Inclusion of quotient ideals. if $I subseteq J$, then $I/qI subseteq J/qJ$?

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Let $mathbbK$ be an algebraic number field.

Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.

Is it true that if $I subseteq J$, then $I/qI subseteq J/qJ$?

ring-theory field-theory algebraic-number-theory ideals

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edited Jul 21 at 13:17

asked Jul 21 at 13:07

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favorite

Let $mathbbK$ be an algebraic number field.

Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.

Is it true that if $I subseteq J$, then $I/qI subseteq J/qJ$?

ring-theory field-theory algebraic-number-theory ideals

share|cite|improve this question

edited Jul 21 at 13:17

asked Jul 21 at 13:07

C.S.

134

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $mathbbK$ be an algebraic number field.

Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.

Is it true that if $I subseteq J$, then $I/qI subseteq J/qJ$?

ring-theory field-theory algebraic-number-theory ideals

share|cite|improve this question

edited Jul 21 at 13:17

asked Jul 21 at 13:07

C.S.

134

Let $mathbbK$ be an algebraic number field.

Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.

Is it true that if $I subseteq J$, then $I/qI subseteq J/qJ$?

ring-theory field-theory algebraic-number-theory ideals

share|cite|improve this question

edited Jul 21 at 13:17

asked Jul 21 at 13:07

C.S.

134

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 13:17

edited Jul 21 at 13:17

edited Jul 21 at 13:17

asked Jul 21 at 13:07

C.S.

134

asked Jul 21 at 13:07

C.S.

134

asked Jul 21 at 13:07

C.S.

134

134

add a comment | 

add a comment | 

2 Answers
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oldest

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0
down vote

Let $mathbbK=mathbbQ$, and $q=5$.

Take $I=4mathbbZ$, and $J=2mathbbZ$; hence $I subseteq J$. But

$I/5I= 4mathbbZ/20mathbbZ= 0,4,8,12,16$, and

$J/5J= 2mathbbZ/10mathbbZ= 0,2,4,6,8$. Clearly we do not have an inclusion in any sens.

Question: can we find always a bijection?

share|cite|improve this answer

answered Jul 21 at 13:27

C.S.

134

add a comment | 

up vote
0
down vote

Note that from $Isubseteq J$ follows $qIsubseteq qJ$ hence we have a group homomorphism $I/qIto J/qJ$.
This is injective if and only if $qI=Icap qJ$ and, in that case, you can identify $I/qI$ with a subgroup of $J/qJ$.

share|cite|improve this answer

edited Jul 21 at 20:42

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

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active

oldest

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active

oldest

votes

up vote
0
down vote

Let $mathbbK=mathbbQ$, and $q=5$.

Take $I=4mathbbZ$, and $J=2mathbbZ$; hence $I subseteq J$. But

$I/5I= 4mathbbZ/20mathbbZ= 0,4,8,12,16$, and

$J/5J= 2mathbbZ/10mathbbZ= 0,2,4,6,8$. Clearly we do not have an inclusion in any sens.

Question: can we find always a bijection?

share|cite|improve this answer

answered Jul 21 at 13:27

C.S.

134

add a comment | 

up vote
0
down vote

Let $mathbbK=mathbbQ$, and $q=5$.

Take $I=4mathbbZ$, and $J=2mathbbZ$; hence $I subseteq J$. But

$I/5I= 4mathbbZ/20mathbbZ= 0,4,8,12,16$, and

$J/5J= 2mathbbZ/10mathbbZ= 0,2,4,6,8$. Clearly we do not have an inclusion in any sens.

Question: can we find always a bijection?

share|cite|improve this answer

answered Jul 21 at 13:27

C.S.

134

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $mathbbK=mathbbQ$, and $q=5$.

Take $I=4mathbbZ$, and $J=2mathbbZ$; hence $I subseteq J$. But

$I/5I= 4mathbbZ/20mathbbZ= 0,4,8,12,16$, and

$J/5J= 2mathbbZ/10mathbbZ= 0,2,4,6,8$. Clearly we do not have an inclusion in any sens.

Question: can we find always a bijection?

share|cite|improve this answer

answered Jul 21 at 13:27

C.S.

134

Let $mathbbK=mathbbQ$, and $q=5$.

Take $I=4mathbbZ$, and $J=2mathbbZ$; hence $I subseteq J$. But

$I/5I= 4mathbbZ/20mathbbZ= 0,4,8,12,16$, and

$J/5J= 2mathbbZ/10mathbbZ= 0,2,4,6,8$. Clearly we do not have an inclusion in any sens.

Question: can we find always a bijection?

share|cite|improve this answer

answered Jul 21 at 13:27

C.S.

134

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 13:27

C.S.

134

answered Jul 21 at 13:27

C.S.

134

answered Jul 21 at 13:27

C.S.

134

134

add a comment | 

add a comment | 

up vote
0
down vote

Note that from $Isubseteq J$ follows $qIsubseteq qJ$ hence we have a group homomorphism $I/qIto J/qJ$.
This is injective if and only if $qI=Icap qJ$ and, in that case, you can identify $I/qI$ with a subgroup of $J/qJ$.

share|cite|improve this answer

edited Jul 21 at 20:42

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

add a comment | 

up vote
0
down vote

Note that from $Isubseteq J$ follows $qIsubseteq qJ$ hence we have a group homomorphism $I/qIto J/qJ$.
This is injective if and only if $qI=Icap qJ$ and, in that case, you can identify $I/qI$ with a subgroup of $J/qJ$.

share|cite|improve this answer

edited Jul 21 at 20:42

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

add a comment | 

up vote
0
down vote

up vote
0
down vote

Note that from $Isubseteq J$ follows $qIsubseteq qJ$ hence we have a group homomorphism $I/qIto J/qJ$.
This is injective if and only if $qI=Icap qJ$ and, in that case, you can identify $I/qI$ with a subgroup of $J/qJ$.

share|cite|improve this answer

edited Jul 21 at 20:42

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

Note that from $Isubseteq J$ follows $qIsubseteq qJ$ hence we have a group homomorphism $I/qIto J/qJ$.
This is injective if and only if $qI=Icap qJ$ and, in that case, you can identify $I/qI$ with a subgroup of $J/qJ$.

share|cite|improve this answer

edited Jul 21 at 20:42

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 20:42

edited Jul 21 at 20:42

edited Jul 21 at 20:42

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

answered Jul 21 at 20:21

Fabio Lucchini

5,62411025

5,62411025

add a comment | 

add a comment | 

 

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