Is a “local enough” class number always equal to one? [duplicate]
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Standard argument for making the class group of a number field trivial
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Let $F$ be a number field, and $mathcal{O}$ its ring of integers. Is there always a finite set $S$ of places of $F$ such that $mathcal{O}_S$ has class number one? Is it a consequence of standard results on class field theory or is it something else?
number-theory algebraic-number-theory class-field-theory
asked Nov 26 '18 at 7:06
TheStudent
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marked as duplicate by Watson, Community♦ Nov 26 '18 at 7:48
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This question already has an answer here:
Standard argument for making the class group of a number field trivial
1 answer
Let $F$ be a number field, and $mathcal{O}$ its ring of integers. Is there always a finite set $S$ of places of $F$ such that $mathcal{O}_S$ has class number one? Is it a consequence of standard results on class field theory or is it something else?
number-theory algebraic-number-theory class-field-theory
asked Nov 26 '18 at 7:06
TheStudent
3086
marked as duplicate by Watson, Community♦ Nov 26 '18 at 7:48
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Standard argument for making the class group of a number field trivial
1 answer
Let $F$ be a number field, and $mathcal{O}$ its ring of integers. Is there always a finite set $S$ of places of $F$ such that $mathcal{O}_S$ has class number one? Is it a consequence of standard results on class field theory or is it something else?
number-theory algebraic-number-theory class-field-theory
asked Nov 26 '18 at 7:06
TheStudent
3086
This question already has an answer here:
Standard argument for making the class group of a number field trivial
1 answer
Let $F$ be a number field, and $mathcal{O}$ its ring of integers. Is there always a finite set $S$ of places of $F$ such that $mathcal{O}_S$ has class number one? Is it a consequence of standard results on class field theory or is it something else?
This question already has an answer here:
Standard argument for making the class group of a number field trivial
1 answer
number-theory algebraic-number-theory class-field-theory
number-theory algebraic-number-theory class-field-theory
asked Nov 26 '18 at 7:06
TheStudent
3086
asked Nov 26 '18 at 7:06
TheStudent
3086
asked Nov 26 '18 at 7:06
TheStudent
3086
asked Nov 26 '18 at 7:06
TheStudent
3086
asked Nov 26 '18 at 7:06
TheStudent
3086
3086
marked as duplicate by Watson, Community♦ Nov 26 '18 at 7:48
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Watson, Community♦ Nov 26 '18 at 7:48
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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1 Answer
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Yes, it's really just the finiteness of the class number. Let $I$
be a non-principal ideal in $cal O$. Then $I^h=(a)$ is principal, where $h$ is the classnumber. Let $S_1={P_1,ldots,P_k}$ be the prime ideals containing $a$,
(and so containing $I$). In the Dedekind domain $mathcal{O}_{S_1}$, the ideal
$(a)$ (and so also $I$) become trivial. The class groups of $mathcal{O}$
surjects onto that of $mathcal{O}_{S_1}$, and the surjection kills
$[I]$. So we get a Dedekind domain with a smaller class number. Now just repeat
until the classnumber is $1$.
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
Yes, it's really just the finiteness of the class number. Let $I$
be a non-principal ideal in $cal O$. Then $I^h=(a)$ is principal, where $h$ is the classnumber. Let $S_1={P_1,ldots,P_k}$ be the prime ideals containing $a$,
(and so containing $I$). In the Dedekind domain $mathcal{O}_{S_1}$, the ideal
$(a)$ (and so also $I$) become trivial. The class groups of $mathcal{O}$
surjects onto that of $mathcal{O}_{S_1}$, and the surjection kills
$[I]$. So we get a Dedekind domain with a smaller class number. Now just repeat
until the classnumber is $1$.
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
add a comment |
Yes, it's really just the finiteness of the class number. Let $I$
be a non-principal ideal in $cal O$. Then $I^h=(a)$ is principal, where $h$ is the classnumber. Let $S_1={P_1,ldots,P_k}$ be the prime ideals containing $a$,
(and so containing $I$). In the Dedekind domain $mathcal{O}_{S_1}$, the ideal
$(a)$ (and so also $I$) become trivial. The class groups of $mathcal{O}$
surjects onto that of $mathcal{O}_{S_1}$, and the surjection kills
$[I]$. So we get a Dedekind domain with a smaller class number. Now just repeat
until the classnumber is $1$.
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
add a comment |
Yes, it's really just the finiteness of the class number. Let $I$
be a non-principal ideal in $cal O$. Then $I^h=(a)$ is principal, where $h$ is the classnumber. Let $S_1={P_1,ldots,P_k}$ be the prime ideals containing $a$,
(and so containing $I$). In the Dedekind domain $mathcal{O}_{S_1}$, the ideal
$(a)$ (and so also $I$) become trivial. The class groups of $mathcal{O}$
surjects onto that of $mathcal{O}_{S_1}$, and the surjection kills
$[I]$. So we get a Dedekind domain with a smaller class number. Now just repeat
until the classnumber is $1$.
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
Yes, it's really just the finiteness of the class number. Let $I$
be a non-principal ideal in $cal O$. Then $I^h=(a)$ is principal, where $h$ is the classnumber. Let $S_1={P_1,ldots,P_k}$ be the prime ideals containing $a$,
(and so containing $I$). In the Dedekind domain $mathcal{O}_{S_1}$, the ideal
$(a)$ (and so also $I$) become trivial. The class groups of $mathcal{O}$
surjects onto that of $mathcal{O}_{S_1}$, and the surjection kills
$[I]$. So we get a Dedekind domain with a smaller class number. Now just repeat
until the classnumber is $1$.
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
answered Nov 26 '18 at 7:23
Lord Shark the Unknown
101k958132
101k958132
add a comment |
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