Visual Group Theory, Lecture 6.6: The fundamental theorem of Galois theory

Visual Group Theory, Lecture 6.6: The fundamental theorem of Galois theory

The fundamental theorem of Galois theory guarantees a remarkable correspondence between the subfield lattice of a polynomial and the subgroup lattice of its Galois group. After illustrating this with a detailed example, we define what it means for a group to be "solvable". Galois proved that a polynomial is solvable by radicals if and only if its Galois group is solvable. We conclude by finding a degree-5 polynomial f(x) whose Galois group acts on the roots by a 5-cycle and by a 2-cycle. Since these two elements generate the (unsolvable) symmetric group S_5, the roots of f(x) are unsolvable by radicals.

[hritizgogoi]

One of the most important playlist in YouTube for any budding Mathematician!

[manidhakal]

One of the best instructor for Abstract algebra in YouTube..

[greg]

I have heard the history of Galois's life 8.5 million times.

[shashikadilshan]

Thank you Professor...You describing beauty of the mathematics..Its Awesome...Thank you again

[gunhasirac]

For anyone being curious of how is it go wrong in writing formulae in radicals, here’s a short explanation. (Look up proof to Abel’s theorem for animated explanation)

For 2, 3, 4th degree of polynomial, the formula always comes by reducing the equation to a lower degree and solve it together with the previous formula, which gives the formula of the form of nested roots, 2nd for square root, 3rd for square root in cubic root, 4th for root in root in root. To “prove” this, we first note that, by permuting roots continuously (say 1 and -1 on a unit circle in complex plane), the roots end up swapped, but the radicals of the corresponding equation remain the same. Take 2nd order polynomial for example, if we choose the proper branch of the square root function in complex plane (or simply consider it on the Riemann surface), we can actually move root 1 to root 2 by changing radicals smoothly (which is the converse of radicals remain the same after swapping roots). As we will demonstrate, the solvability of a group corresponds to swapping the roots around and remain unmoved.

In case of quadratic polynomial, there’s only one possible swap for two roots, so the formula is trivially valid.

In case of cubic polynomial, there’s three possible swaps, 1⇄2, 2⇄3, 1⇄3, and the 2 3-loops. Now if we do 1⇄2 then 2⇄3, and undo 1⇄2 then undo 2⇄3, we ended up with the loop (1 2 3), similarly all the other commutators (elements of the form ABA’B’, where prime denotes inverse) ended up to be an element of the alternating group A3. In the other word the commutator subgroup of S3 is A3. Since there’s 2 roots in formula for cubic polynomial, we generate commutators from A3 again, which ends up only with the identity element. Hence the expression of the formula by radicals represent a unique number.

You can also see in process above why solvability is define in tower of normal subgroups. First commutator subgroup is always normal. Second the quotient group by the commutator subgroup is abelian (commutator subgroup is the non-abelian part of the group). Hence the approach by building tower by consecutive commutator subgroups satisfies the tower condition automatically. The only thing we need now is the tower reduce to the trivial subgroup at the end. And this happens exactly when Sn with n smaller than five and fails otherwise. Hence even we find a formula expression for polynomial of order higher than five, that expression does not always define a unique number as the example in the video.

[ramzihedimay]

Great!!! I really enjoyed watching this very interesting lecture. Congratulation. Many thanks

[PasseScience]

Very good serie! Do you agree with the following phrasing: when looking at the lattices (of the splitting field structure or of the gallois group) there are edges that are "good" and edges that are "bad", I call good the ones that can be used to build a "good chain" between the two extremes of the lattice, with "good chain" beeing the same kind of chain that the one required to call the group "solvable". I dont know if there is a terminology for such good edges. On the other hand we remark that when the poly is solvable by radicals it implies we can build a chain of subfields bettween the 2 extreme of the splitting field lattice. And the counterpart of this chain would be a good one in the Galois group lattice. And so if we can deduces from the "symetries" within the roots enough pieces of information to be sure there is no good chain in the Galois group lattice (based on the fact the Galois group is homomorph to the action group on radicals), it implies that there is no good chain in the splitting field lattice, which thus means the thing cant be solvable by radicals.

[samtux]

Presentation is great!

The theory is hard. Now I see why it wasn't proven till 19 century...

[gsgp]

This video is good, but I need many more examples to fully understand the theory presented in this section of the videos to figure out which polynomials are solvable and which are not.
I want to see the differences between what makes the polynomials solvable and what makes them unsolvable by actual examples.

[Max]

Great series, thank you! IMO, you can go quicker on the earlier sections (6.1-6.4) and spend more time on 6.5 and 6.6 (there was so much good material in this lesson which could benefit from more explanation! Maybe a whole 'nother class?).

[Supware]

Is it correct to conclude that a polynomial is solvable in radicals iff there exists a set of linear transformations on ℂ that together can map any root onto any other?

[yanghwanlim]

Professor! I think that {e} is normal to any group A regardless of A being abelian or not. Is it not? I think that to be solvable, N1 should not be the whole group!

[shacharh]

doesn't solvability require that the series of subgroups doesn't miss any subgroup in between? if A < C normal and C/A is abelian but there is a B with A < B < C s.t. e.g. B/A is not abelian or that B isn't even normal.. that means C is not solvable

[Hans_Magnusson]

So I am in the beginning of the clip, …
The fact that a twenty year old kid can come up with this, is due to that brains are not organized exactly the same for every individual.
On top of that your childhood and where and how you end up expecting it, will make you gifted in different things!

Some will be brilliant mathematicians, and others just another of those Jonses!

[ozzyfromspace]

This is quite clearly beyond me. But here I am anyway

[josecartagena]

On slide 2, r does not fix any element, it is a rotation!!

[hyperduality]

Subfields are dual to subgroups -- the Galois correspondence.
Even is dual to odd.
"Always two there are" -- Yoda.

[thomaskim]

I am sure it is an amazing discovery. However, how does this effect my or an average human life? How does this discovery change our civilization? What is the application of this discovery in the real life?

[Theo_Caro]

At the end he says he did this work as a teenager, but he was in his early twenties.

[terrym]

He should of gotten more sleep. Better for marksmanship.