Galois Theory and Polynomials Discriminant
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I am trying to show for some complex numbers `$l_j , 1 ≤ j ≤ 5$
$∆(x_1, x_2, x_3)=l_1Ã^2_3,3 + l_2Ã_3,1Ã_3,2Ã_3,3 + l_3Ã^3_3,1Ã_3,3+ l_4Ã^3_3,2 + l_5Ã^2_3,1Ã^2_3,2$.
Second, by using the polynomial $x^
3 − 1$, want to show that $l_1 = −27$.
I know the discriminant $∆_y3+py+q = −4p^3 − 27q^
2$ and P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ldots, n$ one has $F(X_Ã(1),ldots, X_Ã(n)) = P(X_1,ldots, X_n)$. But I am not sure how to proceed in the above example.
abstract-algebra galois-theory
edited Aug 25 at 6:18
Empty
7,91042254
asked Aug 25 at 4:30
Homaniac
494110
add a comment |Â
up vote
0
down vote
favorite
I am trying to show for some complex numbers `$l_j , 1 ≤ j ≤ 5$
$∆(x_1, x_2, x_3)=l_1Ã^2_3,3 + l_2Ã_3,1Ã_3,2Ã_3,3 + l_3Ã^3_3,1Ã_3,3+ l_4Ã^3_3,2 + l_5Ã^2_3,1Ã^2_3,2$.
Second, by using the polynomial $x^
3 − 1$, want to show that $l_1 = −27$.
I know the discriminant $∆_y3+py+q = −4p^3 − 27q^
2$ and P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ldots, n$ one has $F(X_Ã(1),ldots, X_Ã(n)) = P(X_1,ldots, X_n)$. But I am not sure how to proceed in the above example.
abstract-algebra galois-theory
edited Aug 25 at 6:18
Empty
7,91042254
asked Aug 25 at 4:30
Homaniac
494110
Could you illustrate how the function would better compute $l_1$ ?
– Homaniac
Aug 26 at 16:02
if $sigma_3,1 = sigma_3,2 = 0$ then $Delta = l_1 sigma_3,3^2$ so you can compute $l_1$ with knowledge of $Delta$ and $sigma_3,3^2$, both of which can be computed directly with the roots. My previous comment was mistaken, $sigma_3,1 = sigma_3,2$ in the case of $x^3-1$.
– Rolf Hoyer
Aug 26 at 18:17
Ahh could you show how the computation works, appreciate it!
– Homaniac
Aug 26 at 18:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to show for some complex numbers `$l_j , 1 ≤ j ≤ 5$
$∆(x_1, x_2, x_3)=l_1Ã^2_3,3 + l_2Ã_3,1Ã_3,2Ã_3,3 + l_3Ã^3_3,1Ã_3,3+ l_4Ã^3_3,2 + l_5Ã^2_3,1Ã^2_3,2$.
Second, by using the polynomial $x^
3 − 1$, want to show that $l_1 = −27$.
I know the discriminant $∆_y3+py+q = −4p^3 − 27q^
2$ and P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ldots, n$ one has $F(X_Ã(1),ldots, X_Ã(n)) = P(X_1,ldots, X_n)$. But I am not sure how to proceed in the above example.
abstract-algebra galois-theory
edited Aug 25 at 6:18
Empty
7,91042254
asked Aug 25 at 4:30
Homaniac
494110
I am trying to show for some complex numbers `$l_j , 1 ≤ j ≤ 5$
$∆(x_1, x_2, x_3)=l_1Ã^2_3,3 + l_2Ã_3,1Ã_3,2Ã_3,3 + l_3Ã^3_3,1Ã_3,3+ l_4Ã^3_3,2 + l_5Ã^2_3,1Ã^2_3,2$.
Second, by using the polynomial $x^
3 − 1$, want to show that $l_1 = −27$.
I know the discriminant $∆_y3+py+q = −4p^3 − 27q^
2$ and P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ldots, n$ one has $F(X_Ã(1),ldots, X_Ã(n)) = P(X_1,ldots, X_n)$. But I am not sure how to proceed in the above example.
abstract-algebra galois-theory
edited Aug 25 at 6:18
Empty
7,91042254
asked Aug 25 at 4:30
Homaniac
494110
edited Aug 25 at 6:18
Empty
7,91042254
edited Aug 25 at 6:18
Empty
7,91042254
edited Aug 25 at 6:18
Empty
7,91042254
7,91042254
asked Aug 25 at 4:30
Homaniac
494110
asked Aug 25 at 4:30
Homaniac
494110
asked Aug 25 at 4:30
Homaniac
494110
494110
Could you illustrate how the function would better compute $l_1$ ?
– Homaniac
Aug 26 at 16:02
if $sigma_3,1 = sigma_3,2 = 0$ then $Delta = l_1 sigma_3,3^2$ so you can compute $l_1$ with knowledge of $Delta$ and $sigma_3,3^2$, both of which can be computed directly with the roots. My previous comment was mistaken, $sigma_3,1 = sigma_3,2$ in the case of $x^3-1$.
– Rolf Hoyer
Aug 26 at 18:17
Ahh could you show how the computation works, appreciate it!
– Homaniac
Aug 26 at 18:24
add a comment |Â
Could you illustrate how the function would better compute $l_1$ ?
– Homaniac
Aug 26 at 16:02
if $sigma_3,1 = sigma_3,2 = 0$ then $Delta = l_1 sigma_3,3^2$ so you can compute $l_1$ with knowledge of $Delta$ and $sigma_3,3^2$, both of which can be computed directly with the roots. My previous comment was mistaken, $sigma_3,1 = sigma_3,2$ in the case of $x^3-1$.
– Rolf Hoyer
Aug 26 at 18:17
Ahh could you show how the computation works, appreciate it!
– Homaniac
Aug 26 at 18:24
Could you illustrate how the function would better compute $l_1$ ?
– Homaniac
Aug 26 at 16:02
Could you illustrate how the function would better compute $l_1$ ?
– Homaniac
Aug 26 at 16:02
if $sigma_3,1 = sigma_3,2 = 0$ then $Delta = l_1 sigma_3,3^2$ so you can compute $l_1$ with knowledge of $Delta$ and $sigma_3,3^2$, both of which can be computed directly with the roots. My previous comment was mistaken, $sigma_3,1 = sigma_3,2$ in the case of $x^3-1$.
– Rolf Hoyer
Aug 26 at 18:17
if $sigma_3,1 = sigma_3,2 = 0$ then $Delta = l_1 sigma_3,3^2$ so you can compute $l_1$ with knowledge of $Delta$ and $sigma_3,3^2$, both of which can be computed directly with the roots. My previous comment was mistaken, $sigma_3,1 = sigma_3,2$ in the case of $x^3-1$.
– Rolf Hoyer
Aug 26 at 18:17
Ahh could you show how the computation works, appreciate it!
– Homaniac
Aug 26 at 18:24
Ahh could you show how the computation works, appreciate it!
– Homaniac
Aug 26 at 18:24
add a comment |Â
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Could you illustrate how the function would better compute $l_1$ ?
– Homaniac
Aug 26 at 16:02
if $sigma_3,1 = sigma_3,2 = 0$ then $Delta = l_1 sigma_3,3^2$ so you can compute $l_1$ with knowledge of $Delta$ and $sigma_3,3^2$, both of which can be computed directly with the roots. My previous comment was mistaken, $sigma_3,1 = sigma_3,2$ in the case of $x^3-1$.
– Rolf Hoyer
Aug 26 at 18:17
Ahh could you show how the computation works, appreciate it!
– Homaniac
Aug 26 at 18:24