Why $mathbb{C}(f(t),g(t))=mathbb{C}(t)$ implies that $gcd(f(t)-a,g(t)-b)=t-c$, for some $a,b,c in mathbb{C}$?












5












$begingroup$


Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:



(1) $deg(f) geq 2$ and $deg(g) geq 2$.



(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.




In this question it was mentioned that in that case, there exist
$a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.




Unfortunately, I do not see why this is true.



Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).



Edit: Just to make sure:




Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?




Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.



Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



W.l.o.g., there exist finitely many $a in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.



This is impossible from the following reason:
Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.



So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



Any hints are welcome!






polynomials commutative-algebra field-theory greatest-common-divisor







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:



    (1) $deg(f) geq 2$ and $deg(g) geq 2$.



    (2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.




    In this question it was mentioned that in that case, there exist
    $a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.




    Unfortunately, I do not see why this is true.



    Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).



    Edit: Just to make sure:




    Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
    $gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?




    Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.



    Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
    such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



    W.l.o.g., there exist finitely many $a in mathbb{C}$
    such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



    By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.



    This is impossible from the following reason:
    Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.



    So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
    $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



    Any hints are welcome!






    polynomials commutative-algebra field-theory greatest-common-divisor







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      2



      $begingroup$


      Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:



      (1) $deg(f) geq 2$ and $deg(g) geq 2$.



      (2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.




      In this question it was mentioned that in that case, there exist
      $a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.




      Unfortunately, I do not see why this is true.



      Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).



      Edit: Just to make sure:




      Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
      $gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?




      Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.



      Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
      such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



      W.l.o.g., there exist finitely many $a in mathbb{C}$
      such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



      By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.



      This is impossible from the following reason:
      Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.



      So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
      $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



      Any hints are welcome!






      polynomials commutative-algebra field-theory greatest-common-divisor







      share|cite|improve this question











      $endgroup$




      Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:



      (1) $deg(f) geq 2$ and $deg(g) geq 2$.



      (2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.




      In this question it was mentioned that in that case, there exist
      $a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.




      Unfortunately, I do not see why this is true.



      Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).



      Edit: Just to make sure:




      Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
      $gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?




      Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.



      Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
      such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



      W.l.o.g., there exist finitely many $a in mathbb{C}$
      such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



      By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.



      This is impossible from the following reason:
      Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.



      So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
      $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.



      Any hints are welcome!






      polynomials commutative-algebra field-theory greatest-common-divisor




      polynomials commutative-algebra field-theory greatest-common-divisor






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 24 '18 at 10:26







      user237522

















      asked Dec 15 '18 at 22:08









      user237522user237522

      2,1631617




      2,1631617






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          We may assume that $f$ and $g$ are monic.
          There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).



          Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.



          Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:51












          • $begingroup$
            I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:54










          • $begingroup$
            Thanks; please, could you slightly elaborate your answer?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:55












          • $begingroup$
            I edited. Is there enough detail?
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:59










          • $begingroup$
            Thank you. I think I now understand your answer, very nice!
            $endgroup$
            – user237522
            Dec 15 '18 at 23:07











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          We may assume that $f$ and $g$ are monic.
          There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).



          Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.



          Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:51












          • $begingroup$
            I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:54










          • $begingroup$
            Thanks; please, could you slightly elaborate your answer?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:55












          • $begingroup$
            I edited. Is there enough detail?
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:59










          • $begingroup$
            Thank you. I think I now understand your answer, very nice!
            $endgroup$
            – user237522
            Dec 15 '18 at 23:07
















          4












          $begingroup$

          We may assume that $f$ and $g$ are monic.
          There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).



          Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.



          Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:51












          • $begingroup$
            I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:54










          • $begingroup$
            Thanks; please, could you slightly elaborate your answer?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:55












          • $begingroup$
            I edited. Is there enough detail?
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:59










          • $begingroup$
            Thank you. I think I now understand your answer, very nice!
            $endgroup$
            – user237522
            Dec 15 '18 at 23:07














          4












          4








          4





          $begingroup$

          We may assume that $f$ and $g$ are monic.
          There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).



          Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.



          Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.






          share|cite|improve this answer











          $endgroup$



          We may assume that $f$ and $g$ are monic.
          There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).



          Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.



          Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 22:59

























          answered Dec 15 '18 at 22:47









          MindlackMindlack

          4,885211




          4,885211












          • $begingroup$
            Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:51












          • $begingroup$
            I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:54










          • $begingroup$
            Thanks; please, could you slightly elaborate your answer?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:55












          • $begingroup$
            I edited. Is there enough detail?
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:59










          • $begingroup$
            Thank you. I think I now understand your answer, very nice!
            $endgroup$
            – user237522
            Dec 15 '18 at 23:07


















          • $begingroup$
            Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:51












          • $begingroup$
            I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:54










          • $begingroup$
            Thanks; please, could you slightly elaborate your answer?
            $endgroup$
            – user237522
            Dec 15 '18 at 22:55












          • $begingroup$
            I edited. Is there enough detail?
            $endgroup$
            – Mindlack
            Dec 15 '18 at 22:59










          • $begingroup$
            Thank you. I think I now understand your answer, very nice!
            $endgroup$
            – user237522
            Dec 15 '18 at 23:07
















          $begingroup$
          Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
          $endgroup$
          – user237522
          Dec 15 '18 at 22:51






          $begingroup$
          Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
          $endgroup$
          – user237522
          Dec 15 '18 at 22:51














          $begingroup$
          I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
          $endgroup$
          – Mindlack
          Dec 15 '18 at 22:54




          $begingroup$
          I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
          $endgroup$
          – Mindlack
          Dec 15 '18 at 22:54












          $begingroup$
          Thanks; please, could you slightly elaborate your answer?
          $endgroup$
          – user237522
          Dec 15 '18 at 22:55






          $begingroup$
          Thanks; please, could you slightly elaborate your answer?
          $endgroup$
          – user237522
          Dec 15 '18 at 22:55














          $begingroup$
          I edited. Is there enough detail?
          $endgroup$
          – Mindlack
          Dec 15 '18 at 22:59




          $begingroup$
          I edited. Is there enough detail?
          $endgroup$
          – Mindlack
          Dec 15 '18 at 22:59












          $begingroup$
          Thank you. I think I now understand your answer, very nice!
          $endgroup$
          – user237522
          Dec 15 '18 at 23:07




          $begingroup$
          Thank you. I think I now understand your answer, very nice!
          $endgroup$
          – user237522
          Dec 15 '18 at 23:07


















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          Puebla de Zaragoza

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          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
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          Musa

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          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
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