Can we find $lambda in mathbb{C}$ such that $deg(gcd(f(t-lambda),g(t-lambda)))=1$?












0












$begingroup$


Let $f=f(t),g=g(t) in mathbb{C}[t]$ with $deg(f) geq 2$ and $deg(g) geq 2$.




Can we find $lambda in mathbb{C}$ such that $gcd(f(t-lambda),g(t-lambda))$ is of degree $1$?




Please notice that a somewhat similar question to the above question has a positive answer, namely, given $f$ and $g$ of degrees $geq 2$, there exists $a,b in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)$ is of degree $1$.



Thank you very much!






algebraic-geometry polynomials commutative-algebra







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $f=f(t),g=g(t) in mathbb{C}[t]$ with $deg(f) geq 2$ and $deg(g) geq 2$.




    Can we find $lambda in mathbb{C}$ such that $gcd(f(t-lambda),g(t-lambda))$ is of degree $1$?




    Please notice that a somewhat similar question to the above question has a positive answer, namely, given $f$ and $g$ of degrees $geq 2$, there exists $a,b in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)$ is of degree $1$.



    Thank you very much!






    algebraic-geometry polynomials commutative-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f=f(t),g=g(t) in mathbb{C}[t]$ with $deg(f) geq 2$ and $deg(g) geq 2$.




      Can we find $lambda in mathbb{C}$ such that $gcd(f(t-lambda),g(t-lambda))$ is of degree $1$?




      Please notice that a somewhat similar question to the above question has a positive answer, namely, given $f$ and $g$ of degrees $geq 2$, there exists $a,b in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)$ is of degree $1$.



      Thank you very much!






      algebraic-geometry polynomials commutative-algebra







      share|cite|improve this question









      $endgroup$




      Let $f=f(t),g=g(t) in mathbb{C}[t]$ with $deg(f) geq 2$ and $deg(g) geq 2$.




      Can we find $lambda in mathbb{C}$ such that $gcd(f(t-lambda),g(t-lambda))$ is of degree $1$?




      Please notice that a somewhat similar question to the above question has a positive answer, namely, given $f$ and $g$ of degrees $geq 2$, there exists $a,b in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)$ is of degree $1$.



      Thank you very much!






      algebraic-geometry polynomials commutative-algebra




      algebraic-geometry polynomials commutative-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 17 '18 at 0:31









      user237522user237522

      2,1671617




      2,1671617






















          1 Answer
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          $begingroup$

          Not really, because if $u=operatorname{gcd}(f,g)$, then $operatorname{gcd}(f(t-lambda),g(t-lambda))=u(t-lambda)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! I see that my question is too trivial..
            $endgroup$
            – user237522
            Dec 17 '18 at 0:40











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Not really, because if $u=operatorname{gcd}(f,g)$, then $operatorname{gcd}(f(t-lambda),g(t-lambda))=u(t-lambda)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! I see that my question is too trivial..
            $endgroup$
            – user237522
            Dec 17 '18 at 0:40
















          2












          $begingroup$

          Not really, because if $u=operatorname{gcd}(f,g)$, then $operatorname{gcd}(f(t-lambda),g(t-lambda))=u(t-lambda)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! I see that my question is too trivial..
            $endgroup$
            – user237522
            Dec 17 '18 at 0:40














          2












          2








          2





          $begingroup$

          Not really, because if $u=operatorname{gcd}(f,g)$, then $operatorname{gcd}(f(t-lambda),g(t-lambda))=u(t-lambda)$.






          share|cite|improve this answer









          $endgroup$



          Not really, because if $u=operatorname{gcd}(f,g)$, then $operatorname{gcd}(f(t-lambda),g(t-lambda))=u(t-lambda)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 0:34









          Saucy O'PathSaucy O'Path

          5,9691627




          5,9691627












          • $begingroup$
            Thank you! I see that my question is too trivial..
            $endgroup$
            – user237522
            Dec 17 '18 at 0:40


















          • $begingroup$
            Thank you! I see that my question is too trivial..
            $endgroup$
            – user237522
            Dec 17 '18 at 0:40
















          $begingroup$
          Thank you! I see that my question is too trivial..
          $endgroup$
          – user237522
          Dec 17 '18 at 0:40




          $begingroup$
          Thank you! I see that my question is too trivial..
          $endgroup$
          – user237522
          Dec 17 '18 at 0:40


















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