Mixing
Number of regular primes in an order is finite
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Given an order ${mathcal O}$ in a number field a regular prime ${mathfrak p} neq 0$ is defined by the condition ${mathcal O}_{mathfrak p}$ being integrally closed. (Neukirch : Algebraic Number Theory, Pg 79). I need to verify that there are only finitely many non-regular primes.
As I have not yet proved that the Picard group of ${mathcal O}$ is finite, I don't see why this is immediate.
commutative-algebra algebraic-number-theory
edited Jan 26 at 23:04
user26857
39.4k124183
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
$endgroup$
add a comment |
$begingroup$
Given an order ${mathcal O}$ in a number field a regular prime ${mathfrak p} neq 0$ is defined by the condition ${mathcal O}_{mathfrak p}$ being integrally closed. (Neukirch : Algebraic Number Theory, Pg 79). I need to verify that there are only finitely many non-regular primes.
As I have not yet proved that the Picard group of ${mathcal O}$ is finite, I don't see why this is immediate.
commutative-algebra algebraic-number-theory
edited Jan 26 at 23:04
user26857
39.4k124183
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
$endgroup$
$begingroup$
The hypothesis that the integral closure ${tilde {mathcal O}}$ of an order $mathcal O$ is a finitely generated $mathcal O$-module proves the fact. However the question is : is it true for any arbitrary order?
$endgroup$
– Siddhartha
Jan 27 at 19:31
add a comment |
$begingroup$
Given an order ${mathcal O}$ in a number field a regular prime ${mathfrak p} neq 0$ is defined by the condition ${mathcal O}_{mathfrak p}$ being integrally closed. (Neukirch : Algebraic Number Theory, Pg 79). I need to verify that there are only finitely many non-regular primes.
As I have not yet proved that the Picard group of ${mathcal O}$ is finite, I don't see why this is immediate.
commutative-algebra algebraic-number-theory
edited Jan 26 at 23:04
user26857
39.4k124183
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
$endgroup$
Given an order ${mathcal O}$ in a number field a regular prime ${mathfrak p} neq 0$ is defined by the condition ${mathcal O}_{mathfrak p}$ being integrally closed. (Neukirch : Algebraic Number Theory, Pg 79). I need to verify that there are only finitely many non-regular primes.
As I have not yet proved that the Picard group of ${mathcal O}$ is finite, I don't see why this is immediate.
commutative-algebra algebraic-number-theory
commutative-algebra algebraic-number-theory
edited Jan 26 at 23:04
user26857
39.4k124183
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
edited Jan 26 at 23:04
user26857
39.4k124183
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
edited Jan 26 at 23:04
user26857
39.4k124183
edited Jan 26 at 23:04
user26857
39.4k124183
edited Jan 26 at 23:04
user26857
39.4k124183
39.4k124183
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
asked Jan 26 at 22:01
SiddharthaSiddhartha
416
416
$begingroup$
The hypothesis that the integral closure ${tilde {mathcal O}}$ of an order $mathcal O$ is a finitely generated $mathcal O$-module proves the fact. However the question is : is it true for any arbitrary order?
$endgroup$
– Siddhartha
Jan 27 at 19:31
add a comment |
$begingroup$
The hypothesis that the integral closure ${tilde {mathcal O}}$ of an order $mathcal O$ is a finitely generated $mathcal O$-module proves the fact. However the question is : is it true for any arbitrary order?
$endgroup$
– Siddhartha
Jan 27 at 19:31
$begingroup$
The hypothesis that the integral closure ${tilde {mathcal O}}$ of an order $mathcal O$ is a finitely generated $mathcal O$-module proves the fact. However the question is : is it true for any arbitrary order?
$endgroup$
– Siddhartha
Jan 27 at 19:31
$begingroup$
The hypothesis that the integral closure ${tilde {mathcal O}}$ of an order $mathcal O$ is a finitely generated $mathcal O$-module proves the fact. However the question is : is it true for any arbitrary order?
$endgroup$
– Siddhartha
Jan 27 at 19:31
add a comment |
1 Answer
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$begingroup$
Here is the answer : As ${mathcal O} subseteq {mathcal O}_K$ is a subring it embeds $mathbb Z$ inside it. Hence the integral closure ${tilde{mathcal O}} subseteq {mathcal O}_K$. However ${mathcal O}_K$ is the full integral closure of $mathbb Z$ inside $K$. Hence ${tilde{mathcal O}} = {mathcal O}_K$.
Once these two are equal, and the order contain a $mathbb Z$-module of rank $n = [K : {mathbb Q}]$, we have that ${tilde{mathcal O}}$ is a free $mathbb Z$-module of rank $n$. Hence it must be a finitely generated ${mathcal O}$-module.
Let ${mathcal O}_K = {mathcal O} omega_1 + dotsc + {mathcal O} omega_n$ where $omega_1, dotsc, omega_n in {mathcal O}_K$. Now write $omega_i = {frac {a_i}{b_i}}$ for $a_i, b_i in {mathcal O}$ (as $K$ is the quotient field of $mathcal O$). By definition, ${mathfrak p} leq {mathcal O}$ is not regular if and only if $b_i notin {mathfrak p}$ for some $i$. But ${mathcal O}$ is Noetherian of dimension $1$, which means there are only finitely many primes $0 neq b_i notin {mathfrak p}$.
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here is the answer : As ${mathcal O} subseteq {mathcal O}_K$ is a subring it embeds $mathbb Z$ inside it. Hence the integral closure ${tilde{mathcal O}} subseteq {mathcal O}_K$. However ${mathcal O}_K$ is the full integral closure of $mathbb Z$ inside $K$. Hence ${tilde{mathcal O}} = {mathcal O}_K$.
Once these two are equal, and the order contain a $mathbb Z$-module of rank $n = [K : {mathbb Q}]$, we have that ${tilde{mathcal O}}$ is a free $mathbb Z$-module of rank $n$. Hence it must be a finitely generated ${mathcal O}$-module.
Let ${mathcal O}_K = {mathcal O} omega_1 + dotsc + {mathcal O} omega_n$ where $omega_1, dotsc, omega_n in {mathcal O}_K$. Now write $omega_i = {frac {a_i}{b_i}}$ for $a_i, b_i in {mathcal O}$ (as $K$ is the quotient field of $mathcal O$). By definition, ${mathfrak p} leq {mathcal O}$ is not regular if and only if $b_i notin {mathfrak p}$ for some $i$. But ${mathcal O}$ is Noetherian of dimension $1$, which means there are only finitely many primes $0 neq b_i notin {mathfrak p}$.
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
$endgroup$
add a comment |
$begingroup$
Here is the answer : As ${mathcal O} subseteq {mathcal O}_K$ is a subring it embeds $mathbb Z$ inside it. Hence the integral closure ${tilde{mathcal O}} subseteq {mathcal O}_K$. However ${mathcal O}_K$ is the full integral closure of $mathbb Z$ inside $K$. Hence ${tilde{mathcal O}} = {mathcal O}_K$.
Once these two are equal, and the order contain a $mathbb Z$-module of rank $n = [K : {mathbb Q}]$, we have that ${tilde{mathcal O}}$ is a free $mathbb Z$-module of rank $n$. Hence it must be a finitely generated ${mathcal O}$-module.
Let ${mathcal O}_K = {mathcal O} omega_1 + dotsc + {mathcal O} omega_n$ where $omega_1, dotsc, omega_n in {mathcal O}_K$. Now write $omega_i = {frac {a_i}{b_i}}$ for $a_i, b_i in {mathcal O}$ (as $K$ is the quotient field of $mathcal O$). By definition, ${mathfrak p} leq {mathcal O}$ is not regular if and only if $b_i notin {mathfrak p}$ for some $i$. But ${mathcal O}$ is Noetherian of dimension $1$, which means there are only finitely many primes $0 neq b_i notin {mathfrak p}$.
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
$endgroup$
add a comment |
$begingroup$
Here is the answer : As ${mathcal O} subseteq {mathcal O}_K$ is a subring it embeds $mathbb Z$ inside it. Hence the integral closure ${tilde{mathcal O}} subseteq {mathcal O}_K$. However ${mathcal O}_K$ is the full integral closure of $mathbb Z$ inside $K$. Hence ${tilde{mathcal O}} = {mathcal O}_K$.
Once these two are equal, and the order contain a $mathbb Z$-module of rank $n = [K : {mathbb Q}]$, we have that ${tilde{mathcal O}}$ is a free $mathbb Z$-module of rank $n$. Hence it must be a finitely generated ${mathcal O}$-module.
Let ${mathcal O}_K = {mathcal O} omega_1 + dotsc + {mathcal O} omega_n$ where $omega_1, dotsc, omega_n in {mathcal O}_K$. Now write $omega_i = {frac {a_i}{b_i}}$ for $a_i, b_i in {mathcal O}$ (as $K$ is the quotient field of $mathcal O$). By definition, ${mathfrak p} leq {mathcal O}$ is not regular if and only if $b_i notin {mathfrak p}$ for some $i$. But ${mathcal O}$ is Noetherian of dimension $1$, which means there are only finitely many primes $0 neq b_i notin {mathfrak p}$.
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
$endgroup$
Here is the answer : As ${mathcal O} subseteq {mathcal O}_K$ is a subring it embeds $mathbb Z$ inside it. Hence the integral closure ${tilde{mathcal O}} subseteq {mathcal O}_K$. However ${mathcal O}_K$ is the full integral closure of $mathbb Z$ inside $K$. Hence ${tilde{mathcal O}} = {mathcal O}_K$.
Once these two are equal, and the order contain a $mathbb Z$-module of rank $n = [K : {mathbb Q}]$, we have that ${tilde{mathcal O}}$ is a free $mathbb Z$-module of rank $n$. Hence it must be a finitely generated ${mathcal O}$-module.
Let ${mathcal O}_K = {mathcal O} omega_1 + dotsc + {mathcal O} omega_n$ where $omega_1, dotsc, omega_n in {mathcal O}_K$. Now write $omega_i = {frac {a_i}{b_i}}$ for $a_i, b_i in {mathcal O}$ (as $K$ is the quotient field of $mathcal O$). By definition, ${mathfrak p} leq {mathcal O}$ is not regular if and only if $b_i notin {mathfrak p}$ for some $i$. But ${mathcal O}$ is Noetherian of dimension $1$, which means there are only finitely many primes $0 neq b_i notin {mathfrak p}$.
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
answered Jan 31 at 19:37
SiddharthaSiddhartha
416
416
add a comment |
add a comment |
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$begingroup$
The hypothesis that the integral closure ${tilde {mathcal O}}$ of an order $mathcal O$ is a finitely generated $mathcal O$-module proves the fact. However the question is : is it true for any arbitrary order?
$endgroup$
– Siddhartha
Jan 27 at 19:31