Can someone give me any intuition on the insolubility of the quintic equation? I can follow the proof, but can't really get a 'feel' for it
think of it like this. If you deny it, your peers will look at you like you're a fucking retard. you don't want to look like a fucking retard. so, intuitively, you should accept it.
have you tried arnol'd topological proof? https://www.youtube.com/watch?v=RhpVSV6iCko
>>8115950
You just have to look at N-th roots differently. N-th roots are solutions to polynomials of the form:
x^n + a
Quintic equations (and beyond) just happen to have roots that can't be expressed via these numbers. But iyou can solve them if you use numbers that are the solutions of these polynomials:
x^5 + x + a
(See: Bring radicals)
There's no insolvability here. It's just that not every algebraic number can be expressed as the root of some "x^n + a" polynomial. Big fucking deal...
>>8115950
which proof did you follow? the standard galois one, where they show S5 is not solvable?
This is at a really informal level, but I think of it like this: unsolvable groups are just too "twisted" to "unwind". FTGT says that for a field extension L over K, subfields are in a gc with subgroups of Aut(K/L), and more importantly splitting fields are in correspondence with the normal subgroups. Take a splitting field M corresponding to N, so that G/N represents automorphisms "modulo the automorphisms fixing M", i.e. two automorphisms t, s are equivalent mod N if they differ only in an automorphism which fixes M anyway. This sort of gives you what roots are "left to solve for" after spitting out the roots in M.
Take the simple cases of when this group is cyclic (such as a simple quadratic extension associated to Z/2) to see what this says about the roots (the pattern of factoring the two parts of the square-root taking are very symmetrical).
When G/N is not solvable, then the symmetries between the roots would be "more complicated" than root-taking - were it able to solve the polynomial - would lead to
this is really informal; just play with some examples more. specifically, take a simple nontrivial solvable one and then find the simplest examples of something with an S5 galois group you can.
>>8115961
kek
>>8117393
I realized immediately I said something totally backwards of what I meant but I am starting teaching now so I'll have to fix it later... sorry haven't done Galois theory in a while
>>8115961
> believe in it because your scientific bible tell you it is true
>>8117495
>re-wording a troll post to be clearer
>>8117502
ah, so you must be that guy/girls who checks all the proofs in complete details by himself/herself!
>>8117532
brainlet
>>8117532
>the point
>your head
>>8117538
Bro please stop embarrassing yourself. The point is that you literally re-worded someone who completely agrees with you as if you were mocking them. I don't care that you're an idiot with stupid beliefs, I just feel bad for you attempting to argue with your own kind.
>>8115950
>insolubility of quintic polynomials
What do you mean? We learned to solve these in high school using synthetic division.
Oh, you mean a closed-form solution
>>8117540
just follow blindly your leader - he can't be wrong
it wouldn't dissolve in aqueous solution under a variety of pH and temperature conditions, and it didn't dissolve in ethanol, dichloromethane, acetone, diethyl ether, tert-butyl alcohol, hexane, benzene, or methyl ethyl ketone either. so I'm gonna say it's insoluble.
>>8117419
basically, i'm still not gonna get specific here, when you take G/N, you are finding the group which rearranges the splitting field, i.e. the field which contains all the roots of the polynomial, modulo the automorphisms which fix a particular intermediate splitting field. Then the 0 element of G/N is things that fix this intermediate splitting field, and the nonzero elements are the ones which permute the roots coming from the intermediate splitting field. This was what I said backwards. This tells you the ways to permute the roots of the polynomial in question "localized" to this intermediate splitting field (ie. how these permutations act on the roots coming from the intermediate field).
G/N must be abelian for the roots coming from this intermediate field to have been formed by radical extensions. see here https://en.wikipedia.org/wiki/Radical_extension. effectively, a field extension is radical iff the group of symmetries you get out (permuting the `new roots obtained by taking a radical') is abelian. It's obvious to see why a radical extension leads to an abelian group of permutations on those roots, just cycle around the primitive roots. The other direction is less obvious, but can be proven see wiki.
Then a polynomial is solvable iff each intermediate splitting (pulling the polynomial apart giving more and more roots) arises from a radical extension (the new roots are nth roots of something from some intermediate field). Once you see that radical extensions are associated with abelian galois groups, the theorem makes sense; the ways to permute the roots "at" any "level of splitting" over the "partially solved" polynomial (i.e. taking the whole galois group mod automorphisms fixing the intermediate splitting field in question) is radical iff it is abelian. otherwise, as I stated before, if G/N is not abelian, then the roots will have more complex symmetries than root-taking could lead to, so the new elements can't just be nth roots of something.
