Proof of a stronger form of Chinese remainder theorem (12.3) in Neukirch
$begingroup$
Let $mathcal O$ be an order in a number field and ${mathfrak a} neq 0$ an ideal in $mathcal O$. Then the theorem shows ${mathcal O}/{mathfrak a} = oplus_{{mathfrak p} supseteq {mathfrak a}} {mathcal O}_{mathfrak p}/ {mathfrak a}{mathcal O}_{mathfrak p}$.
In the proof we set ${{tilde{mathfrak a}}_{mathfrak p}} = {mathcal O} cap {mathfrak a}{mathcal O}_{mathfrak p}$ and it is claimed that using the bijective correspondence between primes and primes after localizing it follows : if ${mathfrak p} supseteq {mathfrak a}$ then $mathfrak p$ is the only prime ideal containing ${{tilde{mathfrak a}}_{mathfrak p}}$.
The correspondence works only among the primes and I don't see why it is directly applicable to ${{tilde{mathfrak a}}_{mathfrak p}}$.
My approach was : if ${mathfrak p} neq {mathfrak q} supseteq {mathfrak a}$ with ${mathfrak q} supseteq {{tilde{mathfrak a}}_{mathfrak p}}$ then we can get a contradiction if ${{tilde{mathfrak a}}_{mathfrak p}} {mathcal O}_{mathfrak q}$ contain a unit of ${mathcal O}_{mathfrak q}$. However, ${mathfrak a} subseteq {mathfrak p} cap {mathfrak q}$ and I don't see how a unit can be constructed.
Another way could be to take the multiplicative set $S = {mathcal O} setminus ({mathfrak p} cup {mathfrak q})$ and relate ${mathcal O}_{mathfrak q}$ to ${mathcal O}_S$. Here again the same problem occurs.
commutative-algebra algebraic-number-theory
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
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add a comment |
$begingroup$
Let $mathcal O$ be an order in a number field and ${mathfrak a} neq 0$ an ideal in $mathcal O$. Then the theorem shows ${mathcal O}/{mathfrak a} = oplus_{{mathfrak p} supseteq {mathfrak a}} {mathcal O}_{mathfrak p}/ {mathfrak a}{mathcal O}_{mathfrak p}$.
In the proof we set ${{tilde{mathfrak a}}_{mathfrak p}} = {mathcal O} cap {mathfrak a}{mathcal O}_{mathfrak p}$ and it is claimed that using the bijective correspondence between primes and primes after localizing it follows : if ${mathfrak p} supseteq {mathfrak a}$ then $mathfrak p$ is the only prime ideal containing ${{tilde{mathfrak a}}_{mathfrak p}}$.
The correspondence works only among the primes and I don't see why it is directly applicable to ${{tilde{mathfrak a}}_{mathfrak p}}$.
My approach was : if ${mathfrak p} neq {mathfrak q} supseteq {mathfrak a}$ with ${mathfrak q} supseteq {{tilde{mathfrak a}}_{mathfrak p}}$ then we can get a contradiction if ${{tilde{mathfrak a}}_{mathfrak p}} {mathcal O}_{mathfrak q}$ contain a unit of ${mathcal O}_{mathfrak q}$. However, ${mathfrak a} subseteq {mathfrak p} cap {mathfrak q}$ and I don't see how a unit can be constructed.
Another way could be to take the multiplicative set $S = {mathcal O} setminus ({mathfrak p} cup {mathfrak q})$ and relate ${mathcal O}_{mathfrak q}$ to ${mathcal O}_S$. Here again the same problem occurs.
commutative-algebra algebraic-number-theory
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
$endgroup$
add a comment |
$begingroup$
Let $mathcal O$ be an order in a number field and ${mathfrak a} neq 0$ an ideal in $mathcal O$. Then the theorem shows ${mathcal O}/{mathfrak a} = oplus_{{mathfrak p} supseteq {mathfrak a}} {mathcal O}_{mathfrak p}/ {mathfrak a}{mathcal O}_{mathfrak p}$.
In the proof we set ${{tilde{mathfrak a}}_{mathfrak p}} = {mathcal O} cap {mathfrak a}{mathcal O}_{mathfrak p}$ and it is claimed that using the bijective correspondence between primes and primes after localizing it follows : if ${mathfrak p} supseteq {mathfrak a}$ then $mathfrak p$ is the only prime ideal containing ${{tilde{mathfrak a}}_{mathfrak p}}$.
The correspondence works only among the primes and I don't see why it is directly applicable to ${{tilde{mathfrak a}}_{mathfrak p}}$.
My approach was : if ${mathfrak p} neq {mathfrak q} supseteq {mathfrak a}$ with ${mathfrak q} supseteq {{tilde{mathfrak a}}_{mathfrak p}}$ then we can get a contradiction if ${{tilde{mathfrak a}}_{mathfrak p}} {mathcal O}_{mathfrak q}$ contain a unit of ${mathcal O}_{mathfrak q}$. However, ${mathfrak a} subseteq {mathfrak p} cap {mathfrak q}$ and I don't see how a unit can be constructed.
Another way could be to take the multiplicative set $S = {mathcal O} setminus ({mathfrak p} cup {mathfrak q})$ and relate ${mathcal O}_{mathfrak q}$ to ${mathcal O}_S$. Here again the same problem occurs.
commutative-algebra algebraic-number-theory
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
$endgroup$
Let $mathcal O$ be an order in a number field and ${mathfrak a} neq 0$ an ideal in $mathcal O$. Then the theorem shows ${mathcal O}/{mathfrak a} = oplus_{{mathfrak p} supseteq {mathfrak a}} {mathcal O}_{mathfrak p}/ {mathfrak a}{mathcal O}_{mathfrak p}$.
In the proof we set ${{tilde{mathfrak a}}_{mathfrak p}} = {mathcal O} cap {mathfrak a}{mathcal O}_{mathfrak p}$ and it is claimed that using the bijective correspondence between primes and primes after localizing it follows : if ${mathfrak p} supseteq {mathfrak a}$ then $mathfrak p$ is the only prime ideal containing ${{tilde{mathfrak a}}_{mathfrak p}}$.
The correspondence works only among the primes and I don't see why it is directly applicable to ${{tilde{mathfrak a}}_{mathfrak p}}$.
My approach was : if ${mathfrak p} neq {mathfrak q} supseteq {mathfrak a}$ with ${mathfrak q} supseteq {{tilde{mathfrak a}}_{mathfrak p}}$ then we can get a contradiction if ${{tilde{mathfrak a}}_{mathfrak p}} {mathcal O}_{mathfrak q}$ contain a unit of ${mathcal O}_{mathfrak q}$. However, ${mathfrak a} subseteq {mathfrak p} cap {mathfrak q}$ and I don't see how a unit can be constructed.
Another way could be to take the multiplicative set $S = {mathcal O} setminus ({mathfrak p} cup {mathfrak q})$ and relate ${mathcal O}_{mathfrak q}$ to ${mathcal O}_S$. Here again the same problem occurs.
commutative-algebra algebraic-number-theory
commutative-algebra algebraic-number-theory
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
asked Jan 15 at 21:28
SiddharthaSiddhartha
416
416
add a comment |
add a comment |
1 Answer
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An order has Krull dimension 1, and in particular non-zero prime ideals are maximal. It follows that $mathcal{O}_{mathfrak{p}}$ has exactly two prime ideals, the maximal ideal $bar{mathfrak{p}}$ corresponding to $mathfrak{p}$ and $(0)$. Since $mathfrak{a}neq 0$, we have $(0)notsupseteq bar{mathfrak{a}}$. Therefore the only possibility is that $bar{mathfrak{a}}subseteq bar{mathfrak{p}}$. And this inclusion holds because we had $mathfrak{a}subseteq mathfrak{p}$.
answered Jan 15 at 23:31
PedroPedro
2,9441721
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$begingroup$
Thanks. This works.
$endgroup$
– Siddhartha
Jan 16 at 11:53
add a comment |
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1 Answer
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1 Answer
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$begingroup$
An order has Krull dimension 1, and in particular non-zero prime ideals are maximal. It follows that $mathcal{O}_{mathfrak{p}}$ has exactly two prime ideals, the maximal ideal $bar{mathfrak{p}}$ corresponding to $mathfrak{p}$ and $(0)$. Since $mathfrak{a}neq 0$, we have $(0)notsupseteq bar{mathfrak{a}}$. Therefore the only possibility is that $bar{mathfrak{a}}subseteq bar{mathfrak{p}}$. And this inclusion holds because we had $mathfrak{a}subseteq mathfrak{p}$.
answered Jan 15 at 23:31
PedroPedro
2,9441721
$endgroup$
$begingroup$
Thanks. This works.
$endgroup$
– Siddhartha
Jan 16 at 11:53
add a comment |
$begingroup$
An order has Krull dimension 1, and in particular non-zero prime ideals are maximal. It follows that $mathcal{O}_{mathfrak{p}}$ has exactly two prime ideals, the maximal ideal $bar{mathfrak{p}}$ corresponding to $mathfrak{p}$ and $(0)$. Since $mathfrak{a}neq 0$, we have $(0)notsupseteq bar{mathfrak{a}}$. Therefore the only possibility is that $bar{mathfrak{a}}subseteq bar{mathfrak{p}}$. And this inclusion holds because we had $mathfrak{a}subseteq mathfrak{p}$.
answered Jan 15 at 23:31
PedroPedro
2,9441721
$endgroup$
$begingroup$
Thanks. This works.
$endgroup$
– Siddhartha
Jan 16 at 11:53
add a comment |
$begingroup$
An order has Krull dimension 1, and in particular non-zero prime ideals are maximal. It follows that $mathcal{O}_{mathfrak{p}}$ has exactly two prime ideals, the maximal ideal $bar{mathfrak{p}}$ corresponding to $mathfrak{p}$ and $(0)$. Since $mathfrak{a}neq 0$, we have $(0)notsupseteq bar{mathfrak{a}}$. Therefore the only possibility is that $bar{mathfrak{a}}subseteq bar{mathfrak{p}}$. And this inclusion holds because we had $mathfrak{a}subseteq mathfrak{p}$.
answered Jan 15 at 23:31
PedroPedro
2,9441721
$endgroup$
An order has Krull dimension 1, and in particular non-zero prime ideals are maximal. It follows that $mathcal{O}_{mathfrak{p}}$ has exactly two prime ideals, the maximal ideal $bar{mathfrak{p}}$ corresponding to $mathfrak{p}$ and $(0)$. Since $mathfrak{a}neq 0$, we have $(0)notsupseteq bar{mathfrak{a}}$. Therefore the only possibility is that $bar{mathfrak{a}}subseteq bar{mathfrak{p}}$. And this inclusion holds because we had $mathfrak{a}subseteq mathfrak{p}$.
answered Jan 15 at 23:31
PedroPedro
2,9441721
answered Jan 15 at 23:31
PedroPedro
2,9441721
answered Jan 15 at 23:31
PedroPedro
2,9441721
answered Jan 15 at 23:31
PedroPedro
2,9441721
2,9441721
$begingroup$
Thanks. This works.
$endgroup$
– Siddhartha
Jan 16 at 11:53
add a comment |
$begingroup$
Thanks. This works.
$endgroup$
– Siddhartha
Jan 16 at 11:53
$begingroup$
Thanks. This works.
$endgroup$
– Siddhartha
Jan 16 at 11:53
$begingroup$
Thanks. This works.
$endgroup$
– Siddhartha
Jan 16 at 11:53
add a comment |
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