Efficiently computing GCDs in $mathbb{Z}[(1+sqrt{-19})/2]$












8












$begingroup$


The ring $mathbb{Z}[(1+sqrt{-19})/2]$ is a PID; hence any two elements have a GCD. How you would compute their GCD?



In a Euclidean domain, you would use the Euclidean algorithm. But $mathbb{Z}[(1+sqrt{-19})/2]$ is not Euclidean.



If you knew prime factorizations of the two elements, you could immediately compute the GCD. But surely factoring in $mathbb{Z}[(1+sqrt{-19})/2]$ is at least as hard as factoring in $mathbb{Z}$. At least for now, we do not know how to do this efficiently.



So, is there an efficient algorithm to compute GCD?



(You may use the de facto interpretation of "efficient" as "polynomial time".)






abstract-algebra ring-theory algorithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is possible to use the Dedekind-Hasse criterion for devising an algorithm; see arxiv.org/pdf/1205.1147.pdf
    $endgroup$
    – franz lemmermeyer
    Feb 4 at 15:28


















8












$begingroup$


The ring $mathbb{Z}[(1+sqrt{-19})/2]$ is a PID; hence any two elements have a GCD. How you would compute their GCD?



In a Euclidean domain, you would use the Euclidean algorithm. But $mathbb{Z}[(1+sqrt{-19})/2]$ is not Euclidean.



If you knew prime factorizations of the two elements, you could immediately compute the GCD. But surely factoring in $mathbb{Z}[(1+sqrt{-19})/2]$ is at least as hard as factoring in $mathbb{Z}$. At least for now, we do not know how to do this efficiently.



So, is there an efficient algorithm to compute GCD?



(You may use the de facto interpretation of "efficient" as "polynomial time".)






abstract-algebra ring-theory algorithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is possible to use the Dedekind-Hasse criterion for devising an algorithm; see arxiv.org/pdf/1205.1147.pdf
    $endgroup$
    – franz lemmermeyer
    Feb 4 at 15:28
















8












8








8





$begingroup$


The ring $mathbb{Z}[(1+sqrt{-19})/2]$ is a PID; hence any two elements have a GCD. How you would compute their GCD?



In a Euclidean domain, you would use the Euclidean algorithm. But $mathbb{Z}[(1+sqrt{-19})/2]$ is not Euclidean.



If you knew prime factorizations of the two elements, you could immediately compute the GCD. But surely factoring in $mathbb{Z}[(1+sqrt{-19})/2]$ is at least as hard as factoring in $mathbb{Z}$. At least for now, we do not know how to do this efficiently.



So, is there an efficient algorithm to compute GCD?



(You may use the de facto interpretation of "efficient" as "polynomial time".)






abstract-algebra ring-theory algorithms







share|cite|improve this question









$endgroup$




The ring $mathbb{Z}[(1+sqrt{-19})/2]$ is a PID; hence any two elements have a GCD. How you would compute their GCD?



In a Euclidean domain, you would use the Euclidean algorithm. But $mathbb{Z}[(1+sqrt{-19})/2]$ is not Euclidean.



If you knew prime factorizations of the two elements, you could immediately compute the GCD. But surely factoring in $mathbb{Z}[(1+sqrt{-19})/2]$ is at least as hard as factoring in $mathbb{Z}$. At least for now, we do not know how to do this efficiently.



So, is there an efficient algorithm to compute GCD?



(You may use the de facto interpretation of "efficient" as "polynomial time".)






abstract-algebra ring-theory algorithms




abstract-algebra ring-theory algorithms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 7 '15 at 3:35









echinodermataechinodermata

2,52911228




2,52911228












  • $begingroup$
    It is possible to use the Dedekind-Hasse criterion for devising an algorithm; see arxiv.org/pdf/1205.1147.pdf
    $endgroup$
    – franz lemmermeyer
    Feb 4 at 15:28




















  • $begingroup$
    It is possible to use the Dedekind-Hasse criterion for devising an algorithm; see arxiv.org/pdf/1205.1147.pdf
    $endgroup$
    – franz lemmermeyer
    Feb 4 at 15:28


















$begingroup$
It is possible to use the Dedekind-Hasse criterion for devising an algorithm; see arxiv.org/pdf/1205.1147.pdf
$endgroup$
– franz lemmermeyer
Feb 4 at 15:28






$begingroup$
It is possible to use the Dedekind-Hasse criterion for devising an algorithm; see arxiv.org/pdf/1205.1147.pdf
$endgroup$
– franz lemmermeyer
Feb 4 at 15:28












2 Answers
2






active

oldest

votes


















4












$begingroup$

Not a complete answer, just trying to draw attention to your question.



Humans such as yourself generally find factorization in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ to be more difficult than factorization in $mathbb{Z}$ because $mathbb{Z}$ is familiar and $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ is unfamiliar.



So if I was bedraggled by human failings, I would see about how to leverage my familiarity with one to my advantage in the other. The way is with norms.



If the norms of two numbers in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ are coprime, then the two numbers must be coprime as well. Example: $$Nleft( frac{3 + sqrt{-19}}{2} right) = 7$$ and $$Nleft( frac{9 + sqrt{-19}}{2} right) = 25,$$ so since $gcd(7, 25) = 1$, this must mean that $$gcdleft( frac{3 + sqrt{-19}}{2}, frac{9 + sqrt{-19}}{2} right) = 1$$ as well.



But if the norms do have a common divisor, then it's a little more difficult, for humans anyway. Example: $$gcdleft( frac{3 + sqrt{-19}}{2}, 11 + sqrt{-19} right) = ???$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The last time I wanted to do this, I used the $O(n^2)$ algorithm described in Agarwal and Frandsen. I think I recall that your specific case is easier because $-19 < 0$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Not a complete answer, just trying to draw attention to your question.



      Humans such as yourself generally find factorization in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ to be more difficult than factorization in $mathbb{Z}$ because $mathbb{Z}$ is familiar and $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ is unfamiliar.



      So if I was bedraggled by human failings, I would see about how to leverage my familiarity with one to my advantage in the other. The way is with norms.



      If the norms of two numbers in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ are coprime, then the two numbers must be coprime as well. Example: $$Nleft( frac{3 + sqrt{-19}}{2} right) = 7$$ and $$Nleft( frac{9 + sqrt{-19}}{2} right) = 25,$$ so since $gcd(7, 25) = 1$, this must mean that $$gcdleft( frac{3 + sqrt{-19}}{2}, frac{9 + sqrt{-19}}{2} right) = 1$$ as well.



      But if the norms do have a common divisor, then it's a little more difficult, for humans anyway. Example: $$gcdleft( frac{3 + sqrt{-19}}{2}, 11 + sqrt{-19} right) = ???$$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Not a complete answer, just trying to draw attention to your question.



        Humans such as yourself generally find factorization in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ to be more difficult than factorization in $mathbb{Z}$ because $mathbb{Z}$ is familiar and $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ is unfamiliar.



        So if I was bedraggled by human failings, I would see about how to leverage my familiarity with one to my advantage in the other. The way is with norms.



        If the norms of two numbers in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ are coprime, then the two numbers must be coprime as well. Example: $$Nleft( frac{3 + sqrt{-19}}{2} right) = 7$$ and $$Nleft( frac{9 + sqrt{-19}}{2} right) = 25,$$ so since $gcd(7, 25) = 1$, this must mean that $$gcdleft( frac{3 + sqrt{-19}}{2}, frac{9 + sqrt{-19}}{2} right) = 1$$ as well.



        But if the norms do have a common divisor, then it's a little more difficult, for humans anyway. Example: $$gcdleft( frac{3 + sqrt{-19}}{2}, 11 + sqrt{-19} right) = ???$$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Not a complete answer, just trying to draw attention to your question.



          Humans such as yourself generally find factorization in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ to be more difficult than factorization in $mathbb{Z}$ because $mathbb{Z}$ is familiar and $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ is unfamiliar.



          So if I was bedraggled by human failings, I would see about how to leverage my familiarity with one to my advantage in the other. The way is with norms.



          If the norms of two numbers in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ are coprime, then the two numbers must be coprime as well. Example: $$Nleft( frac{3 + sqrt{-19}}{2} right) = 7$$ and $$Nleft( frac{9 + sqrt{-19}}{2} right) = 25,$$ so since $gcd(7, 25) = 1$, this must mean that $$gcdleft( frac{3 + sqrt{-19}}{2}, frac{9 + sqrt{-19}}{2} right) = 1$$ as well.



          But if the norms do have a common divisor, then it's a little more difficult, for humans anyway. Example: $$gcdleft( frac{3 + sqrt{-19}}{2}, 11 + sqrt{-19} right) = ???$$






          share|cite|improve this answer









          $endgroup$



          Not a complete answer, just trying to draw attention to your question.



          Humans such as yourself generally find factorization in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ to be more difficult than factorization in $mathbb{Z}$ because $mathbb{Z}$ is familiar and $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ is unfamiliar.



          So if I was bedraggled by human failings, I would see about how to leverage my familiarity with one to my advantage in the other. The way is with norms.



          If the norms of two numbers in $mathbb{Z}left[frac{1 + sqrt{-19}}{2}right]$ are coprime, then the two numbers must be coprime as well. Example: $$Nleft( frac{3 + sqrt{-19}}{2} right) = 7$$ and $$Nleft( frac{9 + sqrt{-19}}{2} right) = 25,$$ so since $gcd(7, 25) = 1$, this must mean that $$gcdleft( frac{3 + sqrt{-19}}{2}, frac{9 + sqrt{-19}}{2} right) = 1$$ as well.



          But if the norms do have a common divisor, then it's a little more difficult, for humans anyway. Example: $$gcdleft( frac{3 + sqrt{-19}}{2}, 11 + sqrt{-19} right) = ???$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 23:26









          The Short OneThe Short One

          5941625




          5941625























              1












              $begingroup$

              The last time I wanted to do this, I used the $O(n^2)$ algorithm described in Agarwal and Frandsen. I think I recall that your specific case is easier because $-19 < 0$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The last time I wanted to do this, I used the $O(n^2)$ algorithm described in Agarwal and Frandsen. I think I recall that your specific case is easier because $-19 < 0$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The last time I wanted to do this, I used the $O(n^2)$ algorithm described in Agarwal and Frandsen. I think I recall that your specific case is easier because $-19 < 0$.






                  share|cite|improve this answer









                  $endgroup$



                  The last time I wanted to do this, I used the $O(n^2)$ algorithm described in Agarwal and Frandsen. I think I recall that your specific case is easier because $-19 < 0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 0:59









                  Eric TowersEric Towers

                  33k22370




                  33k22370






























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