When $langle F(x,y)y+u(x)(x-c), G(x,y)y+v(x)(x-c) rangle$ is a maximal ideal of $mathbb{C}[x,y]$?












0












$begingroup$


Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.



Denote:



$A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,



$B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.




When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?




It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.



Remarks:
1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.




  1. What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?


Any hints and comments are welcome!






algebraic-geometry ring-theory commutative-algebra ideals







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.



    Denote:



    $A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,



    $B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.




    When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?




    It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
    Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.



    Remarks:
    1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.




    1. What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?


    Any hints and comments are welcome!






    algebraic-geometry ring-theory commutative-algebra ideals







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.



      Denote:



      $A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,



      $B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.




      When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?




      It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
      Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.



      Remarks:
      1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.




      1. What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?


      Any hints and comments are welcome!






      algebraic-geometry ring-theory commutative-algebra ideals







      share|cite|improve this question











      $endgroup$




      Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.



      Denote:



      $A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,



      $B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.




      When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?




      It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
      Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.



      Remarks:
      1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.




      1. What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?


      Any hints and comments are welcome!






      algebraic-geometry ring-theory commutative-algebra ideals




      algebraic-geometry ring-theory commutative-algebra ideals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 19:40







      user237522

















      asked Dec 12 '18 at 19:34









      user237522user237522

      2,1601617




      2,1601617






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          This is true if and only if $Fv-Guin Bbb{C}$.



          To see why, let
          $$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
          & vemat.$$
          Then $$bmat A \ B emat = M bmat y \ x-c emat.$$



          Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
          Then the matrix
          $$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
          $$ Ebmat A \ B emat = bmat y \ x-c emat.$$
          Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$



          Expanding this gives
          $$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$



          Then
          $$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
          which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
          $e_{11}F+e_{12}G=1$.
          Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
          tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.



          Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.



          Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
          $e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.



          Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much! Your answer is very nice.
            $endgroup$
            – user237522
            Dec 12 '18 at 21:44










          • $begingroup$
            Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
            $endgroup$
            – user237522
            Dec 13 '18 at 14:38













          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%2f3037127%2fwhen-langle-fx-yyuxx-c-gx-yyvxx-c-rangle-is-a-maximal-ideal-o%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









          1












          $begingroup$

          This is true if and only if $Fv-Guin Bbb{C}$.



          To see why, let
          $$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
          & vemat.$$
          Then $$bmat A \ B emat = M bmat y \ x-c emat.$$



          Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
          Then the matrix
          $$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
          $$ Ebmat A \ B emat = bmat y \ x-c emat.$$
          Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$



          Expanding this gives
          $$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$



          Then
          $$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
          which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
          $e_{11}F+e_{12}G=1$.
          Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
          tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.



          Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.



          Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
          $e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.



          Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much! Your answer is very nice.
            $endgroup$
            – user237522
            Dec 12 '18 at 21:44










          • $begingroup$
            Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
            $endgroup$
            – user237522
            Dec 13 '18 at 14:38


















          1












          $begingroup$

          This is true if and only if $Fv-Guin Bbb{C}$.



          To see why, let
          $$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
          & vemat.$$
          Then $$bmat A \ B emat = M bmat y \ x-c emat.$$



          Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
          Then the matrix
          $$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
          $$ Ebmat A \ B emat = bmat y \ x-c emat.$$
          Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$



          Expanding this gives
          $$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$



          Then
          $$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
          which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
          $e_{11}F+e_{12}G=1$.
          Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
          tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.



          Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.



          Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
          $e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.



          Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much! Your answer is very nice.
            $endgroup$
            – user237522
            Dec 12 '18 at 21:44










          • $begingroup$
            Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
            $endgroup$
            – user237522
            Dec 13 '18 at 14:38
















          1












          1








          1





          $begingroup$

          This is true if and only if $Fv-Guin Bbb{C}$.



          To see why, let
          $$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
          & vemat.$$
          Then $$bmat A \ B emat = M bmat y \ x-c emat.$$



          Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
          Then the matrix
          $$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
          $$ Ebmat A \ B emat = bmat y \ x-c emat.$$
          Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$



          Expanding this gives
          $$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$



          Then
          $$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
          which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
          $e_{11}F+e_{12}G=1$.
          Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
          tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.



          Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.



          Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
          $e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.



          Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.






          share|cite|improve this answer









          $endgroup$



          This is true if and only if $Fv-Guin Bbb{C}$.



          To see why, let
          $$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
          & vemat.$$
          Then $$bmat A \ B emat = M bmat y \ x-c emat.$$



          Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
          Then the matrix
          $$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
          $$ Ebmat A \ B emat = bmat y \ x-c emat.$$
          Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$



          Expanding this gives
          $$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$



          Then
          $$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
          which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
          $e_{11}F+e_{12}G=1$.
          Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
          tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.



          Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.



          Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
          $e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.



          Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 19:58









          jgonjgon

          14.8k22042




          14.8k22042












          • $begingroup$
            Thank you very much! Your answer is very nice.
            $endgroup$
            – user237522
            Dec 12 '18 at 21:44










          • $begingroup$
            Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
            $endgroup$
            – user237522
            Dec 13 '18 at 14:38




















          • $begingroup$
            Thank you very much! Your answer is very nice.
            $endgroup$
            – user237522
            Dec 12 '18 at 21:44










          • $begingroup$
            Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
            $endgroup$
            – user237522
            Dec 13 '18 at 14:38


















          $begingroup$
          Thank you very much! Your answer is very nice.
          $endgroup$
          – user237522
          Dec 12 '18 at 21:44




          $begingroup$
          Thank you very much! Your answer is very nice.
          $endgroup$
          – user237522
          Dec 12 '18 at 21:44












          $begingroup$
          Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
          $endgroup$
          – user237522
          Dec 13 '18 at 14:38






          $begingroup$
          Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
          $endgroup$
          – user237522
          Dec 13 '18 at 14:38




















          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%2f3037127%2fwhen-langle-fx-yyuxx-c-gx-yyvxx-c-rangle-is-a-maximal-ideal-o%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