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Understanding a solution for the fact that $mathcalO_mathbbQ(sqrt[3]2)=mathbbZ[sqrt[3]2]$

Easy way to show that $mathbbZ[sqrt[3]2]$ is the ring of integers of $mathbbQ[sqrt[3]2]$Ideal Class Group is finiteWhat is known on bounds comparing $ [mathcalO_K :mathbbZ[alpha ]] $ to $textdisc(K)$, $alpha$ a unit?Prove that $mathbbA cap mathbbQ(sqrt2,sqrt-3)$ is a PID.Why if $alpha$ is a root of $x^3-5x^2+2$, then $mathcal O_Bbb Q[alpha]=mathbb Z[alpha]$?If $dequiv 2,3pmod 4$, then the ring of integer of $K$ is $mathbb Z[sqrt d]$.Ring of integers of $mathbbQ(sqrt-3,sqrt5)|mathbbQ$ and group of unitsProve that $1, alpha, alpha^2$ is an integral basis of $mathcal O_K$$mathcalO_K/pmathcalO_K$ isomorphic to $mathbbZ[alpha]/pmathbbZ[alpha]$ for some $alpha in mathcalO_K$?$operatornamedisc(mathbbZ[alpha]) = [mathcalO_K:mathbbZ[alpha]]^2operatornamedisc(mathcalO_K)$.Probability of $mathcalO_mathbbQ(alpha)=mathbbZ[alpha]$

3

1

$begingroup$

The problem above has many answers in StackExchange. I'm trying to understand this specific one.

He mentions the formula $textdisc(mathbbZ[alpha])=(mathcalO:mathbbZ[alpha])^2textdisc(mathcalO)$, computes $textdisc(mathbbZ[alpha])=-2^23^3$, then concludes that $6mathcalOsubset mathbbZ[alpha]$ (here $alpha:=sqrt[3]2$).

The only part I don't understand why is "$6mathcalOsubset mathbbZ[alpha]$"

Any hints?

number-theory algebraic-number-theory discriminant

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asked Mar 11 at 19:48

rmdmc89rmdmc89

2,2931923

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3

1

$begingroup$

The problem above has many answers in StackExchange. I'm trying to understand this specific one.

He mentions the formula $textdisc(mathbbZ[alpha])=(mathcalO:mathbbZ[alpha])^2textdisc(mathcalO)$, computes $textdisc(mathbbZ[alpha])=-2^23^3$, then concludes that $6mathcalOsubset mathbbZ[alpha]$ (here $alpha:=sqrt[3]2$).

The only part I don't understand why is "$6mathcalOsubset mathbbZ[alpha]$"

Any hints?

number-theory algebraic-number-theory discriminant

share|cite|improve this question

asked Mar 11 at 19:48

rmdmc89rmdmc89

2,2931923

$endgroup$

add a comment | 

$begingroup$

The problem above has many answers in StackExchange. I'm trying to understand this specific one.

He mentions the formula $textdisc(mathbbZ[alpha])=(mathcalO:mathbbZ[alpha])^2textdisc(mathcalO)$, computes $textdisc(mathbbZ[alpha])=-2^23^3$, then concludes that $6mathcalOsubset mathbbZ[alpha]$ (here $alpha:=sqrt[3]2$).

The only part I don't understand why is "$6mathcalOsubset mathbbZ[alpha]$"

Any hints?

number-theory algebraic-number-theory discriminant

share|cite|improve this question

asked Mar 11 at 19:48

rmdmc89rmdmc89

2,2931923

$endgroup$

share|cite|improve this question

asked Mar 11 at 19:48

rmdmc89rmdmc89

2,2931923

share|cite|improve this question

asked Mar 11 at 19:48

rmdmc89rmdmc89

2,2931923

asked Mar 11 at 19:48

rmdmc89rmdmc89

2,2931923

asked Mar 11 at 19:48

rmdmc89rmdmc89

2,2931923

2 Answers
2

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oldest

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1

$begingroup$

From the identity
$$(mathcalO:BbbZ[alpha])^2cdotoperatornamedisc(mathcalO)
=operatornamedisc(BbbZ[alpha])
=-2^23^3,$$
it follows that $(mathcalO:BbbZ[alpha])$ divides $6$. That means precisely that the order of the quotient of abelian groups $mathcalO/BbbZ[alpha]$ divides $6$. In particular $6equiv0$ in the quotient $mathcalO/BbbZ[alpha]$ and so $6xinBbbZ[alpha]$ for all $xinmathcalO$, meaning that $6mathcalOsubsetBbbZ[alpha]$.

share|cite|improve this answer

edited Mar 11 at 20:28

answered Mar 11 at 20:07

ServaesServaes

28.3k34099

$endgroup$

add a comment | 

1

$begingroup$

From the mentioned formula, you get $(mathcal O:mathbb Z[alpha])^2midmathrmdisc(mathbb Z[alpha])=-2^23^3$. What can you conclude about $(mathcal O:mathbb Z[alpha])$?

share|cite|improve this answer

answered Mar 11 at 19:52

WojowuWojowu

18.9k23173

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1

$begingroup$

From the identity
$$(mathcalO:BbbZ[alpha])^2cdotoperatornamedisc(mathcalO)
=operatornamedisc(BbbZ[alpha])
=-2^23^3,$$
it follows that $(mathcalO:BbbZ[alpha])$ divides $6$. That means precisely that the order of the quotient of abelian groups $mathcalO/BbbZ[alpha]$ divides $6$. In particular $6equiv0$ in the quotient $mathcalO/BbbZ[alpha]$ and so $6xinBbbZ[alpha]$ for all $xinmathcalO$, meaning that $6mathcalOsubsetBbbZ[alpha]$.

share|cite|improve this answer

edited Mar 11 at 20:28

answered Mar 11 at 20:07

ServaesServaes

28.3k34099

$endgroup$

add a comment | 

1

$begingroup$

From the identity
$$(mathcalO:BbbZ[alpha])^2cdotoperatornamedisc(mathcalO)
=operatornamedisc(BbbZ[alpha])
=-2^23^3,$$
it follows that $(mathcalO:BbbZ[alpha])$ divides $6$. That means precisely that the order of the quotient of abelian groups $mathcalO/BbbZ[alpha]$ divides $6$. In particular $6equiv0$ in the quotient $mathcalO/BbbZ[alpha]$ and so $6xinBbbZ[alpha]$ for all $xinmathcalO$, meaning that $6mathcalOsubsetBbbZ[alpha]$.

share|cite|improve this answer

edited Mar 11 at 20:28

answered Mar 11 at 20:07

ServaesServaes

28.3k34099

$endgroup$

add a comment | 

$begingroup$

From the identity
$$(mathcalO:BbbZ[alpha])^2cdotoperatornamedisc(mathcalO)
=operatornamedisc(BbbZ[alpha])
=-2^23^3,$$
it follows that $(mathcalO:BbbZ[alpha])$ divides $6$. That means precisely that the order of the quotient of abelian groups $mathcalO/BbbZ[alpha]$ divides $6$. In particular $6equiv0$ in the quotient $mathcalO/BbbZ[alpha]$ and so $6xinBbbZ[alpha]$ for all $xinmathcalO$, meaning that $6mathcalOsubsetBbbZ[alpha]$.

share|cite|improve this answer

edited Mar 11 at 20:28

answered Mar 11 at 20:07

ServaesServaes

28.3k34099

$endgroup$

share|cite|improve this answer

edited Mar 11 at 20:28

answered Mar 11 at 20:07

ServaesServaes

28.3k34099

answered Mar 11 at 20:07

ServaesServaes

28.3k34099

answered Mar 11 at 20:07

ServaesServaes

28.3k34099

1

$begingroup$

From the mentioned formula, you get $(mathcal O:mathbb Z[alpha])^2midmathrmdisc(mathbb Z[alpha])=-2^23^3$. What can you conclude about $(mathcal O:mathbb Z[alpha])$?

share|cite|improve this answer

answered Mar 11 at 19:52

WojowuWojowu

18.9k23173

$endgroup$

add a comment | 

1

$begingroup$

From the mentioned formula, you get $(mathcal O:mathbb Z[alpha])^2midmathrmdisc(mathbb Z[alpha])=-2^23^3$. What can you conclude about $(mathcal O:mathbb Z[alpha])$?

share|cite|improve this answer

answered Mar 11 at 19:52

WojowuWojowu

18.9k23173

$endgroup$

add a comment | 

$begingroup$

From the mentioned formula, you get $(mathcal O:mathbb Z[alpha])^2midmathrmdisc(mathbb Z[alpha])=-2^23^3$. What can you conclude about $(mathcal O:mathbb Z[alpha])$?

share|cite|improve this answer

answered Mar 11 at 19:52

WojowuWojowu

18.9k23173

$endgroup$

share|cite|improve this answer

answered Mar 11 at 19:52

WojowuWojowu

18.9k23173

answered Mar 11 at 19:52

WojowuWojowu

18.9k23173

answered Mar 11 at 19:52

WojowuWojowu

18.9k23173