$langle 3 rangle$ in $textbfZ[omega]$ ramifies, not splits, right?

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Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.

As you know, $$omega = frac-1 + sqrt-32$$ is a complex cubit root of unity, that is, $omega^3 = 1$. This slightly obscures the fact that $1 + 2 omega = sqrt-3$, and since $(sqrt-3)^2 = -3$, it follows that $langle 3 rangle = langle sqrt-3 rangle^2$. Meaning that $langle 3 rangle$ ramifies, it doesn't split. This even though $$left( frac3 - sqrt-32 right) left( frac3 + sqrt-32 right) = 3.$$

That's because $$left( frac-1 + sqrt-32 right) sqrt-3 = frac-sqrt-3 - 32,$$ or $omega(1 + 2 omega) = omega + 2 omega^2$. Since $omega$ is a unit, the ideals $langle 1 + 2 omega rangle$ and $langle omega + 2 omega^2 rangle$ are in fact the same.

Are these calculations correct? Have I drawn the right conclusion?

algebraic-number-theory

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asked Jul 24 at 20:36

Bill Thomas

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Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.

As you know, $$omega = frac-1 + sqrt-32$$ is a complex cubit root of unity, that is, $omega^3 = 1$. This slightly obscures the fact that $1 + 2 omega = sqrt-3$, and since $(sqrt-3)^2 = -3$, it follows that $langle 3 rangle = langle sqrt-3 rangle^2$. Meaning that $langle 3 rangle$ ramifies, it doesn't split. This even though $$left( frac3 - sqrt-32 right) left( frac3 + sqrt-32 right) = 3.$$

That's because $$left( frac-1 + sqrt-32 right) sqrt-3 = frac-sqrt-3 - 32,$$ or $omega(1 + 2 omega) = omega + 2 omega^2$. Since $omega$ is a unit, the ideals $langle 1 + 2 omega rangle$ and $langle omega + 2 omega^2 rangle$ are in fact the same.

Are these calculations correct? Have I drawn the right conclusion?

algebraic-number-theory

share|cite|improve this question

asked Jul 24 at 20:36

Bill Thomas

366419

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.

As you know, $$omega = frac-1 + sqrt-32$$ is a complex cubit root of unity, that is, $omega^3 = 1$. This slightly obscures the fact that $1 + 2 omega = sqrt-3$, and since $(sqrt-3)^2 = -3$, it follows that $langle 3 rangle = langle sqrt-3 rangle^2$. Meaning that $langle 3 rangle$ ramifies, it doesn't split. This even though $$left( frac3 - sqrt-32 right) left( frac3 + sqrt-32 right) = 3.$$

That's because $$left( frac-1 + sqrt-32 right) sqrt-3 = frac-sqrt-3 - 32,$$ or $omega(1 + 2 omega) = omega + 2 omega^2$. Since $omega$ is a unit, the ideals $langle 1 + 2 omega rangle$ and $langle omega + 2 omega^2 rangle$ are in fact the same.

Are these calculations correct? Have I drawn the right conclusion?

algebraic-number-theory

share|cite|improve this question

asked Jul 24 at 20:36

Bill Thomas

366419

Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.

As you know, $$omega = frac-1 + sqrt-32$$ is a complex cubit root of unity, that is, $omega^3 = 1$. This slightly obscures the fact that $1 + 2 omega = sqrt-3$, and since $(sqrt-3)^2 = -3$, it follows that $langle 3 rangle = langle sqrt-3 rangle^2$. Meaning that $langle 3 rangle$ ramifies, it doesn't split. This even though $$left( frac3 - sqrt-32 right) left( frac3 + sqrt-32 right) = 3.$$

That's because $$left( frac-1 + sqrt-32 right) sqrt-3 = frac-sqrt-3 - 32,$$ or $omega(1 + 2 omega) = omega + 2 omega^2$. Since $omega$ is a unit, the ideals $langle 1 + 2 omega rangle$ and $langle omega + 2 omega^2 rangle$ are in fact the same.

Are these calculations correct? Have I drawn the right conclusion?

algebraic-number-theory

share|cite|improve this question

asked Jul 24 at 20:36

Bill Thomas

366419

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 20:36

Bill Thomas

366419

asked Jul 24 at 20:36

Bill Thomas

366419

asked Jul 24 at 20:36

Bill Thomas

366419

366419

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It seems fine, although I must admit I had trouble understanding what you want.

Anyway, first to conclude that $langle 3 rangle $ ramifies you need to prove that $left langle sqrt-3 right rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.

Then you can establish the following relation

$$left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle = left langle frac3 + sqrt-32 right rangle$$

to conclude that the both ways lead to the same prime ideal factorization of $langle 3 rangle$. As you have already noticed we have that $left langle sqrt-3 right rangle = left langle frac3 + sqrt-32 right rangle$, as $left( frac1 - sqrt-32 right) sqrt-3 = frac3 + sqrt-3 2$

Similarly $left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle$, as $left( frac-1 - sqrt-32 right) sqrt-3 = frac3 - sqrt-3 2$ and the left-most factor is $omega^2$, another unit.

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answered Jul 24 at 21:47

Stefan4024

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Yes your conclusion is correct, and the factorisation
$$left(frac3 - sqrt-32right) left(frac3 + sqrt-32right) = 3$$
is no problem since the factors on the left side are associate. Indeed
$$fracfrac3 - sqrt-32frac3 + sqrt-32=frac(3 - sqrt-3)^2(3 + sqrt-3)(3 - sqrt-3)=frac6-6sqrt-312=frac12-fracsqrt-32,$$
is a unit since it has norm $1$.

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edited Jul 25 at 8:10

answered Jul 24 at 21:51

Bernard

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2 Answers
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2 Answers
2

active

oldest

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active

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active

oldest

votes

up vote
2
down vote

It seems fine, although I must admit I had trouble understanding what you want.

Anyway, first to conclude that $langle 3 rangle $ ramifies you need to prove that $left langle sqrt-3 right rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.

Then you can establish the following relation

$$left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle = left langle frac3 + sqrt-32 right rangle$$

to conclude that the both ways lead to the same prime ideal factorization of $langle 3 rangle$. As you have already noticed we have that $left langle sqrt-3 right rangle = left langle frac3 + sqrt-32 right rangle$, as $left( frac1 - sqrt-32 right) sqrt-3 = frac3 + sqrt-3 2$

Similarly $left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle$, as $left( frac-1 - sqrt-32 right) sqrt-3 = frac3 - sqrt-3 2$ and the left-most factor is $omega^2$, another unit.

share|cite|improve this answer

answered Jul 24 at 21:47

Stefan4024

28k53074

add a comment | 

up vote
2
down vote

It seems fine, although I must admit I had trouble understanding what you want.

Anyway, first to conclude that $langle 3 rangle $ ramifies you need to prove that $left langle sqrt-3 right rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.

Then you can establish the following relation

$$left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle = left langle frac3 + sqrt-32 right rangle$$

to conclude that the both ways lead to the same prime ideal factorization of $langle 3 rangle$. As you have already noticed we have that $left langle sqrt-3 right rangle = left langle frac3 + sqrt-32 right rangle$, as $left( frac1 - sqrt-32 right) sqrt-3 = frac3 + sqrt-3 2$

Similarly $left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle$, as $left( frac-1 - sqrt-32 right) sqrt-3 = frac3 - sqrt-3 2$ and the left-most factor is $omega^2$, another unit.

share|cite|improve this answer

answered Jul 24 at 21:47

Stefan4024

28k53074

add a comment | 

up vote
2
down vote

up vote
2
down vote

It seems fine, although I must admit I had trouble understanding what you want.

Anyway, first to conclude that $langle 3 rangle $ ramifies you need to prove that $left langle sqrt-3 right rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.

Then you can establish the following relation

$$left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle = left langle frac3 + sqrt-32 right rangle$$

to conclude that the both ways lead to the same prime ideal factorization of $langle 3 rangle$. As you have already noticed we have that $left langle sqrt-3 right rangle = left langle frac3 + sqrt-32 right rangle$, as $left( frac1 - sqrt-32 right) sqrt-3 = frac3 + sqrt-3 2$

Similarly $left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle$, as $left( frac-1 - sqrt-32 right) sqrt-3 = frac3 - sqrt-3 2$ and the left-most factor is $omega^2$, another unit.

share|cite|improve this answer

answered Jul 24 at 21:47

Stefan4024

28k53074

It seems fine, although I must admit I had trouble understanding what you want.

Anyway, first to conclude that $langle 3 rangle $ ramifies you need to prove that $left langle sqrt-3 right rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.

Then you can establish the following relation

$$left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle = left langle frac3 + sqrt-32 right rangle$$

to conclude that the both ways lead to the same prime ideal factorization of $langle 3 rangle$. As you have already noticed we have that $left langle sqrt-3 right rangle = left langle frac3 + sqrt-32 right rangle$, as $left( frac1 - sqrt-32 right) sqrt-3 = frac3 + sqrt-3 2$

Similarly $left langle sqrt-3 right rangle = left langle frac3 - sqrt-32 right rangle$, as $left( frac-1 - sqrt-32 right) sqrt-3 = frac3 - sqrt-3 2$ and the left-most factor is $omega^2$, another unit.

share|cite|improve this answer

answered Jul 24 at 21:47

Stefan4024

28k53074

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 21:47

Stefan4024

28k53074

answered Jul 24 at 21:47

Stefan4024

28k53074

answered Jul 24 at 21:47

Stefan4024

28k53074

28k53074

add a comment | 

add a comment | 

up vote
2
down vote

Yes your conclusion is correct, and the factorisation
$$left(frac3 - sqrt-32right) left(frac3 + sqrt-32right) = 3$$
is no problem since the factors on the left side are associate. Indeed
$$fracfrac3 - sqrt-32frac3 + sqrt-32=frac(3 - sqrt-3)^2(3 + sqrt-3)(3 - sqrt-3)=frac6-6sqrt-312=frac12-fracsqrt-32,$$
is a unit since it has norm $1$.

share|cite|improve this answer

edited Jul 25 at 8:10

answered Jul 24 at 21:51

Bernard

110k635103

add a comment | 

up vote
2
down vote

Yes your conclusion is correct, and the factorisation
$$left(frac3 - sqrt-32right) left(frac3 + sqrt-32right) = 3$$
is no problem since the factors on the left side are associate. Indeed
$$fracfrac3 - sqrt-32frac3 + sqrt-32=frac(3 - sqrt-3)^2(3 + sqrt-3)(3 - sqrt-3)=frac6-6sqrt-312=frac12-fracsqrt-32,$$
is a unit since it has norm $1$.

share|cite|improve this answer

edited Jul 25 at 8:10

answered Jul 24 at 21:51

Bernard

110k635103

add a comment | 

up vote
2
down vote

up vote
2
down vote

Yes your conclusion is correct, and the factorisation
$$left(frac3 - sqrt-32right) left(frac3 + sqrt-32right) = 3$$
is no problem since the factors on the left side are associate. Indeed
$$fracfrac3 - sqrt-32frac3 + sqrt-32=frac(3 - sqrt-3)^2(3 + sqrt-3)(3 - sqrt-3)=frac6-6sqrt-312=frac12-fracsqrt-32,$$
is a unit since it has norm $1$.

share|cite|improve this answer

edited Jul 25 at 8:10

answered Jul 24 at 21:51

Bernard

110k635103

Yes your conclusion is correct, and the factorisation
$$left(frac3 - sqrt-32right) left(frac3 + sqrt-32right) = 3$$
is no problem since the factors on the left side are associate. Indeed
$$fracfrac3 - sqrt-32frac3 + sqrt-32=frac(3 - sqrt-3)^2(3 + sqrt-3)(3 - sqrt-3)=frac6-6sqrt-312=frac12-fracsqrt-32,$$
is a unit since it has norm $1$.

share|cite|improve this answer

edited Jul 25 at 8:10

answered Jul 24 at 21:51

Bernard

110k635103

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 8:10

edited Jul 25 at 8:10

edited Jul 25 at 8:10

answered Jul 24 at 21:51

Bernard

110k635103

answered Jul 24 at 21:51

Bernard

110k635103

answered Jul 24 at 21:51

Bernard

110k635103

110k635103

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add a comment | 

 

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