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







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$endgroup$

















    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

      2,1691617




      2,1691617






















          1 Answer
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          $begingroup$

          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













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          $begingroup$

          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

          14.2k22042




          14.2k22042












          • $begingroup$
            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




















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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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