]$.



      Since $G=G(K/mathbb{Q})$ is abelian, it follows that $G(E/mathbb{Q})$ is also abelian. Hence by the inverse of the Lagrange's theorem, for each divisor $d$ of $m$, we have a subgroup $H$ of $G$ of order $m/d$. Let $F$ be the fixed field in $E$ by $H$. Then $[F:mathbb{Q}] = d$. Since characteristic is zero, $F = mathbb{Q}(alpha)$, we have that $hbox{irr}(alpha, mathbb{Q})$ has exactly $d$ real solutions degree $d$ polynomial. One more thing we can require is that the polynomial has zeros on $K cap mathbb{R}$.






      share|cite|improve this answer
























        1












        1








        1






        Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$ and let $K = mathbb{Q}(a+bi)$. Assume that $K$ over $mathbb{Q}$ is a Galois extension whose Galois group is abelian.



        Since $K-mathbb{Q}$ is Galois extension, we have $a-bi in K$ and so $a in K$. Consider an intermediate field $K-E-mathbb{Q}$ where $E = mathbb{Q}(a)$. Consider $x^2-2ax+a^2+b^2 in E[x]$. This is irreducible since $b neq 0$ and $E subseteq mathbb{R}$. This is the minimal polynomial of $a+bi$ over $E$. Therefore $K-E$ is degree $2$. This tells us that the order of the Galois group is even, say $2m$ where $m = [E:mathbb{Q}]$.



        Since $G=G(K/mathbb{Q})$ is abelian, it follows that $G(E/mathbb{Q})$ is also abelian. Hence by the inverse of the Lagrange's theorem, for each divisor $d$ of $m$, we have a subgroup $H$ of $G$ of order $m/d$. Let $F$ be the fixed field in $E$ by $H$. Then $[F:mathbb{Q}] = d$. Since characteristic is zero, $F = mathbb{Q}(alpha)$, we have that $hbox{irr}(alpha, mathbb{Q})$ has exactly $d$ real solutions degree $d$ polynomial. One more thing we can require is that the polynomial has zeros on $K cap mathbb{R}$.






        share|cite|improve this answer












        Let $a$ and $b$ be algebraic real numbers over $mathbb{Q}$ and let $K = mathbb{Q}(a+bi)$. Assume that $K$ over $mathbb{Q}$ is a Galois extension whose Galois group is abelian.



        Since $K-mathbb{Q}$ is Galois extension, we have $a-bi in K$ and so $a in K$. Consider an intermediate field $K-E-mathbb{Q}$ where $E = mathbb{Q}(a)$. Consider $x^2-2ax+a^2+b^2 in E[x]$. This is irreducible since $b neq 0$ and $E subseteq mathbb{R}$. This is the minimal polynomial of $a+bi$ over $E$. Therefore $K-E$ is degree $2$. This tells us that the order of the Galois group is even, say $2m$ where $m = [E:mathbb{Q}]$.



        Since $G=G(K/mathbb{Q})$ is abelian, it follows that $G(E/mathbb{Q})$ is also abelian. Hence by the inverse of the Lagrange's theorem, for each divisor $d$ of $m$, we have a subgroup $H$ of $G$ of order $m/d$. Let $F$ be the fixed field in $E$ by $H$. Then $[F:mathbb{Q}] = d$. Since characteristic is zero, $F = mathbb{Q}(alpha)$, we have that $hbox{irr}(alpha, mathbb{Q})$ has exactly $d$ real solutions degree $d$ polynomial. One more thing we can require is that the polynomial has zeros on $K cap mathbb{R}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 21:18









        seoneo

        492213




        492213






























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