Good way to check if an ideal is principal in cubic extensions












2












$begingroup$


I've recently been tacking the computation of class groups. I've noticed that when dealing with a ramified prime ideal in a quadratic field $mathbb{Q}(sqrt{d}), d in mathbb{Z}$ squarefree, it's very easy to check to see if an ideal is principle, due to being able to make norm arguments $N(a+bsqrt{d}) = a^2 - db^2$. However, in cubic fields this intuition breaks down since I don't have a norm; I was wondering if you could recommend a good plan of attack in these extensions.



Take for example: Let $K = mathbb{Q}(eta), eta^3 = 6$. We know that $Disc(K) = -27*6^2 = 972$, which implies that Minkowski's bound is about $M_k = 8.82$. Now we examine primes




  • Suppose $p = 2$


Then notice that $x^3 - 6 equiv_2 x^3 implies 2mathcal{O}_K = (2, eta)^3 = mathfrak{p}_2^3$.




  • Suppose $p = 3$.


Then again we have $3 mathcal{O}_K = (3, eta)^3 = mathfrak{p}_3^3$.




  • Suppose $p = 5$.


Then we have $x^3 - 6 equiv x^3 + 4 equiv (x-1)(x^2+x+1)$. This implies that $5mathcal{O}_K = (5, eta - 1)(5, eta^2 + eta + 1) = mathfrak{p}_5mathfrak{q}_5$.




  • Suppose $p = 7$.


Then we have $x^3 - 6 equiv x^3 + 1 equiv (x-3)(x-5)(x-6)$. Therefore we can write $7 mathcal{O}_K = (7, eta - 3)(7, eta - 5)(7, eta - 6) = mathfrak{p}_7 mathfrak{q}_7 mathfrak{q}_7'$.



Now that i've done this I want to show each of those individual prime ideals are principle, since I know that this field actually has class number 1.



For example, I'm confused on how to see that $(2, eta) = (eta - 2)$. I mean it's easy to see once I know to try $eta - 2$, but how could I come up with that value besides guess and check. Even more so with $(3, eta) = (-eta^2 - 2eta - 3)$.






abstract-algebra algebraic-number-theory ideal-class-group







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I've recently been tacking the computation of class groups. I've noticed that when dealing with a ramified prime ideal in a quadratic field $mathbb{Q}(sqrt{d}), d in mathbb{Z}$ squarefree, it's very easy to check to see if an ideal is principle, due to being able to make norm arguments $N(a+bsqrt{d}) = a^2 - db^2$. However, in cubic fields this intuition breaks down since I don't have a norm; I was wondering if you could recommend a good plan of attack in these extensions.



    Take for example: Let $K = mathbb{Q}(eta), eta^3 = 6$. We know that $Disc(K) = -27*6^2 = 972$, which implies that Minkowski's bound is about $M_k = 8.82$. Now we examine primes




    • Suppose $p = 2$


    Then notice that $x^3 - 6 equiv_2 x^3 implies 2mathcal{O}_K = (2, eta)^3 = mathfrak{p}_2^3$.




    • Suppose $p = 3$.


    Then again we have $3 mathcal{O}_K = (3, eta)^3 = mathfrak{p}_3^3$.




    • Suppose $p = 5$.


    Then we have $x^3 - 6 equiv x^3 + 4 equiv (x-1)(x^2+x+1)$. This implies that $5mathcal{O}_K = (5, eta - 1)(5, eta^2 + eta + 1) = mathfrak{p}_5mathfrak{q}_5$.




    • Suppose $p = 7$.


    Then we have $x^3 - 6 equiv x^3 + 1 equiv (x-3)(x-5)(x-6)$. Therefore we can write $7 mathcal{O}_K = (7, eta - 3)(7, eta - 5)(7, eta - 6) = mathfrak{p}_7 mathfrak{q}_7 mathfrak{q}_7'$.



    Now that i've done this I want to show each of those individual prime ideals are principle, since I know that this field actually has class number 1.



    For example, I'm confused on how to see that $(2, eta) = (eta - 2)$. I mean it's easy to see once I know to try $eta - 2$, but how could I come up with that value besides guess and check. Even more so with $(3, eta) = (-eta^2 - 2eta - 3)$.






    abstract-algebra algebraic-number-theory ideal-class-group







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I've recently been tacking the computation of class groups. I've noticed that when dealing with a ramified prime ideal in a quadratic field $mathbb{Q}(sqrt{d}), d in mathbb{Z}$ squarefree, it's very easy to check to see if an ideal is principle, due to being able to make norm arguments $N(a+bsqrt{d}) = a^2 - db^2$. However, in cubic fields this intuition breaks down since I don't have a norm; I was wondering if you could recommend a good plan of attack in these extensions.



      Take for example: Let $K = mathbb{Q}(eta), eta^3 = 6$. We know that $Disc(K) = -27*6^2 = 972$, which implies that Minkowski's bound is about $M_k = 8.82$. Now we examine primes




      • Suppose $p = 2$


      Then notice that $x^3 - 6 equiv_2 x^3 implies 2mathcal{O}_K = (2, eta)^3 = mathfrak{p}_2^3$.




      • Suppose $p = 3$.


      Then again we have $3 mathcal{O}_K = (3, eta)^3 = mathfrak{p}_3^3$.




      • Suppose $p = 5$.


      Then we have $x^3 - 6 equiv x^3 + 4 equiv (x-1)(x^2+x+1)$. This implies that $5mathcal{O}_K = (5, eta - 1)(5, eta^2 + eta + 1) = mathfrak{p}_5mathfrak{q}_5$.




      • Suppose $p = 7$.


      Then we have $x^3 - 6 equiv x^3 + 1 equiv (x-3)(x-5)(x-6)$. Therefore we can write $7 mathcal{O}_K = (7, eta - 3)(7, eta - 5)(7, eta - 6) = mathfrak{p}_7 mathfrak{q}_7 mathfrak{q}_7'$.



      Now that i've done this I want to show each of those individual prime ideals are principle, since I know that this field actually has class number 1.



      For example, I'm confused on how to see that $(2, eta) = (eta - 2)$. I mean it's easy to see once I know to try $eta - 2$, but how could I come up with that value besides guess and check. Even more so with $(3, eta) = (-eta^2 - 2eta - 3)$.






      abstract-algebra algebraic-number-theory ideal-class-group







      share|cite|improve this question











      $endgroup$




      I've recently been tacking the computation of class groups. I've noticed that when dealing with a ramified prime ideal in a quadratic field $mathbb{Q}(sqrt{d}), d in mathbb{Z}$ squarefree, it's very easy to check to see if an ideal is principle, due to being able to make norm arguments $N(a+bsqrt{d}) = a^2 - db^2$. However, in cubic fields this intuition breaks down since I don't have a norm; I was wondering if you could recommend a good plan of attack in these extensions.



      Take for example: Let $K = mathbb{Q}(eta), eta^3 = 6$. We know that $Disc(K) = -27*6^2 = 972$, which implies that Minkowski's bound is about $M_k = 8.82$. Now we examine primes




      • Suppose $p = 2$


      Then notice that $x^3 - 6 equiv_2 x^3 implies 2mathcal{O}_K = (2, eta)^3 = mathfrak{p}_2^3$.




      • Suppose $p = 3$.


      Then again we have $3 mathcal{O}_K = (3, eta)^3 = mathfrak{p}_3^3$.




      • Suppose $p = 5$.


      Then we have $x^3 - 6 equiv x^3 + 4 equiv (x-1)(x^2+x+1)$. This implies that $5mathcal{O}_K = (5, eta - 1)(5, eta^2 + eta + 1) = mathfrak{p}_5mathfrak{q}_5$.




      • Suppose $p = 7$.


      Then we have $x^3 - 6 equiv x^3 + 1 equiv (x-3)(x-5)(x-6)$. Therefore we can write $7 mathcal{O}_K = (7, eta - 3)(7, eta - 5)(7, eta - 6) = mathfrak{p}_7 mathfrak{q}_7 mathfrak{q}_7'$.



      Now that i've done this I want to show each of those individual prime ideals are principle, since I know that this field actually has class number 1.



      For example, I'm confused on how to see that $(2, eta) = (eta - 2)$. I mean it's easy to see once I know to try $eta - 2$, but how could I come up with that value besides guess and check. Even more so with $(3, eta) = (-eta^2 - 2eta - 3)$.






      abstract-algebra algebraic-number-theory ideal-class-group




      abstract-algebra algebraic-number-theory ideal-class-group






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 19:14







      TrostAft

















      asked Dec 13 '18 at 16:18









      TrostAftTrostAft

      440412




      440412






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          I figured out the answer.



          The problem was that I was too hasty in declaring that in cubic extensions, the ring of integers does not have a norm. In fact, I noticed that the norm in quadratic extensions was actually the field norm of the element $a + bsqrt{d}$.



          $$
          N_{mathcal{O}_K/mathbb{Z}}(a+bsqrt{d}) = det begin{pmatrix}
          a & bd \
          b & a
          end{pmatrix} = a^2 - db^2
          $$



          So then when we consider this cubic extension, and repeat this process (let $eta$ be such that $eta^3 = 6$.)



          $$
          N_{mathcal{O}_K/mathbb{Z}}(a+beta + ceta^2) = det begin{pmatrix}
          a & 6c & 6b \
          b & a & 6c \
          c & b & a
          end{pmatrix} = a^3 + 6b^3 + 36c^3 - 18abc
          $$



          Then using this we can perform norm arguments like in the quadratic scenario. For example, notice that we can construct an element with norm $2$ by choose $a = 2, b = -1 implies 2 - eta$ is a element of norm $2$, which answers my question.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            I figured out the answer.



            The problem was that I was too hasty in declaring that in cubic extensions, the ring of integers does not have a norm. In fact, I noticed that the norm in quadratic extensions was actually the field norm of the element $a + bsqrt{d}$.



            $$
            N_{mathcal{O}_K/mathbb{Z}}(a+bsqrt{d}) = det begin{pmatrix}
            a & bd \
            b & a
            end{pmatrix} = a^2 - db^2
            $$



            So then when we consider this cubic extension, and repeat this process (let $eta$ be such that $eta^3 = 6$.)



            $$
            N_{mathcal{O}_K/mathbb{Z}}(a+beta + ceta^2) = det begin{pmatrix}
            a & 6c & 6b \
            b & a & 6c \
            c & b & a
            end{pmatrix} = a^3 + 6b^3 + 36c^3 - 18abc
            $$



            Then using this we can perform norm arguments like in the quadratic scenario. For example, notice that we can construct an element with norm $2$ by choose $a = 2, b = -1 implies 2 - eta$ is a element of norm $2$, which answers my question.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              I figured out the answer.



              The problem was that I was too hasty in declaring that in cubic extensions, the ring of integers does not have a norm. In fact, I noticed that the norm in quadratic extensions was actually the field norm of the element $a + bsqrt{d}$.



              $$
              N_{mathcal{O}_K/mathbb{Z}}(a+bsqrt{d}) = det begin{pmatrix}
              a & bd \
              b & a
              end{pmatrix} = a^2 - db^2
              $$



              So then when we consider this cubic extension, and repeat this process (let $eta$ be such that $eta^3 = 6$.)



              $$
              N_{mathcal{O}_K/mathbb{Z}}(a+beta + ceta^2) = det begin{pmatrix}
              a & 6c & 6b \
              b & a & 6c \
              c & b & a
              end{pmatrix} = a^3 + 6b^3 + 36c^3 - 18abc
              $$



              Then using this we can perform norm arguments like in the quadratic scenario. For example, notice that we can construct an element with norm $2$ by choose $a = 2, b = -1 implies 2 - eta$ is a element of norm $2$, which answers my question.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                I figured out the answer.



                The problem was that I was too hasty in declaring that in cubic extensions, the ring of integers does not have a norm. In fact, I noticed that the norm in quadratic extensions was actually the field norm of the element $a + bsqrt{d}$.



                $$
                N_{mathcal{O}_K/mathbb{Z}}(a+bsqrt{d}) = det begin{pmatrix}
                a & bd \
                b & a
                end{pmatrix} = a^2 - db^2
                $$



                So then when we consider this cubic extension, and repeat this process (let $eta$ be such that $eta^3 = 6$.)



                $$
                N_{mathcal{O}_K/mathbb{Z}}(a+beta + ceta^2) = det begin{pmatrix}
                a & 6c & 6b \
                b & a & 6c \
                c & b & a
                end{pmatrix} = a^3 + 6b^3 + 36c^3 - 18abc
                $$



                Then using this we can perform norm arguments like in the quadratic scenario. For example, notice that we can construct an element with norm $2$ by choose $a = 2, b = -1 implies 2 - eta$ is a element of norm $2$, which answers my question.






                share|cite|improve this answer









                $endgroup$



                I figured out the answer.



                The problem was that I was too hasty in declaring that in cubic extensions, the ring of integers does not have a norm. In fact, I noticed that the norm in quadratic extensions was actually the field norm of the element $a + bsqrt{d}$.



                $$
                N_{mathcal{O}_K/mathbb{Z}}(a+bsqrt{d}) = det begin{pmatrix}
                a & bd \
                b & a
                end{pmatrix} = a^2 - db^2
                $$



                So then when we consider this cubic extension, and repeat this process (let $eta$ be such that $eta^3 = 6$.)



                $$
                N_{mathcal{O}_K/mathbb{Z}}(a+beta + ceta^2) = det begin{pmatrix}
                a & 6c & 6b \
                b & a & 6c \
                c & b & a
                end{pmatrix} = a^3 + 6b^3 + 36c^3 - 18abc
                $$



                Then using this we can perform norm arguments like in the quadratic scenario. For example, notice that we can construct an element with norm $2$ by choose $a = 2, b = -1 implies 2 - eta$ is a element of norm $2$, which answers my question.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 19:12









                TrostAftTrostAft

                440412




                440412






























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