$IS mid JS$ implies that $Imid J$ where $I$ and $J$ are ideals of a number ring contained in another number ring $S$

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$K$ and $L$ are number fields and $K subset L$ . Now, if $I$ and $J$ are two ideals in number ring of $K$ and $IS mid JS$ then we have to show that $I mid J$. Here, $S$ is number ring of $L$.

The book suggests to factor $I$ and $J$ into primes of number ring of $K$ and then consider what happens in $S$.
This question has been asked here before:

Algebraic number theory, Marcus, Chapter 3, Question 9

once but it doesn't use the hint the book gives and instead part $(b)$ of the same problem to prove this result.
Can anybody explain what the author means by 'considering what happens in $S$' or give hints but no full solution please

algebraic-number-theory

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asked Jul 15 at 10:09

clear

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$K$ and $L$ are number fields and $K subset L$ . Now, if $I$ and $J$ are two ideals in number ring of $K$ and $IS mid JS$ then we have to show that $I mid J$. Here, $S$ is number ring of $L$.

The book suggests to factor $I$ and $J$ into primes of number ring of $K$ and then consider what happens in $S$.
This question has been asked here before:

Algebraic number theory, Marcus, Chapter 3, Question 9

once but it doesn't use the hint the book gives and instead part $(b)$ of the same problem to prove this result.
Can anybody explain what the author means by 'considering what happens in $S$' or give hints but no full solution please

algebraic-number-theory

share|cite|improve this question

asked Jul 15 at 10:09

clear

1,30221128

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

$K$ and $L$ are number fields and $K subset L$ . Now, if $I$ and $J$ are two ideals in number ring of $K$ and $IS mid JS$ then we have to show that $I mid J$. Here, $S$ is number ring of $L$.

The book suggests to factor $I$ and $J$ into primes of number ring of $K$ and then consider what happens in $S$.
This question has been asked here before:

Algebraic number theory, Marcus, Chapter 3, Question 9

once but it doesn't use the hint the book gives and instead part $(b)$ of the same problem to prove this result.
Can anybody explain what the author means by 'considering what happens in $S$' or give hints but no full solution please

algebraic-number-theory

share|cite|improve this question

asked Jul 15 at 10:09

clear

1,30221128

$K$ and $L$ are number fields and $K subset L$ . Now, if $I$ and $J$ are two ideals in number ring of $K$ and $IS mid JS$ then we have to show that $I mid J$. Here, $S$ is number ring of $L$.

The book suggests to factor $I$ and $J$ into primes of number ring of $K$ and then consider what happens in $S$.
This question has been asked here before:

Algebraic number theory, Marcus, Chapter 3, Question 9

once but it doesn't use the hint the book gives and instead part $(b)$ of the same problem to prove this result.
Can anybody explain what the author means by 'considering what happens in $S$' or give hints but no full solution please

algebraic-number-theory

share|cite|improve this question

asked Jul 15 at 10:09

clear

1,30221128

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 10:09

clear

1,30221128

asked Jul 15 at 10:09

clear

1,30221128

asked Jul 15 at 10:09

clear

1,30221128

1,30221128

add a comment | 

add a comment | 

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Let $I = mathfrak p_1^e_1dotsmathfrak p_s^e_s$, $J = mathfrak q_1^k_1dotsmathfrak q_r^k_r$ in $R$, the ring of integers of $K$. Let $P_1$ be a prime ideal in the factorization of $mathfrak p_1S$

Then we have $P_1 mid mathfrak p_1S mid IS mid JS$. So as $P_1$ is a prime ideal and the factorzation in prime ideals is unique, $P_1$ must appear in the ideal factorization of $JS$. On the other side the factorization of $JS$ is completely determined by the factorization of $J$ in $R$. In other words we first factorize each $mathfrak q_iS$ and just multiply them to get the factorization of $JS$. Therefore $P_1$ must come from a factorization of one of the prime ideal factors of $J$, WLOG let it be $mathfrak q_1S$.

However we know that a prime ideal lies above a unique prime ideal. In particular, $mathfrak p_1S subseteq P_1 implies P_1 cap R = mathfrak p_1$. From above we get $mathfrak p_1 = P_1 cap R = mathfrak q_1$. Furthermore from $IS mid JS$ it's not hard to conclude that $e_1 le k_1$. Now repeat the same for all $mathfrak p_i$

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answered Jul 16 at 14:31

Stefan4024

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1 Answer
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1 Answer
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Let $I = mathfrak p_1^e_1dotsmathfrak p_s^e_s$, $J = mathfrak q_1^k_1dotsmathfrak q_r^k_r$ in $R$, the ring of integers of $K$. Let $P_1$ be a prime ideal in the factorization of $mathfrak p_1S$

Then we have $P_1 mid mathfrak p_1S mid IS mid JS$. So as $P_1$ is a prime ideal and the factorzation in prime ideals is unique, $P_1$ must appear in the ideal factorization of $JS$. On the other side the factorization of $JS$ is completely determined by the factorization of $J$ in $R$. In other words we first factorize each $mathfrak q_iS$ and just multiply them to get the factorization of $JS$. Therefore $P_1$ must come from a factorization of one of the prime ideal factors of $J$, WLOG let it be $mathfrak q_1S$.

However we know that a prime ideal lies above a unique prime ideal. In particular, $mathfrak p_1S subseteq P_1 implies P_1 cap R = mathfrak p_1$. From above we get $mathfrak p_1 = P_1 cap R = mathfrak q_1$. Furthermore from $IS mid JS$ it's not hard to conclude that $e_1 le k_1$. Now repeat the same for all $mathfrak p_i$

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answered Jul 16 at 14:31

Stefan4024

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up vote
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down vote

Let $I = mathfrak p_1^e_1dotsmathfrak p_s^e_s$, $J = mathfrak q_1^k_1dotsmathfrak q_r^k_r$ in $R$, the ring of integers of $K$. Let $P_1$ be a prime ideal in the factorization of $mathfrak p_1S$

Then we have $P_1 mid mathfrak p_1S mid IS mid JS$. So as $P_1$ is a prime ideal and the factorzation in prime ideals is unique, $P_1$ must appear in the ideal factorization of $JS$. On the other side the factorization of $JS$ is completely determined by the factorization of $J$ in $R$. In other words we first factorize each $mathfrak q_iS$ and just multiply them to get the factorization of $JS$. Therefore $P_1$ must come from a factorization of one of the prime ideal factors of $J$, WLOG let it be $mathfrak q_1S$.

However we know that a prime ideal lies above a unique prime ideal. In particular, $mathfrak p_1S subseteq P_1 implies P_1 cap R = mathfrak p_1$. From above we get $mathfrak p_1 = P_1 cap R = mathfrak q_1$. Furthermore from $IS mid JS$ it's not hard to conclude that $e_1 le k_1$. Now repeat the same for all $mathfrak p_i$

share|cite|improve this answer

answered Jul 16 at 14:31

Stefan4024

28.5k53174

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $I = mathfrak p_1^e_1dotsmathfrak p_s^e_s$, $J = mathfrak q_1^k_1dotsmathfrak q_r^k_r$ in $R$, the ring of integers of $K$. Let $P_1$ be a prime ideal in the factorization of $mathfrak p_1S$

Then we have $P_1 mid mathfrak p_1S mid IS mid JS$. So as $P_1$ is a prime ideal and the factorzation in prime ideals is unique, $P_1$ must appear in the ideal factorization of $JS$. On the other side the factorization of $JS$ is completely determined by the factorization of $J$ in $R$. In other words we first factorize each $mathfrak q_iS$ and just multiply them to get the factorization of $JS$. Therefore $P_1$ must come from a factorization of one of the prime ideal factors of $J$, WLOG let it be $mathfrak q_1S$.

However we know that a prime ideal lies above a unique prime ideal. In particular, $mathfrak p_1S subseteq P_1 implies P_1 cap R = mathfrak p_1$. From above we get $mathfrak p_1 = P_1 cap R = mathfrak q_1$. Furthermore from $IS mid JS$ it's not hard to conclude that $e_1 le k_1$. Now repeat the same for all $mathfrak p_i$

share|cite|improve this answer

answered Jul 16 at 14:31

Stefan4024

28.5k53174

Let $I = mathfrak p_1^e_1dotsmathfrak p_s^e_s$, $J = mathfrak q_1^k_1dotsmathfrak q_r^k_r$ in $R$, the ring of integers of $K$. Let $P_1$ be a prime ideal in the factorization of $mathfrak p_1S$

Then we have $P_1 mid mathfrak p_1S mid IS mid JS$. So as $P_1$ is a prime ideal and the factorzation in prime ideals is unique, $P_1$ must appear in the ideal factorization of $JS$. On the other side the factorization of $JS$ is completely determined by the factorization of $J$ in $R$. In other words we first factorize each $mathfrak q_iS$ and just multiply them to get the factorization of $JS$. Therefore $P_1$ must come from a factorization of one of the prime ideal factors of $J$, WLOG let it be $mathfrak q_1S$.

However we know that a prime ideal lies above a unique prime ideal. In particular, $mathfrak p_1S subseteq P_1 implies P_1 cap R = mathfrak p_1$. From above we get $mathfrak p_1 = P_1 cap R = mathfrak q_1$. Furthermore from $IS mid JS$ it's not hard to conclude that $e_1 le k_1$. Now repeat the same for all $mathfrak p_i$

share|cite|improve this answer

answered Jul 16 at 14:31

Stefan4024

28.5k53174

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 14:31

Stefan4024

28.5k53174

answered Jul 16 at 14:31

Stefan4024

28.5k53174

answered Jul 16 at 14:31

Stefan4024

28.5k53174

28.5k53174

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add a comment | 

 

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