a fractional ideal is a product of integral ideals Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A property of different in Dedekind domainsWhy is the Ideal Norm Multiplicative?Maximal ideals of commutative Artinian ringsProving factorization into prime ideals in a Dedekind domainFunctional equation for ray class zeta functionsChapter 3, Theorem 15 on Marcus' Number FieldsShow that $mathcalI(V)$ is the product ideal of $k=mathbbF_2$Ideal class defined in terms of linear fractional ideal is chern class?Norm compatibility of ideals and idelesAbout the definition of Fractional Ideals in Number Fields
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a fractional ideal is a product of integral ideals
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A property of different in Dedekind domainsWhy is the Ideal Norm Multiplicative?Maximal ideals of commutative Artinian ringsProving factorization into prime ideals in a Dedekind domainFunctional equation for ray class zeta functionsChapter 3, Theorem 15 on Marcus' Number FieldsShow that $mathcalI(V)$ is the product ideal of $k=mathbbF_2$Ideal class defined in terms of linear fractional ideal is chern class?Norm compatibility of ideals and idelesAbout the definition of Fractional Ideals in Number Fields
$begingroup$
Let $K$ be a finite dimensional extension field of $mathbbQ$ and let $Rsubset K$ be the ring of integers in $K$. If $Isubset R$ is an ideal and $kin K$, we call $kI$ a fractional ideal. In the book "Number Fields" by Marcus (exercise 31c chapter 3), it's claimed that a fractional ideal $kI$ can be written as a product of prime ideals $mathfrakp_isubset R$:
$$kI = mathfrakp_1^m_1mathfrakp_2^m_2...mathfrakp_k^m_k.$$
I've also seen the same thing asserted in other places.
This makes no sense since the right hand side is contained in $R$ and the left hand side is not necessarily contained in $R$. Am I missing something? If not, what is meant here? I guess maybe just that they are equal in the class group?
abstract-algebra number-theory
asked Mar 26 at 8:01
bartbart
83
$endgroup$
add a comment |
$begingroup$
Let $K$ be a finite dimensional extension field of $mathbbQ$ and let $Rsubset K$ be the ring of integers in $K$. If $Isubset R$ is an ideal and $kin K$, we call $kI$ a fractional ideal. In the book "Number Fields" by Marcus (exercise 31c chapter 3), it's claimed that a fractional ideal $kI$ can be written as a product of prime ideals $mathfrakp_isubset R$:
$$kI = mathfrakp_1^m_1mathfrakp_2^m_2...mathfrakp_k^m_k.$$
I've also seen the same thing asserted in other places.
This makes no sense since the right hand side is contained in $R$ and the left hand side is not necessarily contained in $R$. Am I missing something? If not, what is meant here? I guess maybe just that they are equal in the class group?
abstract-algebra number-theory
asked Mar 26 at 8:01
bartbart
83
$endgroup$
$begingroup$
Try an actual example, even with $K = mathbf Q$: $(1/5)mathbf Z$ is a fractional ideal and it is $(5mathbf Z)^-1$. You won't be able to get a factorization with nonnegative exponents.
$endgroup$
– KCd
Mar 26 at 12:35
add a comment |
$begingroup$
Let $K$ be a finite dimensional extension field of $mathbbQ$ and let $Rsubset K$ be the ring of integers in $K$. If $Isubset R$ is an ideal and $kin K$, we call $kI$ a fractional ideal. In the book "Number Fields" by Marcus (exercise 31c chapter 3), it's claimed that a fractional ideal $kI$ can be written as a product of prime ideals $mathfrakp_isubset R$:
$$kI = mathfrakp_1^m_1mathfrakp_2^m_2...mathfrakp_k^m_k.$$
I've also seen the same thing asserted in other places.
This makes no sense since the right hand side is contained in $R$ and the left hand side is not necessarily contained in $R$. Am I missing something? If not, what is meant here? I guess maybe just that they are equal in the class group?
abstract-algebra number-theory
asked Mar 26 at 8:01
bartbart
83
$endgroup$
Let $K$ be a finite dimensional extension field of $mathbbQ$ and let $Rsubset K$ be the ring of integers in $K$. If $Isubset R$ is an ideal and $kin K$, we call $kI$ a fractional ideal. In the book "Number Fields" by Marcus (exercise 31c chapter 3), it's claimed that a fractional ideal $kI$ can be written as a product of prime ideals $mathfrakp_isubset R$:
$$kI = mathfrakp_1^m_1mathfrakp_2^m_2...mathfrakp_k^m_k.$$
I've also seen the same thing asserted in other places.
This makes no sense since the right hand side is contained in $R$ and the left hand side is not necessarily contained in $R$. Am I missing something? If not, what is meant here? I guess maybe just that they are equal in the class group?
abstract-algebra number-theory
abstract-algebra number-theory
asked Mar 26 at 8:01
bartbart
83
asked Mar 26 at 8:01
bartbart
83
edited Mar 26 at 8:12
bart
asked Mar 26 at 8:01
bartbart
83
asked Mar 26 at 8:01
bartbart
83
asked Mar 26 at 8:01
bartbart
83
83
$begingroup$
Try an actual example, even with $K = mathbf Q$: $(1/5)mathbf Z$ is a fractional ideal and it is $(5mathbf Z)^-1$. You won't be able to get a factorization with nonnegative exponents.
$endgroup$
– KCd
Mar 26 at 12:35
add a comment |
$begingroup$
Try an actual example, even with $K = mathbf Q$: $(1/5)mathbf Z$ is a fractional ideal and it is $(5mathbf Z)^-1$. You won't be able to get a factorization with nonnegative exponents.
$endgroup$
– KCd
Mar 26 at 12:35
$begingroup$
Try an actual example, even with $K = mathbf Q$: $(1/5)mathbf Z$ is a fractional ideal and it is $(5mathbf Z)^-1$. You won't be able to get a factorization with nonnegative exponents.
$endgroup$
– KCd
Mar 26 at 12:35
$begingroup$
Try an actual example, even with $K = mathbf Q$: $(1/5)mathbf Z$ is a fractional ideal and it is $(5mathbf Z)^-1$. You won't be able to get a factorization with nonnegative exponents.
$endgroup$
– KCd
Mar 26 at 12:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The right hand side is not contained in $R$, because the exponents $m_i$ are allowed to be negative!
One actually defines a fractional ideal $I$ as an $R$-submodule of $K$ such that for some $k in K$ we have $kI subset R$.
With this in mind, one sees what $mathfrak p_i^m_i$ means when the $m_i$ are negative i.e. how the module structure shows up here.
The fractional ideals have an obvious multiplication, via thinking of the modules as just ideals and multiplying them. It turns out that in the given setting (and not in general) that if we take $R$ as the identity , the fractional ideals form a group under this multiplication i.e. every element has an inverse.
The ideals $mathfrak p_i$ may be in $R$, and are clearly fractional ideals, taking $k=1$. However, their inverses in the group of fractional ideals of $R$ under multiplication, are also fractional ideals of $R$ but not necessarily contained in $R$ : they will be ideals in $K$, however. For example, the ideal $2mathbb Z$ is a fractional ideal of $mathbb Q$ over $mathbb Q$, contained in $mathbb Z$, but its inverse is $frac 12 mathbb Z$, which is not contained in $mathbb Z$ (of course, one easily checks that the product of these ideals is $mathbb Z$).
The equality is therefore an equality of fractional ideals i.e. equality as $R$-submodules of $K$.
What is also true is that now if you quotient out by the principal ideals, then the equivalence classes will also be equal. So that means that $[kI]_C = [mathfrak p_1]_C^m_1...[p_k]_C^m_k$ will be true, where $[cdot]_C$ is the equivalence class in the class group of the given ideal.
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
add a comment |
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$begingroup$
The right hand side is not contained in $R$, because the exponents $m_i$ are allowed to be negative!
One actually defines a fractional ideal $I$ as an $R$-submodule of $K$ such that for some $k in K$ we have $kI subset R$.
With this in mind, one sees what $mathfrak p_i^m_i$ means when the $m_i$ are negative i.e. how the module structure shows up here.
The fractional ideals have an obvious multiplication, via thinking of the modules as just ideals and multiplying them. It turns out that in the given setting (and not in general) that if we take $R$ as the identity , the fractional ideals form a group under this multiplication i.e. every element has an inverse.
The ideals $mathfrak p_i$ may be in $R$, and are clearly fractional ideals, taking $k=1$. However, their inverses in the group of fractional ideals of $R$ under multiplication, are also fractional ideals of $R$ but not necessarily contained in $R$ : they will be ideals in $K$, however. For example, the ideal $2mathbb Z$ is a fractional ideal of $mathbb Q$ over $mathbb Q$, contained in $mathbb Z$, but its inverse is $frac 12 mathbb Z$, which is not contained in $mathbb Z$ (of course, one easily checks that the product of these ideals is $mathbb Z$).
The equality is therefore an equality of fractional ideals i.e. equality as $R$-submodules of $K$.
What is also true is that now if you quotient out by the principal ideals, then the equivalence classes will also be equal. So that means that $[kI]_C = [mathfrak p_1]_C^m_1...[p_k]_C^m_k$ will be true, where $[cdot]_C$ is the equivalence class in the class group of the given ideal.
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
add a comment |
$begingroup$
The right hand side is not contained in $R$, because the exponents $m_i$ are allowed to be negative!
One actually defines a fractional ideal $I$ as an $R$-submodule of $K$ such that for some $k in K$ we have $kI subset R$.
With this in mind, one sees what $mathfrak p_i^m_i$ means when the $m_i$ are negative i.e. how the module structure shows up here.
The fractional ideals have an obvious multiplication, via thinking of the modules as just ideals and multiplying them. It turns out that in the given setting (and not in general) that if we take $R$ as the identity , the fractional ideals form a group under this multiplication i.e. every element has an inverse.
The ideals $mathfrak p_i$ may be in $R$, and are clearly fractional ideals, taking $k=1$. However, their inverses in the group of fractional ideals of $R$ under multiplication, are also fractional ideals of $R$ but not necessarily contained in $R$ : they will be ideals in $K$, however. For example, the ideal $2mathbb Z$ is a fractional ideal of $mathbb Q$ over $mathbb Q$, contained in $mathbb Z$, but its inverse is $frac 12 mathbb Z$, which is not contained in $mathbb Z$ (of course, one easily checks that the product of these ideals is $mathbb Z$).
The equality is therefore an equality of fractional ideals i.e. equality as $R$-submodules of $K$.
What is also true is that now if you quotient out by the principal ideals, then the equivalence classes will also be equal. So that means that $[kI]_C = [mathfrak p_1]_C^m_1...[p_k]_C^m_k$ will be true, where $[cdot]_C$ is the equivalence class in the class group of the given ideal.
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
add a comment |
$begingroup$
The right hand side is not contained in $R$, because the exponents $m_i$ are allowed to be negative!
One actually defines a fractional ideal $I$ as an $R$-submodule of $K$ such that for some $k in K$ we have $kI subset R$.
With this in mind, one sees what $mathfrak p_i^m_i$ means when the $m_i$ are negative i.e. how the module structure shows up here.
The fractional ideals have an obvious multiplication, via thinking of the modules as just ideals and multiplying them. It turns out that in the given setting (and not in general) that if we take $R$ as the identity , the fractional ideals form a group under this multiplication i.e. every element has an inverse.
The ideals $mathfrak p_i$ may be in $R$, and are clearly fractional ideals, taking $k=1$. However, their inverses in the group of fractional ideals of $R$ under multiplication, are also fractional ideals of $R$ but not necessarily contained in $R$ : they will be ideals in $K$, however. For example, the ideal $2mathbb Z$ is a fractional ideal of $mathbb Q$ over $mathbb Q$, contained in $mathbb Z$, but its inverse is $frac 12 mathbb Z$, which is not contained in $mathbb Z$ (of course, one easily checks that the product of these ideals is $mathbb Z$).
The equality is therefore an equality of fractional ideals i.e. equality as $R$-submodules of $K$.
What is also true is that now if you quotient out by the principal ideals, then the equivalence classes will also be equal. So that means that $[kI]_C = [mathfrak p_1]_C^m_1...[p_k]_C^m_k$ will be true, where $[cdot]_C$ is the equivalence class in the class group of the given ideal.
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
The right hand side is not contained in $R$, because the exponents $m_i$ are allowed to be negative!
One actually defines a fractional ideal $I$ as an $R$-submodule of $K$ such that for some $k in K$ we have $kI subset R$.
With this in mind, one sees what $mathfrak p_i^m_i$ means when the $m_i$ are negative i.e. how the module structure shows up here.
The fractional ideals have an obvious multiplication, via thinking of the modules as just ideals and multiplying them. It turns out that in the given setting (and not in general) that if we take $R$ as the identity , the fractional ideals form a group under this multiplication i.e. every element has an inverse.
The ideals $mathfrak p_i$ may be in $R$, and are clearly fractional ideals, taking $k=1$. However, their inverses in the group of fractional ideals of $R$ under multiplication, are also fractional ideals of $R$ but not necessarily contained in $R$ : they will be ideals in $K$, however. For example, the ideal $2mathbb Z$ is a fractional ideal of $mathbb Q$ over $mathbb Q$, contained in $mathbb Z$, but its inverse is $frac 12 mathbb Z$, which is not contained in $mathbb Z$ (of course, one easily checks that the product of these ideals is $mathbb Z$).
The equality is therefore an equality of fractional ideals i.e. equality as $R$-submodules of $K$.
What is also true is that now if you quotient out by the principal ideals, then the equivalence classes will also be equal. So that means that $[kI]_C = [mathfrak p_1]_C^m_1...[p_k]_C^m_k$ will be true, where $[cdot]_C$ is the equivalence class in the class group of the given ideal.
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
edited Mar 26 at 8:38
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
answered Mar 26 at 8:27
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
40.6k33678
add a comment |
add a comment |
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$begingroup$
Try an actual example, even with $K = mathbf Q$: $(1/5)mathbf Z$ is a fractional ideal and it is $(5mathbf Z)^-1$. You won't be able to get a factorization with nonnegative exponents.
$endgroup$
– KCd
Mar 26 at 12:35