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.
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.
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.
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.
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.
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.
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