Finding the minimal polynomial for each algebraic element over $mathbb{Q}$.
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
edited Nov 25 at 19:58
Key Flex
7,46941232
asked Nov 25 at 19:55
numericalorange
1,719311
add a comment |
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
edited Nov 25 at 19:58
Key Flex
7,46941232
asked Nov 25 at 19:55
numericalorange
1,719311
add a comment |
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
edited Nov 25 at 19:58
Key Flex
7,46941232
asked Nov 25 at 19:55
numericalorange
1,719311
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
edited Nov 25 at 19:58
Key Flex
7,46941232
asked Nov 25 at 19:55
numericalorange
1,719311
edited Nov 25 at 19:58
Key Flex
7,46941232
asked Nov 25 at 19:55
numericalorange
1,719311
edited Nov 25 at 19:58
Key Flex
7,46941232
edited Nov 25 at 19:58
Key Flex
7,46941232
edited Nov 25 at 19:58
Key Flex
7,46941232
7,46941232
asked Nov 25 at 19:55
numericalorange
1,719311
asked Nov 25 at 19:55
numericalorange
1,719311
asked Nov 25 at 19:55
numericalorange
1,719311
1,719311
add a comment |
add a comment |
1 Answer
1
active
oldest
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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.
answered Nov 25 at 20:30
dan_fulea
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 theNin the code. (Setting it tonwould have overwritten something in my sage session, sorry...)
– dan_fulea
Nov 25 at 20:49
add a comment |
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1 Answer
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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.
answered Nov 25 at 20:30
dan_fulea
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 theNin the code. (Setting it tonwould have overwritten something in my sage session, sorry...)
– dan_fulea
Nov 25 at 20:49
add a comment |
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.
answered Nov 25 at 20:30
dan_fulea
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 theNin the code. (Setting it tonwould have overwritten something in my sage session, sorry...)
– dan_fulea
Nov 25 at 20:49
add a comment |
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.
answered Nov 25 at 20:30
dan_fulea
6,2301312
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.
answered Nov 25 at 20:30
dan_fulea
6,2301312
answered Nov 25 at 20:30
dan_fulea
6,2301312
answered Nov 25 at 20:30
dan_fulea
6,2301312
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 theNin the code. (Setting it tonwould have overwritten something in my sage session, sorry...)
– dan_fulea
Nov 25 at 20:49
add a comment |
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 theNin the code. (Setting it tonwould 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
add a comment |
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