Mixing
Localization correspondence
$begingroup$
This is taken from Neukirch's algebraic number theory Proposition 12.3.
Proposition (12.3).
If $aneq 0$ is an ideal of an order (one dimensional Noetherian integral domain) $o$, then:
$o/a = oplus_{p}o_p/ao_p = oplus_{asubseteq p}o_p/ao_p$
My question is regarding the first line of the proof:
Let $tilde{a}_p = ocap ao_p$.
As I understand it, $o_p$ is the localization of $o$ at the prime $p$. Therefore, there are two cases:
$a notsubset p$: $tilde{a}_p = o_p$. As is stated firmly in the book.
$a subseteq p$: $tilde{a}_p = a$.
However, regarding case #2, the author implicitly says that $asubseteq tilde{a}_p subseteq p$, and moreover, that whereas $a$ may be contained in several prime ideals $p$, $tilde{a}_p$ is contained only in $p$. If I am not completely going schizophrenic as we speak, then I believe that if case #2 holds, then $tilde{a}_p$ is the same thing as $a$, hence if there are several prime ideals containing $a$, then all of these prime ideals should contain $tilde{a}_p$. The author is implying somehow that under certain circumstances, $a$ is not the same thing as $tilde{a}_p$, and I am completely lost.
Thanks in advance.
number-theory algebraic-number-theory ideals localization
asked Jan 17 at 11:04
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
This is taken from Neukirch's algebraic number theory Proposition 12.3.
Proposition (12.3).
If $aneq 0$ is an ideal of an order (one dimensional Noetherian integral domain) $o$, then:
$o/a = oplus_{p}o_p/ao_p = oplus_{asubseteq p}o_p/ao_p$
My question is regarding the first line of the proof:
Let $tilde{a}_p = ocap ao_p$.
As I understand it, $o_p$ is the localization of $o$ at the prime $p$. Therefore, there are two cases:
$a notsubset p$: $tilde{a}_p = o_p$. As is stated firmly in the book.
$a subseteq p$: $tilde{a}_p = a$.
However, regarding case #2, the author implicitly says that $asubseteq tilde{a}_p subseteq p$, and moreover, that whereas $a$ may be contained in several prime ideals $p$, $tilde{a}_p$ is contained only in $p$. If I am not completely going schizophrenic as we speak, then I believe that if case #2 holds, then $tilde{a}_p$ is the same thing as $a$, hence if there are several prime ideals containing $a$, then all of these prime ideals should contain $tilde{a}_p$. The author is implying somehow that under certain circumstances, $a$ is not the same thing as $tilde{a}_p$, and I am completely lost.
Thanks in advance.
number-theory algebraic-number-theory ideals localization
asked Jan 17 at 11:04
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
This is taken from Neukirch's algebraic number theory Proposition 12.3.
Proposition (12.3).
If $aneq 0$ is an ideal of an order (one dimensional Noetherian integral domain) $o$, then:
$o/a = oplus_{p}o_p/ao_p = oplus_{asubseteq p}o_p/ao_p$
My question is regarding the first line of the proof:
Let $tilde{a}_p = ocap ao_p$.
As I understand it, $o_p$ is the localization of $o$ at the prime $p$. Therefore, there are two cases:
$a notsubset p$: $tilde{a}_p = o_p$. As is stated firmly in the book.
$a subseteq p$: $tilde{a}_p = a$.
However, regarding case #2, the author implicitly says that $asubseteq tilde{a}_p subseteq p$, and moreover, that whereas $a$ may be contained in several prime ideals $p$, $tilde{a}_p$ is contained only in $p$. If I am not completely going schizophrenic as we speak, then I believe that if case #2 holds, then $tilde{a}_p$ is the same thing as $a$, hence if there are several prime ideals containing $a$, then all of these prime ideals should contain $tilde{a}_p$. The author is implying somehow that under certain circumstances, $a$ is not the same thing as $tilde{a}_p$, and I am completely lost.
Thanks in advance.
number-theory algebraic-number-theory ideals localization
asked Jan 17 at 11:04
kindasortakindasorta
9810
$endgroup$
This is taken from Neukirch's algebraic number theory Proposition 12.3.
Proposition (12.3).
If $aneq 0$ is an ideal of an order (one dimensional Noetherian integral domain) $o$, then:
$o/a = oplus_{p}o_p/ao_p = oplus_{asubseteq p}o_p/ao_p$
My question is regarding the first line of the proof:
Let $tilde{a}_p = ocap ao_p$.
As I understand it, $o_p$ is the localization of $o$ at the prime $p$. Therefore, there are two cases:
$a notsubset p$: $tilde{a}_p = o_p$. As is stated firmly in the book.
$a subseteq p$: $tilde{a}_p = a$.
However, regarding case #2, the author implicitly says that $asubseteq tilde{a}_p subseteq p$, and moreover, that whereas $a$ may be contained in several prime ideals $p$, $tilde{a}_p$ is contained only in $p$. If I am not completely going schizophrenic as we speak, then I believe that if case #2 holds, then $tilde{a}_p$ is the same thing as $a$, hence if there are several prime ideals containing $a$, then all of these prime ideals should contain $tilde{a}_p$. The author is implying somehow that under certain circumstances, $a$ is not the same thing as $tilde{a}_p$, and I am completely lost.
Thanks in advance.
number-theory algebraic-number-theory ideals localization
number-theory algebraic-number-theory ideals localization
asked Jan 17 at 11:04
kindasortakindasorta
9810
asked Jan 17 at 11:04
kindasortakindasorta
9810
edited Jan 17 at 14:35
kindasorta
asked Jan 17 at 11:04
kindasortakindasorta
9810
asked Jan 17 at 11:04
kindasortakindasorta
9810
asked Jan 17 at 11:04
kindasortakindasorta
9810
9810
add a comment |
add a comment |
1 Answer
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$begingroup$
After digging a bit through Atiyah-McDonald, I realized that the correspondence theorem between ideals and their extensions in the localization is a correspondence theorem for prime ideals. It does not hold in general for all ideals, and this was my misunderstanding.
answered Jan 18 at 17:14
kindasortakindasorta
9810
$endgroup$
add a comment |
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$begingroup$
After digging a bit through Atiyah-McDonald, I realized that the correspondence theorem between ideals and their extensions in the localization is a correspondence theorem for prime ideals. It does not hold in general for all ideals, and this was my misunderstanding.
answered Jan 18 at 17:14
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
After digging a bit through Atiyah-McDonald, I realized that the correspondence theorem between ideals and their extensions in the localization is a correspondence theorem for prime ideals. It does not hold in general for all ideals, and this was my misunderstanding.
answered Jan 18 at 17:14
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
After digging a bit through Atiyah-McDonald, I realized that the correspondence theorem between ideals and their extensions in the localization is a correspondence theorem for prime ideals. It does not hold in general for all ideals, and this was my misunderstanding.
answered Jan 18 at 17:14
kindasortakindasorta
9810
$endgroup$
After digging a bit through Atiyah-McDonald, I realized that the correspondence theorem between ideals and their extensions in the localization is a correspondence theorem for prime ideals. It does not hold in general for all ideals, and this was my misunderstanding.
answered Jan 18 at 17:14
kindasortakindasorta
9810
answered Jan 18 at 17:14
kindasortakindasorta
9810
answered Jan 18 at 17:14
kindasortakindasorta
9810
answered Jan 18 at 17:14
kindasortakindasorta
9810
9810
add a comment |
add a comment |
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