Mixing
Let $K=Bbb Q (e^{frac{2 pi i}{7}})$ and $alpha in K - Bbb Q$. Then show that, $Bbb Q(alpha)=K$
$begingroup$
Let $K=Bbb Q (e^{frac{2 pi i}{7}})$ and $alpha in K - Bbb Q$. Then show that, $Bbb Q(alpha)=K$ .
My attempt:
I have computed the minimal polynomial of$e^{frac{2 pi i}{7}}$ over $Bbb Q$ to be $x^6+x^5+x^4+x^3+x^2+x+1 $ and thus, $[K:Bbb Q]=6$
Since, $alpha in K - Bbb Q$ , it follows that, $Bbb Q(alpha)$ is a subfield of $K$ containing $Bbb Q$, moreover, $[Bbb Q(alpha):Bbb Q] | [K:Bbb Q] implies [Bbb Q(alpha):Bbb Q] =$1 or 2 or 3 or 6 .Since, $alpha notin Bbb Q$, $[Bbb Q(alpha):Bbb Q] ne 1$ .
Since, for $beta = e^{frac{2 pi i}{7}}$, ${1,beta,beta^2,beta^3 ,beta^4,beta^5}$ is a basis of $K$ over $Bbb Q$. then, $exists a_j in Bbb Q , 0le j le 5$ with at least one of $a_j ne 0 $ for $j ge 1$ so that, $alpha=a_0 +a_1 beta+dots+a_5 beta^5$ .
How to proceed with then proof. Thanks in advance for help!
Note that this question is also present in here, here etc. but they use galois theory, but I only know Field theory and am also trying to solve the question using only that.
field-theory cyclotomic-fields
asked Jan 22 at 2:15
CoherentCoherent
1,197623
$endgroup$
|
show 1 more comment
$begingroup$
Let $K=Bbb Q (e^{frac{2 pi i}{7}})$ and $alpha in K - Bbb Q$. Then show that, $Bbb Q(alpha)=K$ .
My attempt:
I have computed the minimal polynomial of$e^{frac{2 pi i}{7}}$ over $Bbb Q$ to be $x^6+x^5+x^4+x^3+x^2+x+1 $ and thus, $[K:Bbb Q]=6$
Since, $alpha in K - Bbb Q$ , it follows that, $Bbb Q(alpha)$ is a subfield of $K$ containing $Bbb Q$, moreover, $[Bbb Q(alpha):Bbb Q] | [K:Bbb Q] implies [Bbb Q(alpha):Bbb Q] =$1 or 2 or 3 or 6 .Since, $alpha notin Bbb Q$, $[Bbb Q(alpha):Bbb Q] ne 1$ .
Since, for $beta = e^{frac{2 pi i}{7}}$, ${1,beta,beta^2,beta^3 ,beta^4,beta^5}$ is a basis of $K$ over $Bbb Q$. then, $exists a_j in Bbb Q , 0le j le 5$ with at least one of $a_j ne 0 $ for $j ge 1$ so that, $alpha=a_0 +a_1 beta+dots+a_5 beta^5$ .
How to proceed with then proof. Thanks in advance for help!
Note that this question is also present in here, here etc. but they use galois theory, but I only know Field theory and am also trying to solve the question using only that.
field-theory cyclotomic-fields
asked Jan 22 at 2:15
CoherentCoherent
1,197623
$endgroup$
1
$begingroup$
Let $alpha=beta+beta^{-1}$. Then ${bf Q}(alpha)$ has degree 3. There are also examples with degree 2.
$endgroup$
– Gerry Myerson
Jan 22 at 3:40
1
$begingroup$
@GerryMyerson so is my question wrong? Anyway would you considering giving an answer.
$endgroup$
– Coherent
Jan 22 at 4:34
$begingroup$
OK, I've given an answer – does it work for you?
$endgroup$
– Gerry Myerson
Jan 24 at 0:06
$begingroup$
Are you still here, Coherent?
$endgroup$
– Gerry Myerson
Jan 25 at 2:56
$begingroup$
@GerryMyerson In the linked questions, it's proved that there is no intermediate field between $Q$ and $Q(e^{frac{2 pi i}{7}})$ other than those two itself, so I am also waiting for a few other answers. If I find you answer helpful, I'll definitely accept it, you don't need to worry about it.I'm not a new member to the site.
$endgroup$
– Coherent
Jan 25 at 3:05
|
show 1 more comment
$begingroup$
Let $K=Bbb Q (e^{frac{2 pi i}{7}})$ and $alpha in K - Bbb Q$. Then show that, $Bbb Q(alpha)=K$ .
My attempt:
I have computed the minimal polynomial of$e^{frac{2 pi i}{7}}$ over $Bbb Q$ to be $x^6+x^5+x^4+x^3+x^2+x+1 $ and thus, $[K:Bbb Q]=6$
Since, $alpha in K - Bbb Q$ , it follows that, $Bbb Q(alpha)$ is a subfield of $K$ containing $Bbb Q$, moreover, $[Bbb Q(alpha):Bbb Q] | [K:Bbb Q] implies [Bbb Q(alpha):Bbb Q] =$1 or 2 or 3 or 6 .Since, $alpha notin Bbb Q$, $[Bbb Q(alpha):Bbb Q] ne 1$ .
Since, for $beta = e^{frac{2 pi i}{7}}$, ${1,beta,beta^2,beta^3 ,beta^4,beta^5}$ is a basis of $K$ over $Bbb Q$. then, $exists a_j in Bbb Q , 0le j le 5$ with at least one of $a_j ne 0 $ for $j ge 1$ so that, $alpha=a_0 +a_1 beta+dots+a_5 beta^5$ .
How to proceed with then proof. Thanks in advance for help!
Note that this question is also present in here, here etc. but they use galois theory, but I only know Field theory and am also trying to solve the question using only that.
field-theory cyclotomic-fields
asked Jan 22 at 2:15
CoherentCoherent
1,197623
$endgroup$
Let $K=Bbb Q (e^{frac{2 pi i}{7}})$ and $alpha in K - Bbb Q$. Then show that, $Bbb Q(alpha)=K$ .
My attempt:
I have computed the minimal polynomial of$e^{frac{2 pi i}{7}}$ over $Bbb Q$ to be $x^6+x^5+x^4+x^3+x^2+x+1 $ and thus, $[K:Bbb Q]=6$
Since, $alpha in K - Bbb Q$ , it follows that, $Bbb Q(alpha)$ is a subfield of $K$ containing $Bbb Q$, moreover, $[Bbb Q(alpha):Bbb Q] | [K:Bbb Q] implies [Bbb Q(alpha):Bbb Q] =$1 or 2 or 3 or 6 .Since, $alpha notin Bbb Q$, $[Bbb Q(alpha):Bbb Q] ne 1$ .
Since, for $beta = e^{frac{2 pi i}{7}}$, ${1,beta,beta^2,beta^3 ,beta^4,beta^5}$ is a basis of $K$ over $Bbb Q$. then, $exists a_j in Bbb Q , 0le j le 5$ with at least one of $a_j ne 0 $ for $j ge 1$ so that, $alpha=a_0 +a_1 beta+dots+a_5 beta^5$ .
How to proceed with then proof. Thanks in advance for help!
Note that this question is also present in here, here etc. but they use galois theory, but I only know Field theory and am also trying to solve the question using only that.
field-theory cyclotomic-fields
field-theory cyclotomic-fields
asked Jan 22 at 2:15
CoherentCoherent
1,197623
asked Jan 22 at 2:15
CoherentCoherent
1,197623
asked Jan 22 at 2:15
CoherentCoherent
1,197623
asked Jan 22 at 2:15
CoherentCoherent
1,197623
asked Jan 22 at 2:15
CoherentCoherent
1,197623
1,197623
1
$begingroup$
Let $alpha=beta+beta^{-1}$. Then ${bf Q}(alpha)$ has degree 3. There are also examples with degree 2.
$endgroup$
– Gerry Myerson
Jan 22 at 3:40
1
$begingroup$
@GerryMyerson so is my question wrong? Anyway would you considering giving an answer.
$endgroup$
– Coherent
Jan 22 at 4:34
$begingroup$
OK, I've given an answer – does it work for you?
$endgroup$
– Gerry Myerson
Jan 24 at 0:06
$begingroup$
Are you still here, Coherent?
$endgroup$
– Gerry Myerson
Jan 25 at 2:56
$begingroup$
@GerryMyerson In the linked questions, it's proved that there is no intermediate field between $Q$ and $Q(e^{frac{2 pi i}{7}})$ other than those two itself, so I am also waiting for a few other answers. If I find you answer helpful, I'll definitely accept it, you don't need to worry about it.I'm not a new member to the site.
$endgroup$
– Coherent
Jan 25 at 3:05
|
show 1 more comment
1
$begingroup$
Let $alpha=beta+beta^{-1}$. Then ${bf Q}(alpha)$ has degree 3. There are also examples with degree 2.
$endgroup$
– Gerry Myerson
Jan 22 at 3:40
1
$begingroup$
@GerryMyerson so is my question wrong? Anyway would you considering giving an answer.
$endgroup$
– Coherent
Jan 22 at 4:34
$begingroup$
OK, I've given an answer – does it work for you?
$endgroup$
– Gerry Myerson
Jan 24 at 0:06
$begingroup$
Are you still here, Coherent?
$endgroup$
– Gerry Myerson
Jan 25 at 2:56
$begingroup$
@GerryMyerson In the linked questions, it's proved that there is no intermediate field between $Q$ and $Q(e^{frac{2 pi i}{7}})$ other than those two itself, so I am also waiting for a few other answers. If I find you answer helpful, I'll definitely accept it, you don't need to worry about it.I'm not a new member to the site.
$endgroup$
– Coherent
Jan 25 at 3:05
1
1
$begingroup$
Let $alpha=beta+beta^{-1}$. Then ${bf Q}(alpha)$ has degree 3. There are also examples with degree 2.
$endgroup$
– Gerry Myerson
Jan 22 at 3:40
$begingroup$
Let $alpha=beta+beta^{-1}$. Then ${bf Q}(alpha)$ has degree 3. There are also examples with degree 2.
$endgroup$
– Gerry Myerson
Jan 22 at 3:40
1
1
$begingroup$
@GerryMyerson so is my question wrong? Anyway would you considering giving an answer.
$endgroup$
– Coherent
Jan 22 at 4:34
$begingroup$
@GerryMyerson so is my question wrong? Anyway would you considering giving an answer.
$endgroup$
– Coherent
Jan 22 at 4:34
$begingroup$
OK, I've given an answer – does it work for you?
$endgroup$
– Gerry Myerson
Jan 24 at 0:06
$begingroup$
OK, I've given an answer – does it work for you?
$endgroup$
– Gerry Myerson
Jan 24 at 0:06
$begingroup$
Are you still here, Coherent?
$endgroup$
– Gerry Myerson
Jan 25 at 2:56
$begingroup$
Are you still here, Coherent?
$endgroup$
– Gerry Myerson
Jan 25 at 2:56
$begingroup$
@GerryMyerson In the linked questions, it's proved that there is no intermediate field between $Q$ and $Q(e^{frac{2 pi i}{7}})$ other than those two itself, so I am also waiting for a few other answers. If I find you answer helpful, I'll definitely accept it, you don't need to worry about it.I'm not a new member to the site.
$endgroup$
– Coherent
Jan 25 at 3:05
$begingroup$
@GerryMyerson In the linked questions, it's proved that there is no intermediate field between $Q$ and $Q(e^{frac{2 pi i}{7}})$ other than those two itself, so I am also waiting for a few other answers. If I find you answer helpful, I'll definitely accept it, you don't need to worry about it.I'm not a new member to the site.
$endgroup$
– Coherent
Jan 25 at 3:05
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $beta=e^{2pi i/7}$, let $K={bf Q}(beta)$, let $alpha=beta+beta^{-1}$. Then $$alpha^3=beta^3+3beta+3beta^{-1}+beta^{-3}=beta^3+3beta+3beta^6+beta^4$$ and $$alpha^2=beta^2+2+beta^{-2}=beta^2+2+beta^5$$ [thanks to Robert Lewis for drawing my attention to a typo here, which I have corrected] so $$alpha^3+alpha^2-2alpha-1=1+beta+beta^2+beta^3+beta^4+beta^5+beta^6=0$$ It's easily seen that $x^3+x^2-2x-1$ is irreducible over the rationals, so it is the minimal polynomial for $alpha$, so ${bf Q}(alpha)$ is a proper subfield of $K$, of degree 3.
Similarly, one can show that $gamma=beta+beta^2+beta^4$ is of degree two over the rationals.
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
If $alpha = beta + beta^{-1}$, does not $alpha^2 = beta^2 + beta^{-1} + 2$? Where does $alpha^2 = beta^2 + 2beta^{-1} + 2$ come from?
$endgroup$
– Robert Lewis
Jan 22 at 6:52
add a comment |
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$begingroup$
Let $beta=e^{2pi i/7}$, let $K={bf Q}(beta)$, let $alpha=beta+beta^{-1}$. Then $$alpha^3=beta^3+3beta+3beta^{-1}+beta^{-3}=beta^3+3beta+3beta^6+beta^4$$ and $$alpha^2=beta^2+2+beta^{-2}=beta^2+2+beta^5$$ [thanks to Robert Lewis for drawing my attention to a typo here, which I have corrected] so $$alpha^3+alpha^2-2alpha-1=1+beta+beta^2+beta^3+beta^4+beta^5+beta^6=0$$ It's easily seen that $x^3+x^2-2x-1$ is irreducible over the rationals, so it is the minimal polynomial for $alpha$, so ${bf Q}(alpha)$ is a proper subfield of $K$, of degree 3.
Similarly, one can show that $gamma=beta+beta^2+beta^4$ is of degree two over the rationals.
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
If $alpha = beta + beta^{-1}$, does not $alpha^2 = beta^2 + beta^{-1} + 2$? Where does $alpha^2 = beta^2 + 2beta^{-1} + 2$ come from?
$endgroup$
– Robert Lewis
Jan 22 at 6:52
add a comment |
$begingroup$
Let $beta=e^{2pi i/7}$, let $K={bf Q}(beta)$, let $alpha=beta+beta^{-1}$. Then $$alpha^3=beta^3+3beta+3beta^{-1}+beta^{-3}=beta^3+3beta+3beta^6+beta^4$$ and $$alpha^2=beta^2+2+beta^{-2}=beta^2+2+beta^5$$ [thanks to Robert Lewis for drawing my attention to a typo here, which I have corrected] so $$alpha^3+alpha^2-2alpha-1=1+beta+beta^2+beta^3+beta^4+beta^5+beta^6=0$$ It's easily seen that $x^3+x^2-2x-1$ is irreducible over the rationals, so it is the minimal polynomial for $alpha$, so ${bf Q}(alpha)$ is a proper subfield of $K$, of degree 3.
Similarly, one can show that $gamma=beta+beta^2+beta^4$ is of degree two over the rationals.
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
If $alpha = beta + beta^{-1}$, does not $alpha^2 = beta^2 + beta^{-1} + 2$? Where does $alpha^2 = beta^2 + 2beta^{-1} + 2$ come from?
$endgroup$
– Robert Lewis
Jan 22 at 6:52
add a comment |
$begingroup$
Let $beta=e^{2pi i/7}$, let $K={bf Q}(beta)$, let $alpha=beta+beta^{-1}$. Then $$alpha^3=beta^3+3beta+3beta^{-1}+beta^{-3}=beta^3+3beta+3beta^6+beta^4$$ and $$alpha^2=beta^2+2+beta^{-2}=beta^2+2+beta^5$$ [thanks to Robert Lewis for drawing my attention to a typo here, which I have corrected] so $$alpha^3+alpha^2-2alpha-1=1+beta+beta^2+beta^3+beta^4+beta^5+beta^6=0$$ It's easily seen that $x^3+x^2-2x-1$ is irreducible over the rationals, so it is the minimal polynomial for $alpha$, so ${bf Q}(alpha)$ is a proper subfield of $K$, of degree 3.
Similarly, one can show that $gamma=beta+beta^2+beta^4$ is of degree two over the rationals.
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
Let $beta=e^{2pi i/7}$, let $K={bf Q}(beta)$, let $alpha=beta+beta^{-1}$. Then $$alpha^3=beta^3+3beta+3beta^{-1}+beta^{-3}=beta^3+3beta+3beta^6+beta^4$$ and $$alpha^2=beta^2+2+beta^{-2}=beta^2+2+beta^5$$ [thanks to Robert Lewis for drawing my attention to a typo here, which I have corrected] so $$alpha^3+alpha^2-2alpha-1=1+beta+beta^2+beta^3+beta^4+beta^5+beta^6=0$$ It's easily seen that $x^3+x^2-2x-1$ is irreducible over the rationals, so it is the minimal polynomial for $alpha$, so ${bf Q}(alpha)$ is a proper subfield of $K$, of degree 3.
Similarly, one can show that $gamma=beta+beta^2+beta^4$ is of degree two over the rationals.
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
edited Jan 22 at 10:57
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
answered Jan 22 at 6:45
Gerry MyersonGerry Myerson
147k8149302
147k8149302
$begingroup$
If $alpha = beta + beta^{-1}$, does not $alpha^2 = beta^2 + beta^{-1} + 2$? Where does $alpha^2 = beta^2 + 2beta^{-1} + 2$ come from?
$endgroup$
– Robert Lewis
Jan 22 at 6:52
add a comment |
$begingroup$
If $alpha = beta + beta^{-1}$, does not $alpha^2 = beta^2 + beta^{-1} + 2$? Where does $alpha^2 = beta^2 + 2beta^{-1} + 2$ come from?
$endgroup$
– Robert Lewis
Jan 22 at 6:52
$begingroup$
If $alpha = beta + beta^{-1}$, does not $alpha^2 = beta^2 + beta^{-1} + 2$? Where does $alpha^2 = beta^2 + 2beta^{-1} + 2$ come from?
$endgroup$
– Robert Lewis
Jan 22 at 6:52
$begingroup$
If $alpha = beta + beta^{-1}$, does not $alpha^2 = beta^2 + beta^{-1} + 2$? Where does $alpha^2 = beta^2 + 2beta^{-1} + 2$ come from?
$endgroup$
– Robert Lewis
Jan 22 at 6:52
add a comment |
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$begingroup$
Let $alpha=beta+beta^{-1}$. Then ${bf Q}(alpha)$ has degree 3. There are also examples with degree 2.
$endgroup$
– Gerry Myerson
Jan 22 at 3:40
1
$begingroup$
@GerryMyerson so is my question wrong? Anyway would you considering giving an answer.
$endgroup$
– Coherent
Jan 22 at 4:34
$begingroup$
OK, I've given an answer – does it work for you?
$endgroup$
– Gerry Myerson
Jan 24 at 0:06
$begingroup$
Are you still here, Coherent?
$endgroup$
– Gerry Myerson
Jan 25 at 2:56
$begingroup$
@GerryMyerson In the linked questions, it's proved that there is no intermediate field between $Q$ and $Q(e^{frac{2 pi i}{7}})$ other than those two itself, so I am also waiting for a few other answers. If I find you answer helpful, I'll definitely accept it, you don't need to worry about it.I'm not a new member to the site.
$endgroup$
– Coherent
Jan 25 at 3:05