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?






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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







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    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.


















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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







      share|cite|improve this question














      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






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      asked Nov 26 '18 at 7:06









      TheStudent

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      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


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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 $1$.






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            1 Answer
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            1 Answer
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            active

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            active

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            active

            oldest

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            2














            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 $1$.






            share|cite|improve this answer


























              2














              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 $1$.






              share|cite|improve this answer
























                2












                2








                2






                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 $1$.






                share|cite|improve this answer












                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 $1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 '18 at 7:23









                Lord Shark the Unknown

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