$Asubset Bsubset K$, $K$ 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.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020038%2fa-subset-b-subset-k-k-field-of-fractions-of-a-b-flat-over-a-implie%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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.






    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020038%2fa-subset-b-subset-k-k-field-of-fractions-of-a-b-flat-over-a-implie%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more