$Asubset Bsubset K$, $K$ field of fractions of $A$, $B$ flat over $A$ $implies B=S^{-1}A ?$












2












$begingroup$


Let $Asubset Bsubset K$ be domains such that $K$ is the field of fractions of $A$, and $B$ is flat over $A$.



Is there a multiplicative subset $S$ of $A$ such that $B=S^{-1}A ?$






commutative-algebra integral-domain







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    An integral domain is Prufer iff every overring is flat. In particular, every overring of a Dedekind domain is flat. A noetherian domain has the QR-property (each overring is a quotient ring, or a fraction ring if you prefer) iff it is a Dedekind domain with torsion class group. This shows that a Dedekind domain whose class group is not torsion provides a counterexample.
    $endgroup$
    – user26857
    Nov 30 '18 at 17:26






  • 1




    $begingroup$
    @user26857 - Thanks for your awesome comment! Would you mind giving some references?
    $endgroup$
    – Pierre-Yves Gaillard
    Nov 30 '18 at 17:34






  • 1




    $begingroup$
    1) en.wikipedia.org/wiki/Pr%C3%BCfer_domain; 2) link.springer.com/article/10.1007%2FBF01361178
    $endgroup$
    – user26857
    Nov 30 '18 at 17:38










  • $begingroup$
    @user26857 just wanna note that you can get a class of non-Noetherian examples by the same token, since Goldman's theorem extends to generalized Dedekind domains (i.e. Prüfer domains $D$ in which $minSpec(D/I)$ is finite for each ideal $I$ and each prime is idempotent).
    $endgroup$
    – Badam Baplan
    Nov 30 '18 at 18:22


















2












$begingroup$


Let $Asubset Bsubset K$ be domains such that $K$ is the field of fractions of $A$, and $B$ is flat over $A$.



Is there a multiplicative subset $S$ of $A$ such that $B=S^{-1}A ?$






commutative-algebra integral-domain







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    An integral domain is Prufer iff every overring is flat. In particular, every overring of a Dedekind domain is flat. A noetherian domain has the QR-property (each overring is a quotient ring, or a fraction ring if you prefer) iff it is a Dedekind domain with torsion class group. This shows that a Dedekind domain whose class group is not torsion provides a counterexample.
    $endgroup$
    – user26857
    Nov 30 '18 at 17:26






  • 1




    $begingroup$
    @user26857 - Thanks for your awesome comment! Would you mind giving some references?
    $endgroup$
    – Pierre-Yves Gaillard
    Nov 30 '18 at 17:34






  • 1




    $begingroup$
    1) en.wikipedia.org/wiki/Pr%C3%BCfer_domain; 2) link.springer.com/article/10.1007%2FBF01361178
    $endgroup$
    – user26857
    Nov 30 '18 at 17:38










  • $begingroup$
    @user26857 just wanna note that you can get a class of non-Noetherian examples by the same token, since Goldman's theorem extends to generalized Dedekind domains (i.e. Prüfer domains $D$ in which $minSpec(D/I)$ is finite for each ideal $I$ and each prime is idempotent).
    $endgroup$
    – Badam Baplan
    Nov 30 '18 at 18:22
















2












2








2


2



$begingroup$


Let $Asubset Bsubset K$ be domains such that $K$ is the field of fractions of $A$, and $B$ is flat over $A$.



Is there a multiplicative subset $S$ of $A$ such that $B=S^{-1}A ?$






commutative-algebra integral-domain







share|cite|improve this question









$endgroup$




Let $Asubset Bsubset K$ be domains such that $K$ is the field of fractions of $A$, and $B$ is flat over $A$.



Is there a multiplicative subset $S$ of $A$ such that $B=S^{-1}A ?$






commutative-algebra integral-domain




commutative-algebra integral-domain






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 12:28









Pierre-Yves GaillardPierre-Yves Gaillard

13.2k23181




13.2k23181








  • 1




    $begingroup$
    An integral domain is Prufer iff every overring is flat. In particular, every overring of a Dedekind domain is flat. A noetherian domain has the QR-property (each overring is a quotient ring, or a fraction ring if you prefer) iff it is a Dedekind domain with torsion class group. This shows that a Dedekind domain whose class group is not torsion provides a counterexample.
    $endgroup$
    – user26857
    Nov 30 '18 at 17:26






  • 1




    $begingroup$
    @user26857 - Thanks for your awesome comment! Would you mind giving some references?
    $endgroup$
    – Pierre-Yves Gaillard
    Nov 30 '18 at 17:34






  • 1




    $begingroup$
    1) en.wikipedia.org/wiki/Pr%C3%BCfer_domain; 2) link.springer.com/article/10.1007%2FBF01361178
    $endgroup$
    – user26857
    Nov 30 '18 at 17:38










  • $begingroup$
    @user26857 just wanna note that you can get a class of non-Noetherian examples by the same token, since Goldman's theorem extends to generalized Dedekind domains (i.e. Prüfer domains $D$ in which $minSpec(D/I)$ is finite for each ideal $I$ and each prime is idempotent).
    $endgroup$
    – Badam Baplan
    Nov 30 '18 at 18:22
















  • 1




    $begingroup$
    An integral domain is Prufer iff every overring is flat. In particular, every overring of a Dedekind domain is flat. A noetherian domain has the QR-property (each overring is a quotient ring, or a fraction ring if you prefer) iff it is a Dedekind domain with torsion class group. This shows that a Dedekind domain whose class group is not torsion provides a counterexample.
    $endgroup$
    – user26857
    Nov 30 '18 at 17:26






  • 1




    $begingroup$
    @user26857 - Thanks for your awesome comment! Would you mind giving some references?
    $endgroup$
    – Pierre-Yves Gaillard
    Nov 30 '18 at 17:34






  • 1




    $begingroup$
    1) en.wikipedia.org/wiki/Pr%C3%BCfer_domain; 2) link.springer.com/article/10.1007%2FBF01361178
    $endgroup$
    – user26857
    Nov 30 '18 at 17:38










  • $begingroup$
    @user26857 just wanna note that you can get a class of non-Noetherian examples by the same token, since Goldman's theorem extends to generalized Dedekind domains (i.e. Prüfer domains $D$ in which $minSpec(D/I)$ is finite for each ideal $I$ and each prime is idempotent).
    $endgroup$
    – Badam Baplan
    Nov 30 '18 at 18:22










1




1




$begingroup$
An integral domain is Prufer iff every overring is flat. In particular, every overring of a Dedekind domain is flat. A noetherian domain has the QR-property (each overring is a quotient ring, or a fraction ring if you prefer) iff it is a Dedekind domain with torsion class group. This shows that a Dedekind domain whose class group is not torsion provides a counterexample.
$endgroup$
– user26857
Nov 30 '18 at 17:26




$begingroup$
An integral domain is Prufer iff every overring is flat. In particular, every overring of a Dedekind domain is flat. A noetherian domain has the QR-property (each overring is a quotient ring, or a fraction ring if you prefer) iff it is a Dedekind domain with torsion class group. This shows that a Dedekind domain whose class group is not torsion provides a counterexample.
$endgroup$
– user26857
Nov 30 '18 at 17:26




1




1




$begingroup$
@user26857 - Thanks for your awesome comment! Would you mind giving some references?
$endgroup$
– Pierre-Yves Gaillard
Nov 30 '18 at 17:34




$begingroup$
@user26857 - Thanks for your awesome comment! Would you mind giving some references?
$endgroup$
– Pierre-Yves Gaillard
Nov 30 '18 at 17:34




1




1




$begingroup$
1) en.wikipedia.org/wiki/Pr%C3%BCfer_domain; 2) link.springer.com/article/10.1007%2FBF01361178
$endgroup$
– user26857
Nov 30 '18 at 17:38




$begingroup$
1) en.wikipedia.org/wiki/Pr%C3%BCfer_domain; 2) link.springer.com/article/10.1007%2FBF01361178
$endgroup$
– user26857
Nov 30 '18 at 17:38












$begingroup$
@user26857 just wanna note that you can get a class of non-Noetherian examples by the same token, since Goldman's theorem extends to generalized Dedekind domains (i.e. Prüfer domains $D$ in which $minSpec(D/I)$ is finite for each ideal $I$ and each prime is idempotent).
$endgroup$
– Badam Baplan
Nov 30 '18 at 18:22






$begingroup$
@user26857 just wanna note that you can get a class of non-Noetherian examples by the same token, since Goldman's theorem extends to generalized Dedekind domains (i.e. Prüfer domains $D$ in which $minSpec(D/I)$ is finite for each ideal $I$ and each prime is idempotent).
$endgroup$
– Badam Baplan
Nov 30 '18 at 18:22












1 Answer
1






active

oldest

votes


















4












$begingroup$

No, flat overrings are not necessarily localizations (although the converse of course holds). But there are important similarities, enough to motivate their being called "generalized quotient rings."



You need to generalize the notion of a localization to properly characterize flat overrings as quotient-like structures. The appropriate generalization is to take a multiplicatively closed collection of ideals, $mathfrak{I}$, and to define the generalized transform of $R$ with respect to $mathfrak{I}$ as $R_{mathfrak{I}}=bigcup_{Ain mathfrak{I}}A^{-1}$. Here, by $A^{-1}$, I mean the quotient ideal $(R :_K A)$. There's no need to limit our discussion to integral domains.




Theorem: Given $A subset B subset K$, $B$ is flat over $A$ iff there exists a multiplicatively closed collection of ideals $mathfrak{I}$ in $A$ such that $B = A_{mathfrak{I}}$ and each $I in mathfrak{I}$ generates $B$ as a $B$-module.




This is Theorem 1.3 in this paper of Arnold and Brewer.



Akiba gives an example (see beginning of section 2) showing that a flat overring need not be a localization.




Let C be a non-singular plane cubic curve defined over a field $k_0$
and let $P$ be a generic point of C over $k_0$, and let $k$ be a field containing
$k_0(P)$. Then the homogeneous coordinate ring $R_0=k[x, y, z]$
of $C$ over $k$ is normal.



Let $A=k[x, y, z]_{(x,y,z)}$, and set $mathfrak{I} = {mathfrak{p}^n mid n in mathbb{N} }$ where $mathfrak{p}$ is the homogeneous prime ideal of $A$ corresponding to $P$. Let $B= A_mathfrak{I}$.




Then $A$ is a local domain and $B$ is a flat overring of $A$ such that no non-unit of $A$ becomes a unit in $B$. For our purposes, this shows that $B$ is not a localization of $A$. For Akiba's purposes, it showed the stronger statement that $B$ is not a generalized transform of $A$ with respect to any collection of invertible ideals.






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    4












    $begingroup$

    No, flat overrings are not necessarily localizations (although the converse of course holds). But there are important similarities, enough to motivate their being called "generalized quotient rings."



    You need to generalize the notion of a localization to properly characterize flat overrings as quotient-like structures. The appropriate generalization is to take a multiplicatively closed collection of ideals, $mathfrak{I}$, and to define the generalized transform of $R$ with respect to $mathfrak{I}$ as $R_{mathfrak{I}}=bigcup_{Ain mathfrak{I}}A^{-1}$. Here, by $A^{-1}$, I mean the quotient ideal $(R :_K A)$. There's no need to limit our discussion to integral domains.




    Theorem: Given $A subset B subset K$, $B$ is flat over $A$ iff there exists a multiplicatively closed collection of ideals $mathfrak{I}$ in $A$ such that $B = A_{mathfrak{I}}$ and each $I in mathfrak{I}$ generates $B$ as a $B$-module.




    This is Theorem 1.3 in this paper of Arnold and Brewer.



    Akiba gives an example (see beginning of section 2) showing that a flat overring need not be a localization.




    Let C be a non-singular plane cubic curve defined over a field $k_0$
    and let $P$ be a generic point of C over $k_0$, and let $k$ be a field containing
    $k_0(P)$. Then the homogeneous coordinate ring $R_0=k[x, y, z]$
    of $C$ over $k$ is normal.



    Let $A=k[x, y, z]_{(x,y,z)}$, and set $mathfrak{I} = {mathfrak{p}^n mid n in mathbb{N} }$ where $mathfrak{p}$ is the homogeneous prime ideal of $A$ corresponding to $P$. Let $B= A_mathfrak{I}$.




    Then $A$ is a local domain and $B$ is a flat overring of $A$ such that no non-unit of $A$ becomes a unit in $B$. For our purposes, this shows that $B$ is not a localization of $A$. For Akiba's purposes, it showed the stronger statement that $B$ is not a generalized transform of $A$ with respect to any collection of invertible ideals.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      No, flat overrings are not necessarily localizations (although the converse of course holds). But there are important similarities, enough to motivate their being called "generalized quotient rings."



      You need to generalize the notion of a localization to properly characterize flat overrings as quotient-like structures. The appropriate generalization is to take a multiplicatively closed collection of ideals, $mathfrak{I}$, and to define the generalized transform of $R$ with respect to $mathfrak{I}$ as $R_{mathfrak{I}}=bigcup_{Ain mathfrak{I}}A^{-1}$. Here, by $A^{-1}$, I mean the quotient ideal $(R :_K A)$. There's no need to limit our discussion to integral domains.




      Theorem: Given $A subset B subset K$, $B$ is flat over $A$ iff there exists a multiplicatively closed collection of ideals $mathfrak{I}$ in $A$ such that $B = A_{mathfrak{I}}$ and each $I in mathfrak{I}$ generates $B$ as a $B$-module.




      This is Theorem 1.3 in this paper of Arnold and Brewer.



      Akiba gives an example (see beginning of section 2) showing that a flat overring need not be a localization.




      Let C be a non-singular plane cubic curve defined over a field $k_0$
      and let $P$ be a generic point of C over $k_0$, and let $k$ be a field containing
      $k_0(P)$. Then the homogeneous coordinate ring $R_0=k[x, y, z]$
      of $C$ over $k$ is normal.



      Let $A=k[x, y, z]_{(x,y,z)}$, and set $mathfrak{I} = {mathfrak{p}^n mid n in mathbb{N} }$ where $mathfrak{p}$ is the homogeneous prime ideal of $A$ corresponding to $P$. Let $B= A_mathfrak{I}$.




      Then $A$ is a local domain and $B$ is a flat overring of $A$ such that no non-unit of $A$ becomes a unit in $B$. For our purposes, this shows that $B$ is not a localization of $A$. For Akiba's purposes, it showed the stronger statement that $B$ is not a generalized transform of $A$ with respect to any collection of invertible ideals.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        No, flat overrings are not necessarily localizations (although the converse of course holds). But there are important similarities, enough to motivate their being called "generalized quotient rings."



        You need to generalize the notion of a localization to properly characterize flat overrings as quotient-like structures. The appropriate generalization is to take a multiplicatively closed collection of ideals, $mathfrak{I}$, and to define the generalized transform of $R$ with respect to $mathfrak{I}$ as $R_{mathfrak{I}}=bigcup_{Ain mathfrak{I}}A^{-1}$. Here, by $A^{-1}$, I mean the quotient ideal $(R :_K A)$. There's no need to limit our discussion to integral domains.




        Theorem: Given $A subset B subset K$, $B$ is flat over $A$ iff there exists a multiplicatively closed collection of ideals $mathfrak{I}$ in $A$ such that $B = A_{mathfrak{I}}$ and each $I in mathfrak{I}$ generates $B$ as a $B$-module.




        This is Theorem 1.3 in this paper of Arnold and Brewer.



        Akiba gives an example (see beginning of section 2) showing that a flat overring need not be a localization.




        Let C be a non-singular plane cubic curve defined over a field $k_0$
        and let $P$ be a generic point of C over $k_0$, and let $k$ be a field containing
        $k_0(P)$. Then the homogeneous coordinate ring $R_0=k[x, y, z]$
        of $C$ over $k$ is normal.



        Let $A=k[x, y, z]_{(x,y,z)}$, and set $mathfrak{I} = {mathfrak{p}^n mid n in mathbb{N} }$ where $mathfrak{p}$ is the homogeneous prime ideal of $A$ corresponding to $P$. Let $B= A_mathfrak{I}$.




        Then $A$ is a local domain and $B$ is a flat overring of $A$ such that no non-unit of $A$ becomes a unit in $B$. For our purposes, this shows that $B$ is not a localization of $A$. For Akiba's purposes, it showed the stronger statement that $B$ is not a generalized transform of $A$ with respect to any collection of invertible ideals.






        share|cite|improve this answer











        $endgroup$



        No, flat overrings are not necessarily localizations (although the converse of course holds). But there are important similarities, enough to motivate their being called "generalized quotient rings."



        You need to generalize the notion of a localization to properly characterize flat overrings as quotient-like structures. The appropriate generalization is to take a multiplicatively closed collection of ideals, $mathfrak{I}$, and to define the generalized transform of $R$ with respect to $mathfrak{I}$ as $R_{mathfrak{I}}=bigcup_{Ain mathfrak{I}}A^{-1}$. Here, by $A^{-1}$, I mean the quotient ideal $(R :_K A)$. There's no need to limit our discussion to integral domains.




        Theorem: Given $A subset B subset K$, $B$ is flat over $A$ iff there exists a multiplicatively closed collection of ideals $mathfrak{I}$ in $A$ such that $B = A_{mathfrak{I}}$ and each $I in mathfrak{I}$ generates $B$ as a $B$-module.




        This is Theorem 1.3 in this paper of Arnold and Brewer.



        Akiba gives an example (see beginning of section 2) showing that a flat overring need not be a localization.




        Let C be a non-singular plane cubic curve defined over a field $k_0$
        and let $P$ be a generic point of C over $k_0$, and let $k$ be a field containing
        $k_0(P)$. Then the homogeneous coordinate ring $R_0=k[x, y, z]$
        of $C$ over $k$ is normal.



        Let $A=k[x, y, z]_{(x,y,z)}$, and set $mathfrak{I} = {mathfrak{p}^n mid n in mathbb{N} }$ where $mathfrak{p}$ is the homogeneous prime ideal of $A$ corresponding to $P$. Let $B= A_mathfrak{I}$.




        Then $A$ is a local domain and $B$ is a flat overring of $A$ such that no non-unit of $A$ becomes a unit in $B$. For our purposes, this shows that $B$ is not a localization of $A$. For Akiba's purposes, it showed the stronger statement that $B$ is not a generalized transform of $A$ with respect to any collection of invertible ideals.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 17:24

























        answered Nov 30 '18 at 16:47









        Badam BaplanBadam Baplan

        4,476722




        4,476722






























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