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An Upper and Lower Bound for a Field Extension of $mathbbQ$ by a Complex Root
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Consider the polynomial $f(x) = x^3 + zeta x + sqrt3$ in $mathbbC[x],$ where $zeta$ is a primitive third root of unity. Given a root $alpha$ of $f(x)$ in $mathbbC,$ prove that $4 leq [mathbbQ(alpha) : mathbbQ] leq 12.$
We note first that $[mathbbQ(sqrt3, zeta) : mathbbQ] = 4.$ Either $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x]$ or not. Given that $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x],$ we have that $[mathbbQ(sqrt3, zeta, alpha) : mathbbQ] = 12.$ We note that $mathbbQ(alpha) subseteq mathbbQ(sqrt3, zeta, alpha),$ from which we conclude that $[mathbbQ(alpha) : mathbbQ] leq 12.$ Getting the lower bound is proving to be more difficult than this. I would like to say that if $f(x)$ is reducible in $mathbbQ(sqrt3, zeta)[x],$ then $[mathbbQ(alpha) : mathbbQ] geq 4,$ but I am not convinced. Can anyone provide a push in the right direction?
extension-field roots-of-unity separable-extension
asked Aug 2 at 1:25
Dylan_Carlo_Beck
342210
add a comment |Â
up vote
1
down vote
favorite
Consider the polynomial $f(x) = x^3 + zeta x + sqrt3$ in $mathbbC[x],$ where $zeta$ is a primitive third root of unity. Given a root $alpha$ of $f(x)$ in $mathbbC,$ prove that $4 leq [mathbbQ(alpha) : mathbbQ] leq 12.$
We note first that $[mathbbQ(sqrt3, zeta) : mathbbQ] = 4.$ Either $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x]$ or not. Given that $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x],$ we have that $[mathbbQ(sqrt3, zeta, alpha) : mathbbQ] = 12.$ We note that $mathbbQ(alpha) subseteq mathbbQ(sqrt3, zeta, alpha),$ from which we conclude that $[mathbbQ(alpha) : mathbbQ] leq 12.$ Getting the lower bound is proving to be more difficult than this. I would like to say that if $f(x)$ is reducible in $mathbbQ(sqrt3, zeta)[x],$ then $[mathbbQ(alpha) : mathbbQ] geq 4,$ but I am not convinced. Can anyone provide a push in the right direction?
extension-field roots-of-unity separable-extension
asked Aug 2 at 1:25
Dylan_Carlo_Beck
342210
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the polynomial $f(x) = x^3 + zeta x + sqrt3$ in $mathbbC[x],$ where $zeta$ is a primitive third root of unity. Given a root $alpha$ of $f(x)$ in $mathbbC,$ prove that $4 leq [mathbbQ(alpha) : mathbbQ] leq 12.$
We note first that $[mathbbQ(sqrt3, zeta) : mathbbQ] = 4.$ Either $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x]$ or not. Given that $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x],$ we have that $[mathbbQ(sqrt3, zeta, alpha) : mathbbQ] = 12.$ We note that $mathbbQ(alpha) subseteq mathbbQ(sqrt3, zeta, alpha),$ from which we conclude that $[mathbbQ(alpha) : mathbbQ] leq 12.$ Getting the lower bound is proving to be more difficult than this. I would like to say that if $f(x)$ is reducible in $mathbbQ(sqrt3, zeta)[x],$ then $[mathbbQ(alpha) : mathbbQ] geq 4,$ but I am not convinced. Can anyone provide a push in the right direction?
extension-field roots-of-unity separable-extension
asked Aug 2 at 1:25
Dylan_Carlo_Beck
342210
Consider the polynomial $f(x) = x^3 + zeta x + sqrt3$ in $mathbbC[x],$ where $zeta$ is a primitive third root of unity. Given a root $alpha$ of $f(x)$ in $mathbbC,$ prove that $4 leq [mathbbQ(alpha) : mathbbQ] leq 12.$
We note first that $[mathbbQ(sqrt3, zeta) : mathbbQ] = 4.$ Either $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x]$ or not. Given that $f(x)$ is irreducible in $mathbbQ(sqrt3, zeta)[x],$ we have that $[mathbbQ(sqrt3, zeta, alpha) : mathbbQ] = 12.$ We note that $mathbbQ(alpha) subseteq mathbbQ(sqrt3, zeta, alpha),$ from which we conclude that $[mathbbQ(alpha) : mathbbQ] leq 12.$ Getting the lower bound is proving to be more difficult than this. I would like to say that if $f(x)$ is reducible in $mathbbQ(sqrt3, zeta)[x],$ then $[mathbbQ(alpha) : mathbbQ] geq 4,$ but I am not convinced. Can anyone provide a push in the right direction?
extension-field roots-of-unity separable-extension
asked Aug 2 at 1:25
Dylan_Carlo_Beck
342210
asked Aug 2 at 1:25
Dylan_Carlo_Beck
342210
asked Aug 2 at 1:25
Dylan_Carlo_Beck
342210
asked Aug 2 at 1:25
Dylan_Carlo_Beck
342210
342210
add a comment |Â
add a comment |Â
1 Answer
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If $f$ is reducible in $K=mathbbQ(zeta,sqrt3)$, it either splits completely or it decomposes into a linear and a quadratic factors. In the first case $alphain K$. In the second case, $alpha$ is either the roots of the linear factor and again $alphain K$, or is a root of the second factor and the extension requires a further quadratic extension of $K$. So, the lower bound is when $alphain K$.
Now, if $alphain K$, we need to show that $alpha$ is not in an (strict) intermediate extension between $Ksupset Hsupset mathbbQ$. If that is the case then $[H:mathbbQ]=2$, the only divisor of $4=[K:mathbbQ]$ that is not $1$ or $4$. Then $alpha$ is a root of a quadratic polynomial with rational coefficients. This polynomial $x^2+ax+b$ must divide $f$. The condition that the remainder is $0$ gives you $zeta$ and $sqrt3$ expressed as a rational numbers:
$$fracx^3+zeta x+sqrt3x^2+ax+b=(x-a)+frac(zeta-b+a)x+(sqrt3+ab)x^2+ax+b$$
Therefore
$$beginalignzeta&=b-a\sqrt3&=-abendalign$$
which contradicts that $zeta$ and $sqrt3$ are not rational. Therefore, $alpha$ cannot be in a field properly contained in $K=mathbbQ(zeta,sqrt3)$.
Aside:
The minimal polynomial of $alpha$ turns out to be $x^12-2x^10+ 3 x^8-8x^6+7x^4+3x^2+9$. Therefore, actually $[mathbbQ(alpha):mathbbQ]=12$
answered Aug 2 at 2:15
spiralstotheleft
30716
Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $mathbbQ(alpha) = mathbbQ(sqrt3, zeta),$ but as you correctly pointed out, we simply need to show that $alpha$ is not contained in a degree two extension of $mathbbQ.$ Thank you very much for your time.
– Dylan_Carlo_Beck
Aug 2 at 6:06
Can you elaborate on how you obtain the minimal polynomial? I am fascinated.
– Dylan_Carlo_Beck
Aug 2 at 6:08
@Dylan_Carlo_Beck That was done by a computer. $zeta$ and $sqrt3$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand.
– spiralstotheleft
Aug 2 at 10:30
I just typed up my solution to the problem, and I noticed that we should have $zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out.
– Dylan_Carlo_Beck
Aug 2 at 18:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $f$ is reducible in $K=mathbbQ(zeta,sqrt3)$, it either splits completely or it decomposes into a linear and a quadratic factors. In the first case $alphain K$. In the second case, $alpha$ is either the roots of the linear factor and again $alphain K$, or is a root of the second factor and the extension requires a further quadratic extension of $K$. So, the lower bound is when $alphain K$.
Now, if $alphain K$, we need to show that $alpha$ is not in an (strict) intermediate extension between $Ksupset Hsupset mathbbQ$. If that is the case then $[H:mathbbQ]=2$, the only divisor of $4=[K:mathbbQ]$ that is not $1$ or $4$. Then $alpha$ is a root of a quadratic polynomial with rational coefficients. This polynomial $x^2+ax+b$ must divide $f$. The condition that the remainder is $0$ gives you $zeta$ and $sqrt3$ expressed as a rational numbers:
$$fracx^3+zeta x+sqrt3x^2+ax+b=(x-a)+frac(zeta-b+a)x+(sqrt3+ab)x^2+ax+b$$
Therefore
$$beginalignzeta&=b-a\sqrt3&=-abendalign$$
which contradicts that $zeta$ and $sqrt3$ are not rational. Therefore, $alpha$ cannot be in a field properly contained in $K=mathbbQ(zeta,sqrt3)$.
Aside:
The minimal polynomial of $alpha$ turns out to be $x^12-2x^10+ 3 x^8-8x^6+7x^4+3x^2+9$. Therefore, actually $[mathbbQ(alpha):mathbbQ]=12$
answered Aug 2 at 2:15
spiralstotheleft
30716
Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $mathbbQ(alpha) = mathbbQ(sqrt3, zeta),$ but as you correctly pointed out, we simply need to show that $alpha$ is not contained in a degree two extension of $mathbbQ.$ Thank you very much for your time.
– Dylan_Carlo_Beck
Aug 2 at 6:06
Can you elaborate on how you obtain the minimal polynomial? I am fascinated.
– Dylan_Carlo_Beck
Aug 2 at 6:08
@Dylan_Carlo_Beck That was done by a computer. $zeta$ and $sqrt3$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand.
– spiralstotheleft
Aug 2 at 10:30
I just typed up my solution to the problem, and I noticed that we should have $zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out.
– Dylan_Carlo_Beck
Aug 2 at 18:35
add a comment |Â
up vote
2
down vote
accepted
If $f$ is reducible in $K=mathbbQ(zeta,sqrt3)$, it either splits completely or it decomposes into a linear and a quadratic factors. In the first case $alphain K$. In the second case, $alpha$ is either the roots of the linear factor and again $alphain K$, or is a root of the second factor and the extension requires a further quadratic extension of $K$. So, the lower bound is when $alphain K$.
Now, if $alphain K$, we need to show that $alpha$ is not in an (strict) intermediate extension between $Ksupset Hsupset mathbbQ$. If that is the case then $[H:mathbbQ]=2$, the only divisor of $4=[K:mathbbQ]$ that is not $1$ or $4$. Then $alpha$ is a root of a quadratic polynomial with rational coefficients. This polynomial $x^2+ax+b$ must divide $f$. The condition that the remainder is $0$ gives you $zeta$ and $sqrt3$ expressed as a rational numbers:
$$fracx^3+zeta x+sqrt3x^2+ax+b=(x-a)+frac(zeta-b+a)x+(sqrt3+ab)x^2+ax+b$$
Therefore
$$beginalignzeta&=b-a\sqrt3&=-abendalign$$
which contradicts that $zeta$ and $sqrt3$ are not rational. Therefore, $alpha$ cannot be in a field properly contained in $K=mathbbQ(zeta,sqrt3)$.
Aside:
The minimal polynomial of $alpha$ turns out to be $x^12-2x^10+ 3 x^8-8x^6+7x^4+3x^2+9$. Therefore, actually $[mathbbQ(alpha):mathbbQ]=12$
answered Aug 2 at 2:15
spiralstotheleft
30716
Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $mathbbQ(alpha) = mathbbQ(sqrt3, zeta),$ but as you correctly pointed out, we simply need to show that $alpha$ is not contained in a degree two extension of $mathbbQ.$ Thank you very much for your time.
– Dylan_Carlo_Beck
Aug 2 at 6:06
Can you elaborate on how you obtain the minimal polynomial? I am fascinated.
– Dylan_Carlo_Beck
Aug 2 at 6:08
@Dylan_Carlo_Beck That was done by a computer. $zeta$ and $sqrt3$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand.
– spiralstotheleft
Aug 2 at 10:30
I just typed up my solution to the problem, and I noticed that we should have $zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out.
– Dylan_Carlo_Beck
Aug 2 at 18:35
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $f$ is reducible in $K=mathbbQ(zeta,sqrt3)$, it either splits completely or it decomposes into a linear and a quadratic factors. In the first case $alphain K$. In the second case, $alpha$ is either the roots of the linear factor and again $alphain K$, or is a root of the second factor and the extension requires a further quadratic extension of $K$. So, the lower bound is when $alphain K$.
Now, if $alphain K$, we need to show that $alpha$ is not in an (strict) intermediate extension between $Ksupset Hsupset mathbbQ$. If that is the case then $[H:mathbbQ]=2$, the only divisor of $4=[K:mathbbQ]$ that is not $1$ or $4$. Then $alpha$ is a root of a quadratic polynomial with rational coefficients. This polynomial $x^2+ax+b$ must divide $f$. The condition that the remainder is $0$ gives you $zeta$ and $sqrt3$ expressed as a rational numbers:
$$fracx^3+zeta x+sqrt3x^2+ax+b=(x-a)+frac(zeta-b+a)x+(sqrt3+ab)x^2+ax+b$$
Therefore
$$beginalignzeta&=b-a\sqrt3&=-abendalign$$
which contradicts that $zeta$ and $sqrt3$ are not rational. Therefore, $alpha$ cannot be in a field properly contained in $K=mathbbQ(zeta,sqrt3)$.
Aside:
The minimal polynomial of $alpha$ turns out to be $x^12-2x^10+ 3 x^8-8x^6+7x^4+3x^2+9$. Therefore, actually $[mathbbQ(alpha):mathbbQ]=12$
answered Aug 2 at 2:15
spiralstotheleft
30716
If $f$ is reducible in $K=mathbbQ(zeta,sqrt3)$, it either splits completely or it decomposes into a linear and a quadratic factors. In the first case $alphain K$. In the second case, $alpha$ is either the roots of the linear factor and again $alphain K$, or is a root of the second factor and the extension requires a further quadratic extension of $K$. So, the lower bound is when $alphain K$.
Now, if $alphain K$, we need to show that $alpha$ is not in an (strict) intermediate extension between $Ksupset Hsupset mathbbQ$. If that is the case then $[H:mathbbQ]=2$, the only divisor of $4=[K:mathbbQ]$ that is not $1$ or $4$. Then $alpha$ is a root of a quadratic polynomial with rational coefficients. This polynomial $x^2+ax+b$ must divide $f$. The condition that the remainder is $0$ gives you $zeta$ and $sqrt3$ expressed as a rational numbers:
$$fracx^3+zeta x+sqrt3x^2+ax+b=(x-a)+frac(zeta-b+a)x+(sqrt3+ab)x^2+ax+b$$
Therefore
$$beginalignzeta&=b-a\sqrt3&=-abendalign$$
which contradicts that $zeta$ and $sqrt3$ are not rational. Therefore, $alpha$ cannot be in a field properly contained in $K=mathbbQ(zeta,sqrt3)$.
Aside:
The minimal polynomial of $alpha$ turns out to be $x^12-2x^10+ 3 x^8-8x^6+7x^4+3x^2+9$. Therefore, actually $[mathbbQ(alpha):mathbbQ]=12$
answered Aug 2 at 2:15
spiralstotheleft
30716
edited Aug 2 at 2:36
answered Aug 2 at 2:15
spiralstotheleft
30716
answered Aug 2 at 2:15
spiralstotheleft
30716
answered Aug 2 at 2:15
spiralstotheleft
30716
30716
Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $mathbbQ(alpha) = mathbbQ(sqrt3, zeta),$ but as you correctly pointed out, we simply need to show that $alpha$ is not contained in a degree two extension of $mathbbQ.$ Thank you very much for your time.
– Dylan_Carlo_Beck
Aug 2 at 6:06
Can you elaborate on how you obtain the minimal polynomial? I am fascinated.
– Dylan_Carlo_Beck
Aug 2 at 6:08
@Dylan_Carlo_Beck That was done by a computer. $zeta$ and $sqrt3$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand.
– spiralstotheleft
Aug 2 at 10:30
I just typed up my solution to the problem, and I noticed that we should have $zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out.
– Dylan_Carlo_Beck
Aug 2 at 18:35
add a comment |Â
Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $mathbbQ(alpha) = mathbbQ(sqrt3, zeta),$ but as you correctly pointed out, we simply need to show that $alpha$ is not contained in a degree two extension of $mathbbQ.$ Thank you very much for your time.
– Dylan_Carlo_Beck
Aug 2 at 6:06
Can you elaborate on how you obtain the minimal polynomial? I am fascinated.
– Dylan_Carlo_Beck
Aug 2 at 6:08
@Dylan_Carlo_Beck That was done by a computer. $zeta$ and $sqrt3$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand.
– spiralstotheleft
Aug 2 at 10:30
I just typed up my solution to the problem, and I noticed that we should have $zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out.
– Dylan_Carlo_Beck
Aug 2 at 18:35
Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $mathbbQ(alpha) = mathbbQ(sqrt3, zeta),$ but as you correctly pointed out, we simply need to show that $alpha$ is not contained in a degree two extension of $mathbbQ.$ Thank you very much for your time.
– Dylan_Carlo_Beck
Aug 2 at 6:06
Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $mathbbQ(alpha) = mathbbQ(sqrt3, zeta),$ but as you correctly pointed out, we simply need to show that $alpha$ is not contained in a degree two extension of $mathbbQ.$ Thank you very much for your time.
– Dylan_Carlo_Beck
Aug 2 at 6:06
Can you elaborate on how you obtain the minimal polynomial? I am fascinated.
– Dylan_Carlo_Beck
Aug 2 at 6:08
Can you elaborate on how you obtain the minimal polynomial? I am fascinated.
– Dylan_Carlo_Beck
Aug 2 at 6:08
@Dylan_Carlo_Beck That was done by a computer. $zeta$ and $sqrt3$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand.
– spiralstotheleft
Aug 2 at 10:30
@Dylan_Carlo_Beck That was done by a computer. $zeta$ and $sqrt3$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand.
– spiralstotheleft
Aug 2 at 10:30
I just typed up my solution to the problem, and I noticed that we should have $zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out.
– Dylan_Carlo_Beck
Aug 2 at 18:35
I just typed up my solution to the problem, and I noticed that we should have $zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out.
– Dylan_Carlo_Beck
Aug 2 at 18:35
add a comment |Â
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