I remenber that Galois proved that a irreducible polynomial of degree p prime is solvable if, and only if, its roots can be expressed in terms of rational polynomials of the others 2 roots, and this gives as a directly corolary the results that you showed here. This also implies that if G is the Galois group of the irreducible polynomial in question, then |G| is less or equal to p(p-1). If think that this result is kind of ironic, since one of the reasons why Poison discarded the work of Galois was because it lacked a pratical criteria to identify if a given polynomial was solvable.
Of course!! The moment you qualify it as a _root of 1,_ you qualify - it - as expressed as a radical, ain't it? 😊 To cut through the dirty possibilities, one thus expresses the definition (of expressible as a radical) as: _belonging to an extension of the form k(α)/k, where α^n ∈ k for some n ≥ 0._ -to include repeated field operations, and one √ -operation. To include multiple application of various √ -operations, one is naturally led to a tower of these. I hope this answers your other question too.
..And, working over ℚ, say, if you'd like to fix a _canonical_ choice of an n-th root (cos (2π/n) + _i_ sin (2π/n) for roots of unity-α=1, the unique real root otherwise if n is odd or α ≥ 0, ζ_2n times n-th root of -α if α < 0.. etc.), you're still lead to the n-th root of unity case, modulo which, the answer's yes, it is all same as the above definition.
Another question, if a real solution to a polynomial equation of integer coefficients is expressed as "n-th √ of (something involving i)" plus its complex conjugate, is this form considered an expression in terms of radicals?
Yes. Even, in cubic case, if P = x^3 + px + q is irreducible (no rational solution) with 3 real roots (delta < 0), Galois theory shows that roots cannot be expressed other than using 3rd radicals involving complex numbers plus its conjugate (in Galois theory terms, « there is no radical extension containing splitting field of P and included in Real Numbers field ». Such an extension must be ‘over’ real numbers field then involving complex Numbers). One of the reasons of discovery of « sophisticated quantities » (complex numbers today) by Cardano/Bombelli in 16th century when trying to use Cardano cubic formula in the case above which provides negative Numbers under 2nd radicals.
Isn't there a little gap in the argument when he aptly points out that ‘n-th roots of all the conjugates need to be adjoined in order to get a normal extension,’ then pretends adjoining a single one of them gives a galois, thus normal, extension? well, gotta confess, haven't thought about it very carefully.. -great talk otherwise.
so magically you at around 12:09.... turns 5 into zeta and both symbols look the same....is there an explanation why or is that sloppiness on your part?
Subgroups (quantum, discrete) are dual to subfields (classical, continuous) -- the Galois correspondence. All commutators are cyclic permutations. "Always two there are" -- Yoda.
I remenber that Galois proved that a irreducible polynomial of degree p prime is solvable if, and only if, its roots can be expressed in terms of rational polynomials of the others 2 roots, and this gives as a directly corolary the results that you showed here. This also implies that if G is the Galois group of the irreducible polynomial in question, then |G| is less or equal to p(p-1).
If think that this result is kind of ironic, since one of the reasons why Poison discarded the work of Galois was because it lacked a pratical criteria to identify if a given polynomial was solvable.
Will there be lectures on Jacobson's theory to draw parallels with Galois one?
Excellent!
Are roots of 1 always expressible in terms of radicals?
Of course!! The moment you qualify it as a _root of 1,_ you qualify - it - as expressed as a radical, ain't it? 😊
To cut through the dirty possibilities, one thus expresses the definition (of expressible as a radical) as: _belonging to an extension of the form k(α)/k, where α^n ∈ k for some n ≥ 0._ -to include repeated field operations, and one √ -operation. To include multiple application of various √ -operations, one is naturally led to a tower of these.
I hope this answers your other question too.
..And, working over ℚ, say, if you'd like to fix a _canonical_ choice of an n-th root (cos (2π/n) + _i_ sin (2π/n) for roots of unity-α=1, the unique real root otherwise if n is odd or α ≥ 0, ζ_2n times n-th root of -α if α < 0.. etc.), you're still lead to the n-th root of unity case, modulo which, the answer's yes, it is all same as the above definition.
Another question, if a real solution to a polynomial equation of integer coefficients is expressed as "n-th √ of (something involving i)" plus its complex conjugate, is this form considered an expression in terms of radicals?
Yes. Even, in cubic case, if P = x^3 + px + q is irreducible (no rational solution) with 3 real roots (delta < 0), Galois theory shows that roots cannot be expressed other than using 3rd radicals involving complex numbers plus its conjugate (in Galois theory terms, « there is no radical extension containing splitting field of P and included in Real Numbers field ». Such an extension must be ‘over’ real numbers field then involving complex Numbers). One of the reasons of discovery of « sophisticated quantities » (complex numbers today) by Cardano/Bombelli in 16th century when trying to use Cardano cubic formula in the case above which provides negative Numbers under 2nd radicals.
Isn't there a little gap in the argument when he aptly points out that ‘n-th roots of all the conjugates need to be adjoined in order to get a normal extension,’ then pretends adjoining a single one of them gives a galois, thus normal, extension? well, gotta confess, haven't thought about it very carefully.. -great talk otherwise.
so magically you at around 12:09.... turns 5 into zeta and both symbols look the same....is there an explanation why or is that sloppiness on your part?
Subgroups (quantum, discrete) are dual to subfields (classical, continuous) -- the Galois correspondence.
All commutators are cyclic permutations.
"Always two there are" -- Yoda.
Yeee