Calculating the Hilbert class field group of $mathbb{Q}(sqrt[3]{19})$
$begingroup$
In Marcus book “Number Fields” there is an exercise which asks to calculate the Hilbert class field group of $mathbb{Q}(sqrt[3]{19})$
I know that the ring of integers is $R=mathcal{O}_K=mathbb{Z}[alpha,beta]$, where $alpha=sqrt[3]{19}$ and $beta=frac{alpha^2+alpha+1}{3}.$
After some calculations I have that $disc{K}=-3cdot19^2$ and that the indexes of $alpha$ and $beta$ are, respectively $3,2.$ The Minkowski bound say that I only have to consider the following ideals that I have factored using Kummer:
$$2R=(2,alpha-1)(2,3beta)$$
$$3R=(3,beta)^2(3,beta-1)$$
$$5R=(5,alpha+1)(5,alpha^2-alpha+1)$$
$$7R=(7).$$
Using a preceding exercise, I know that
$$N(a+balpha+calpha^2)=a^3+19b^3+19^2c^3-57abc.$$
I observe that the congruence $x^3equiv mmod19$ can be satisfied only if $mequiv0,1,7,8,11,12,18.$ I can deduce that there aren’t elements of norm 2,5, and a similar arguments holds for the primes over $3.$
However, there exist elements of norm $18,45$ respectively $alpha-1$ and $alpha-4$.
With direct computation, I have found that
$$alpha-4in(3beta)(3,beta-1)(5,alpha+1)$$
I would like to calculate the norm of the prime ideal $(5,alpha+1)$ I know it is a power of $5,$ more precisely $5$ or $25.$ How can I calculate this norm? For example, if I would like to calculate it computing
$$|(5,alpha+1)/(5,alpha+1)^2|,$$ how can I proceed?
If it is $5$, in this way I should conclude that the class field group is cyclic with the class of $(3,beta)$ as generator.
Finally, how can I prove that the order of the group is divisible by $3$? I think I have to match the powers of the ideal class $(3,beta)$ in the right way, but I’m not completely sure.
group-theory algebraic-number-theory class-field-theory
asked Jun 28 '18 at 1:40
MatPMatP
1367
$endgroup$
add a comment |
$begingroup$
In Marcus book “Number Fields” there is an exercise which asks to calculate the Hilbert class field group of $mathbb{Q}(sqrt[3]{19})$
I know that the ring of integers is $R=mathcal{O}_K=mathbb{Z}[alpha,beta]$, where $alpha=sqrt[3]{19}$ and $beta=frac{alpha^2+alpha+1}{3}.$
After some calculations I have that $disc{K}=-3cdot19^2$ and that the indexes of $alpha$ and $beta$ are, respectively $3,2.$ The Minkowski bound say that I only have to consider the following ideals that I have factored using Kummer:
$$2R=(2,alpha-1)(2,3beta)$$
$$3R=(3,beta)^2(3,beta-1)$$
$$5R=(5,alpha+1)(5,alpha^2-alpha+1)$$
$$7R=(7).$$
Using a preceding exercise, I know that
$$N(a+balpha+calpha^2)=a^3+19b^3+19^2c^3-57abc.$$
I observe that the congruence $x^3equiv mmod19$ can be satisfied only if $mequiv0,1,7,8,11,12,18.$ I can deduce that there aren’t elements of norm 2,5, and a similar arguments holds for the primes over $3.$
However, there exist elements of norm $18,45$ respectively $alpha-1$ and $alpha-4$.
With direct computation, I have found that
$$alpha-4in(3beta)(3,beta-1)(5,alpha+1)$$
I would like to calculate the norm of the prime ideal $(5,alpha+1)$ I know it is a power of $5,$ more precisely $5$ or $25.$ How can I calculate this norm? For example, if I would like to calculate it computing
$$|(5,alpha+1)/(5,alpha+1)^2|,$$ how can I proceed?
If it is $5$, in this way I should conclude that the class field group is cyclic with the class of $(3,beta)$ as generator.
Finally, how can I prove that the order of the group is divisible by $3$? I think I have to match the powers of the ideal class $(3,beta)$ in the right way, but I’m not completely sure.
group-theory algebraic-number-theory class-field-theory
asked Jun 28 '18 at 1:40
MatPMatP
1367
$endgroup$
$begingroup$
The residue class group clearly only has $5$ elements.
$endgroup$
– franz lemmermeyer
Jan 8 at 21:27
add a comment |
$begingroup$
In Marcus book “Number Fields” there is an exercise which asks to calculate the Hilbert class field group of $mathbb{Q}(sqrt[3]{19})$
I know that the ring of integers is $R=mathcal{O}_K=mathbb{Z}[alpha,beta]$, where $alpha=sqrt[3]{19}$ and $beta=frac{alpha^2+alpha+1}{3}.$
After some calculations I have that $disc{K}=-3cdot19^2$ and that the indexes of $alpha$ and $beta$ are, respectively $3,2.$ The Minkowski bound say that I only have to consider the following ideals that I have factored using Kummer:
$$2R=(2,alpha-1)(2,3beta)$$
$$3R=(3,beta)^2(3,beta-1)$$
$$5R=(5,alpha+1)(5,alpha^2-alpha+1)$$
$$7R=(7).$$
Using a preceding exercise, I know that
$$N(a+balpha+calpha^2)=a^3+19b^3+19^2c^3-57abc.$$
I observe that the congruence $x^3equiv mmod19$ can be satisfied only if $mequiv0,1,7,8,11,12,18.$ I can deduce that there aren’t elements of norm 2,5, and a similar arguments holds for the primes over $3.$
However, there exist elements of norm $18,45$ respectively $alpha-1$ and $alpha-4$.
With direct computation, I have found that
$$alpha-4in(3beta)(3,beta-1)(5,alpha+1)$$
I would like to calculate the norm of the prime ideal $(5,alpha+1)$ I know it is a power of $5,$ more precisely $5$ or $25.$ How can I calculate this norm? For example, if I would like to calculate it computing
$$|(5,alpha+1)/(5,alpha+1)^2|,$$ how can I proceed?
If it is $5$, in this way I should conclude that the class field group is cyclic with the class of $(3,beta)$ as generator.
Finally, how can I prove that the order of the group is divisible by $3$? I think I have to match the powers of the ideal class $(3,beta)$ in the right way, but I’m not completely sure.
group-theory algebraic-number-theory class-field-theory
asked Jun 28 '18 at 1:40
MatPMatP
1367
$endgroup$
In Marcus book “Number Fields” there is an exercise which asks to calculate the Hilbert class field group of $mathbb{Q}(sqrt[3]{19})$
I know that the ring of integers is $R=mathcal{O}_K=mathbb{Z}[alpha,beta]$, where $alpha=sqrt[3]{19}$ and $beta=frac{alpha^2+alpha+1}{3}.$
After some calculations I have that $disc{K}=-3cdot19^2$ and that the indexes of $alpha$ and $beta$ are, respectively $3,2.$ The Minkowski bound say that I only have to consider the following ideals that I have factored using Kummer:
$$2R=(2,alpha-1)(2,3beta)$$
$$3R=(3,beta)^2(3,beta-1)$$
$$5R=(5,alpha+1)(5,alpha^2-alpha+1)$$
$$7R=(7).$$
Using a preceding exercise, I know that
$$N(a+balpha+calpha^2)=a^3+19b^3+19^2c^3-57abc.$$
I observe that the congruence $x^3equiv mmod19$ can be satisfied only if $mequiv0,1,7,8,11,12,18.$ I can deduce that there aren’t elements of norm 2,5, and a similar arguments holds for the primes over $3.$
However, there exist elements of norm $18,45$ respectively $alpha-1$ and $alpha-4$.
With direct computation, I have found that
$$alpha-4in(3beta)(3,beta-1)(5,alpha+1)$$
I would like to calculate the norm of the prime ideal $(5,alpha+1)$ I know it is a power of $5,$ more precisely $5$ or $25.$ How can I calculate this norm? For example, if I would like to calculate it computing
$$|(5,alpha+1)/(5,alpha+1)^2|,$$ how can I proceed?
If it is $5$, in this way I should conclude that the class field group is cyclic with the class of $(3,beta)$ as generator.
Finally, how can I prove that the order of the group is divisible by $3$? I think I have to match the powers of the ideal class $(3,beta)$ in the right way, but I’m not completely sure.
group-theory algebraic-number-theory class-field-theory
group-theory algebraic-number-theory class-field-theory
asked Jun 28 '18 at 1:40
MatPMatP
1367
asked Jun 28 '18 at 1:40
MatPMatP
1367
edited Jan 8 at 21:20
MatP
asked Jun 28 '18 at 1:40
MatPMatP
1367
asked Jun 28 '18 at 1:40
MatPMatP
1367
asked Jun 28 '18 at 1:40
MatPMatP
1367
1367
$begingroup$
The residue class group clearly only has $5$ elements.
$endgroup$
– franz lemmermeyer
Jan 8 at 21:27
add a comment |
$begingroup$
The residue class group clearly only has $5$ elements.
$endgroup$
– franz lemmermeyer
Jan 8 at 21:27
$begingroup$
The residue class group clearly only has $5$ elements.
$endgroup$
– franz lemmermeyer
Jan 8 at 21:27
$begingroup$
The residue class group clearly only has $5$ elements.
$endgroup$
– franz lemmermeyer
Jan 8 at 21:27
add a comment |
1 Answer
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$begingroup$
The norm of the ideal $Isubsetmathcal{O}_K$ is the order of the quotient
$$mathcal{O}_K/I=Bbb{Z}[alpha,beta]/(5,alpha+1)congBbb{F}_5[beta],$$
which is easily verified to be $5$ because $alpha=sqrt[3]{19}=4inBbb{F}_5$.
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The norm of the ideal $Isubsetmathcal{O}_K$ is the order of the quotient
$$mathcal{O}_K/I=Bbb{Z}[alpha,beta]/(5,alpha+1)congBbb{F}_5[beta],$$
which is easily verified to be $5$ because $alpha=sqrt[3]{19}=4inBbb{F}_5$.
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
$endgroup$
add a comment |
$begingroup$
The norm of the ideal $Isubsetmathcal{O}_K$ is the order of the quotient
$$mathcal{O}_K/I=Bbb{Z}[alpha,beta]/(5,alpha+1)congBbb{F}_5[beta],$$
which is easily verified to be $5$ because $alpha=sqrt[3]{19}=4inBbb{F}_5$.
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
$endgroup$
add a comment |
$begingroup$
The norm of the ideal $Isubsetmathcal{O}_K$ is the order of the quotient
$$mathcal{O}_K/I=Bbb{Z}[alpha,beta]/(5,alpha+1)congBbb{F}_5[beta],$$
which is easily verified to be $5$ because $alpha=sqrt[3]{19}=4inBbb{F}_5$.
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
$endgroup$
The norm of the ideal $Isubsetmathcal{O}_K$ is the order of the quotient
$$mathcal{O}_K/I=Bbb{Z}[alpha,beta]/(5,alpha+1)congBbb{F}_5[beta],$$
which is easily verified to be $5$ because $alpha=sqrt[3]{19}=4inBbb{F}_5$.
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
answered Jan 26 at 13:36
ServaesServaes
29.4k342101
29.4k342101
add a comment |
add a comment |
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$begingroup$
The residue class group clearly only has $5$ elements.
$endgroup$
– franz lemmermeyer
Jan 8 at 21:27