Finding the minimal polynomial for each algebraic element over $mathbb{Q}$.












0














Can someone check whether the following are the correct minimal polynomials for each root?



For root $sqrt{3}+sqrt[3]{5}$, I got $p(x)=x^{6}-9x^{4}-10x^{3}+27x^{2}-90x - 27$.



For root $costheta +isintheta$, where $theta =frac{2pi}{n}$ for $ngeq 1$, I got $p(x)=x^{2}-2xcostheta +1$.



For root $sqrt{sqrt[3]{2}-i}$, I got $p(x)=x^{12}-15x^{8}-4x^{6}+3x^{4}+12x^{2}+5$.



I tried using a calculator to see whether the equations become zero if I substitute the roots, but no calculator I can find supports such lengthy equations.






abstract-algebra field-theory minimal-polynomials







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    0














    Can someone check whether the following are the correct minimal polynomials for each root?



    For root $sqrt{3}+sqrt[3]{5}$, I got $p(x)=x^{6}-9x^{4}-10x^{3}+27x^{2}-90x - 27$.



    For root $costheta +isintheta$, where $theta =frac{2pi}{n}$ for $ngeq 1$, I got $p(x)=x^{2}-2xcostheta +1$.



    For root $sqrt{sqrt[3]{2}-i}$, I got $p(x)=x^{12}-15x^{8}-4x^{6}+3x^{4}+12x^{2}+5$.



    I tried using a calculator to see whether the equations become zero if I substitute the roots, but no calculator I can find supports such lengthy equations.






    abstract-algebra field-theory minimal-polynomials







    share|cite|improve this question



























      0












      0








      0







      Can someone check whether the following are the correct minimal polynomials for each root?



      For root $sqrt{3}+sqrt[3]{5}$, I got $p(x)=x^{6}-9x^{4}-10x^{3}+27x^{2}-90x - 27$.



      For root $costheta +isintheta$, where $theta =frac{2pi}{n}$ for $ngeq 1$, I got $p(x)=x^{2}-2xcostheta +1$.



      For root $sqrt{sqrt[3]{2}-i}$, I got $p(x)=x^{12}-15x^{8}-4x^{6}+3x^{4}+12x^{2}+5$.



      I tried using a calculator to see whether the equations become zero if I substitute the roots, but no calculator I can find supports such lengthy equations.






      abstract-algebra field-theory minimal-polynomials







      share|cite|improve this question















      Can someone check whether the following are the correct minimal polynomials for each root?



      For root $sqrt{3}+sqrt[3]{5}$, I got $p(x)=x^{6}-9x^{4}-10x^{3}+27x^{2}-90x - 27$.



      For root $costheta +isintheta$, where $theta =frac{2pi}{n}$ for $ngeq 1$, I got $p(x)=x^{2}-2xcostheta +1$.



      For root $sqrt{sqrt[3]{2}-i}$, I got $p(x)=x^{12}-15x^{8}-4x^{6}+3x^{4}+12x^{2}+5$.



      I tried using a calculator to see whether the equations become zero if I substitute the roots, but no calculator I can find supports such lengthy equations.






      abstract-algebra field-theory minimal-polynomials




      abstract-algebra field-theory minimal-polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 at 19:58









      Key Flex

      7,46941232




      7,46941232










      asked Nov 25 at 19:55









      numericalorange

      1,719311




      1,719311






















          1 Answer
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          2














          Just use sage. Here is the code, just checking:



          First polynomial is ok:



          sage: (3^(1/2) + 5^(1/3) ).minpoly()
          
x^6 - 9*x^4 - 10*x^3 + 27*x^2 - 90*x - 2


          In fact, sage can work explicitly in the tower of fields:



          sage: K.<a> = QuadraticField(3)
          
sage: R.<X> = PolynomialRing(K)
          
sage: L.<b> = K.extension(X^2+3)
          
sage: S.<Y> = PolynomialRing(L)
          
sage: M.<c> = L.extension(x^3-5)
          
sage: a^2, b^2, c^3
          
(3, -3, 5)
          
sage: third_roots_of_one = [ 1, (-1+b)/2, (-1-b)/2 ]
          
sage: secnd_roots_of_one = [ 1, -1 ]
          
sage: T.<Z> = PolynomialRing(M)
          
sage: prod( [ Z - (r2*a + r3*c) 
          
....:         for r2 in secnd_roots_of_one
          
....:         for r3 in third_roots_of_one ] )
          

          
Z^6 - 9*Z^4 - 10*Z^3 + 27*Z^2 - 90*Z - 2


          For the second question, if we need the minimal polynomial over $Bbb Q$, the result should be a cyclotomic polynomial. For instance, sage again:



          sage: N = 9
          
sage: exp( 2*pi*i/N ).minpoly()
          
x^6 + x^3 + 1
          
sage: cyclotomic_polynomial(N)
          
x^6 + x^3 + 1


          The third one:



          sage: sqrt( 2^(1/3) - i ).minpoly()
          
x^12 + 3*x^8 - 4*x^6 + 3*x^4 + 12*x^2 + 5


          Again, one can manually / humanly build the tower of fields, by starting with $Bbb Q$, and adjoining $i$, then $2^{1/3}$ and its conjugates, then the square root we need.






          share|cite|improve this answer





















          • Thanks so much for helping me check my mistakes! May I ask what does it mean $N=9$ for the second one?
            – numericalorange
            Nov 25 at 20:42








          • 1




            In case of $n=9$, the algebraic element $$a=cosfrac{2pi}n+isinfrac{2pi}n$$ is a primitive $9$.th root of unity. So i was expecting the cyclotomic polynomial $Phi_9$ of degree $varphi(9)=9left(1-frac 13right)=6$ to be the answer. (The polynomial of degree two in the OP is the minimal polynomial over $Bbb R$.) So the posted $n$ corresponds to the N in the code. (Setting it to n would have overwritten something in my sage session, sorry...)
            – dan_fulea
            Nov 25 at 20:49













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          1 Answer
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          1 Answer
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          2














          Just use sage. Here is the code, just checking:



          First polynomial is ok:



          sage: (3^(1/2) + 5^(1/3) ).minpoly()
          
x^6 - 9*x^4 - 10*x^3 + 27*x^2 - 90*x - 2


          In fact, sage can work explicitly in the tower of fields:



          sage: K.<a> = QuadraticField(3)
          
sage: R.<X> = PolynomialRing(K)
          
sage: L.<b> = K.extension(X^2+3)
          
sage: S.<Y> = PolynomialRing(L)
          
sage: M.<c> = L.extension(x^3-5)
          
sage: a^2, b^2, c^3
          
(3, -3, 5)
          
sage: third_roots_of_one = [ 1, (-1+b)/2, (-1-b)/2 ]
          
sage: secnd_roots_of_one = [ 1, -1 ]
          
sage: T.<Z> = PolynomialRing(M)
          
sage: prod( [ Z - (r2*a + r3*c) 
          
....:         for r2 in secnd_roots_of_one
          
....:         for r3 in third_roots_of_one ] )
          

          
Z^6 - 9*Z^4 - 10*Z^3 + 27*Z^2 - 90*Z - 2


          For the second question, if we need the minimal polynomial over $Bbb Q$, the result should be a cyclotomic polynomial. For instance, sage again:



          sage: N = 9
          
sage: exp( 2*pi*i/N ).minpoly()
          
x^6 + x^3 + 1
          
sage: cyclotomic_polynomial(N)
          
x^6 + x^3 + 1


          The third one:



          sage: sqrt( 2^(1/3) - i ).minpoly()
          
x^12 + 3*x^8 - 4*x^6 + 3*x^4 + 12*x^2 + 5


          Again, one can manually / humanly build the tower of fields, by starting with $Bbb Q$, and adjoining $i$, then $2^{1/3}$ and its conjugates, then the square root we need.






          share|cite|improve this answer





















          • Thanks so much for helping me check my mistakes! May I ask what does it mean $N=9$ for the second one?
            – numericalorange
            Nov 25 at 20:42








          • 1




            In case of $n=9$, the algebraic element $$a=cosfrac{2pi}n+isinfrac{2pi}n$$ is a primitive $9$.th root of unity. So i was expecting the cyclotomic polynomial $Phi_9$ of degree $varphi(9)=9left(1-frac 13right)=6$ to be the answer. (The polynomial of degree two in the OP is the minimal polynomial over $Bbb R$.) So the posted $n$ corresponds to the N in the code. (Setting it to n would have overwritten something in my sage session, sorry...)
            – dan_fulea
            Nov 25 at 20:49


















          2














          Just use sage. Here is the code, just checking:



          First polynomial is ok:



          sage: (3^(1/2) + 5^(1/3) ).minpoly()
          
x^6 - 9*x^4 - 10*x^3 + 27*x^2 - 90*x - 2


          In fact, sage can work explicitly in the tower of fields:



          sage: K.<a> = QuadraticField(3)
          
sage: R.<X> = PolynomialRing(K)
          
sage: L.<b> = K.extension(X^2+3)
          
sage: S.<Y> = PolynomialRing(L)
          
sage: M.<c> = L.extension(x^3-5)
          
sage: a^2, b^2, c^3
          
(3, -3, 5)
          
sage: third_roots_of_one = [ 1, (-1+b)/2, (-1-b)/2 ]
          
sage: secnd_roots_of_one = [ 1, -1 ]
          
sage: T.<Z> = PolynomialRing(M)
          
sage: prod( [ Z - (r2*a + r3*c) 
          
....:         for r2 in secnd_roots_of_one
          
....:         for r3 in third_roots_of_one ] )
          

          
Z^6 - 9*Z^4 - 10*Z^3 + 27*Z^2 - 90*Z - 2


          For the second question, if we need the minimal polynomial over $Bbb Q$, the result should be a cyclotomic polynomial. For instance, sage again:



          sage: N = 9
          
sage: exp( 2*pi*i/N ).minpoly()
          
x^6 + x^3 + 1
          
sage: cyclotomic_polynomial(N)
          
x^6 + x^3 + 1


          The third one:



          sage: sqrt( 2^(1/3) - i ).minpoly()
          
x^12 + 3*x^8 - 4*x^6 + 3*x^4 + 12*x^2 + 5


          Again, one can manually / humanly build the tower of fields, by starting with $Bbb Q$, and adjoining $i$, then $2^{1/3}$ and its conjugates, then the square root we need.






          share|cite|improve this answer





















          • Thanks so much for helping me check my mistakes! May I ask what does it mean $N=9$ for the second one?
            – numericalorange
            Nov 25 at 20:42








          • 1




            In case of $n=9$, the algebraic element $$a=cosfrac{2pi}n+isinfrac{2pi}n$$ is a primitive $9$.th root of unity. So i was expecting the cyclotomic polynomial $Phi_9$ of degree $varphi(9)=9left(1-frac 13right)=6$ to be the answer. (The polynomial of degree two in the OP is the minimal polynomial over $Bbb R$.) So the posted $n$ corresponds to the N in the code. (Setting it to n would have overwritten something in my sage session, sorry...)
            – dan_fulea
            Nov 25 at 20:49
















          2












          2








          2






          Just use sage. Here is the code, just checking:



          First polynomial is ok:



          sage: (3^(1/2) + 5^(1/3) ).minpoly()
          
x^6 - 9*x^4 - 10*x^3 + 27*x^2 - 90*x - 2


          In fact, sage can work explicitly in the tower of fields:



          sage: K.<a> = QuadraticField(3)
          
sage: R.<X> = PolynomialRing(K)
          
sage: L.<b> = K.extension(X^2+3)
          
sage: S.<Y> = PolynomialRing(L)
          
sage: M.<c> = L.extension(x^3-5)
          
sage: a^2, b^2, c^3
          
(3, -3, 5)
          
sage: third_roots_of_one = [ 1, (-1+b)/2, (-1-b)/2 ]
          
sage: secnd_roots_of_one = [ 1, -1 ]
          
sage: T.<Z> = PolynomialRing(M)
          
sage: prod( [ Z - (r2*a + r3*c) 
          
....:         for r2 in secnd_roots_of_one
          
....:         for r3 in third_roots_of_one ] )
          

          
Z^6 - 9*Z^4 - 10*Z^3 + 27*Z^2 - 90*Z - 2


          For the second question, if we need the minimal polynomial over $Bbb Q$, the result should be a cyclotomic polynomial. For instance, sage again:



          sage: N = 9
          
sage: exp( 2*pi*i/N ).minpoly()
          
x^6 + x^3 + 1
          
sage: cyclotomic_polynomial(N)
          
x^6 + x^3 + 1


          The third one:



          sage: sqrt( 2^(1/3) - i ).minpoly()
          
x^12 + 3*x^8 - 4*x^6 + 3*x^4 + 12*x^2 + 5


          Again, one can manually / humanly build the tower of fields, by starting with $Bbb Q$, and adjoining $i$, then $2^{1/3}$ and its conjugates, then the square root we need.






          share|cite|improve this answer












          Just use sage. Here is the code, just checking:



          First polynomial is ok:



          sage: (3^(1/2) + 5^(1/3) ).minpoly()
          
x^6 - 9*x^4 - 10*x^3 + 27*x^2 - 90*x - 2


          In fact, sage can work explicitly in the tower of fields:



          sage: K.<a> = QuadraticField(3)
          
sage: R.<X> = PolynomialRing(K)
          
sage: L.<b> = K.extension(X^2+3)
          
sage: S.<Y> = PolynomialRing(L)
          
sage: M.<c> = L.extension(x^3-5)
          
sage: a^2, b^2, c^3
          
(3, -3, 5)
          
sage: third_roots_of_one = [ 1, (-1+b)/2, (-1-b)/2 ]
          
sage: secnd_roots_of_one = [ 1, -1 ]
          
sage: T.<Z> = PolynomialRing(M)
          
sage: prod( [ Z - (r2*a + r3*c) 
          
....:         for r2 in secnd_roots_of_one
          
....:         for r3 in third_roots_of_one ] )
          

          
Z^6 - 9*Z^4 - 10*Z^3 + 27*Z^2 - 90*Z - 2


          For the second question, if we need the minimal polynomial over $Bbb Q$, the result should be a cyclotomic polynomial. For instance, sage again:



          sage: N = 9
          
sage: exp( 2*pi*i/N ).minpoly()
          
x^6 + x^3 + 1
          
sage: cyclotomic_polynomial(N)
          
x^6 + x^3 + 1


          The third one:



          sage: sqrt( 2^(1/3) - i ).minpoly()
          
x^12 + 3*x^8 - 4*x^6 + 3*x^4 + 12*x^2 + 5


          Again, one can manually / humanly build the tower of fields, by starting with $Bbb Q$, and adjoining $i$, then $2^{1/3}$ and its conjugates, then the square root we need.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 20:30









          dan_fulea

          6,2301312




          6,2301312












          • Thanks so much for helping me check my mistakes! May I ask what does it mean $N=9$ for the second one?
            – numericalorange
            Nov 25 at 20:42








          • 1




            In case of $n=9$, the algebraic element $$a=cosfrac{2pi}n+isinfrac{2pi}n$$ is a primitive $9$.th root of unity. So i was expecting the cyclotomic polynomial $Phi_9$ of degree $varphi(9)=9left(1-frac 13right)=6$ to be the answer. (The polynomial of degree two in the OP is the minimal polynomial over $Bbb R$.) So the posted $n$ corresponds to the N in the code. (Setting it to n would have overwritten something in my sage session, sorry...)
            – dan_fulea
            Nov 25 at 20:49




















          • Thanks so much for helping me check my mistakes! May I ask what does it mean $N=9$ for the second one?
            – numericalorange
            Nov 25 at 20:42








          • 1




            In case of $n=9$, the algebraic element $$a=cosfrac{2pi}n+isinfrac{2pi}n$$ is a primitive $9$.th root of unity. So i was expecting the cyclotomic polynomial $Phi_9$ of degree $varphi(9)=9left(1-frac 13right)=6$ to be the answer. (The polynomial of degree two in the OP is the minimal polynomial over $Bbb R$.) So the posted $n$ corresponds to the N in the code. (Setting it to n would have overwritten something in my sage session, sorry...)
            – dan_fulea
            Nov 25 at 20:49


















          Thanks so much for helping me check my mistakes! May I ask what does it mean $N=9$ for the second one?
          – numericalorange
          Nov 25 at 20:42






          Thanks so much for helping me check my mistakes! May I ask what does it mean $N=9$ for the second one?
          – numericalorange
          Nov 25 at 20:42






          1




          1




          In case of $n=9$, the algebraic element $$a=cosfrac{2pi}n+isinfrac{2pi}n$$ is a primitive $9$.th root of unity. So i was expecting the cyclotomic polynomial $Phi_9$ of degree $varphi(9)=9left(1-frac 13right)=6$ to be the answer. (The polynomial of degree two in the OP is the minimal polynomial over $Bbb R$.) So the posted $n$ corresponds to the N in the code. (Setting it to n would have overwritten something in my sage session, sorry...)
          – dan_fulea
          Nov 25 at 20:49






          In case of $n=9$, the algebraic element $$a=cosfrac{2pi}n+isinfrac{2pi}n$$ is a primitive $9$.th root of unity. So i was expecting the cyclotomic polynomial $Phi_9$ of degree $varphi(9)=9left(1-frac 13right)=6$ to be the answer. (The polynomial of degree two in the OP is the minimal polynomial over $Bbb R$.) So the posted $n$ corresponds to the N in the code. (Setting it to n would have overwritten something in my sage session, sorry...)
          – dan_fulea
          Nov 25 at 20:49




















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