# What is the intuition behind the proof of Abel-Ruffini theorem in abstract algebra?

Is there a way to explain this proof in Wikipedia without knowing the abstract algebra too much or deep function experience? In addition, I don't how the abstract algebra work even after I look at an introductory abstract algebra book, if you can't take your time to explain the whole thing, may you give some preliminary reference for your explanation?

Or may you give hints in a way that is a list of fact would be helpful to the proof and verify it for me please.

• No. The AR theorem is fundamentally an algebraic fact, so there won't be intuition present if you look at it from a different angle. – anon Nov 16 '11 at 0:08
• @anon - Then may you explain how does abstract algebra work in the proof? or give the intuition behind the abstract algebra in that proof? Help even hints would be appreciated to you – Victor Nov 16 '11 at 0:14
• Reference: A one-year upper-divison or one year first-year-graduate course sequence in abstract algebra. Any textbook thereof. – Arturo Magidin Nov 16 '11 at 1:15
• Dear Victor, This answer may help a little: math.stackexchange.com/a/792/221 Regards, – Matt E Dec 19 '11 at 7:03
• A short explanation of Arnold's proof that only requires knowledge of complex numbers is given here youtu.be/RhpVSV6iCko – user118083 Dec 29 '13 at 2:09

## 1 Answer

It takes a semester to prove it. Here's an outline. It may not mean much to you, but it should give you some idea of what you have to learn.

Let $p$ be an irreducible polynomial of degree 5 with rational coefficients.

Let $a,b,c,d,e$ be the roots of $p$ (in, say, the complex numbers).

Let $K$ be the smallest field containing the rationals and those five roots (so $K$ contains all results of the four arithmetic operations carried out on rationals and on those five roots).

An automorphism of $K$ is a function $g:K\to K$ such that $g(x+y)=g(x)+g(y)$ and $g(xy)=g(x)g(y)$ for all $x,y$ in $K$, and such that if $g(x)=0$ then $x=0$. The set of all automorphisms of $K$ is finite and forms a group under composition. Let $G$ be the group of all automorphisms of $K$.

There is a one-one correspondence between the subgroups of $G$ and the subfields of $K$. What's more, this correspondence converts useful information about subgroups into useful information about subfields, and vice versa. In particular, radical extensions (roughly speaking, subfields where you can write all the elements in terms of square roots, cube roots, fifth roots, etc., and the four arithmetic operations) correspond to solvable groups (roughly, groups $H$ such that you can find a chain of subgroups $1\subset H_1\subset H_2\subset\cdots\subset H$ with $H_{k+1}/H_k$ commutative for all $k$).

Back to our polynomial $p$. For some such $p$, $G$ turns out to be $S_5$, the group of permutations on five letters. That group turns out not to be solvable; its only non-trivial normal subgroup is $A_5$, the alternating group on five letters, and $A_5$ is non-commutative and has no normal subgroups at all (other than itself and the one-element group) so you can't find the chain of subgroups you need for solvability. So $G$ is not solvable, so $K$ is not radical, so there is no expression for $a,b,c,d,e$ in terms of square roots, cube roots, etc., and arithemtic operations.

• "It takes a semester to prove it." is only correct as for the proof using Galois theory. The original proof is shorter, more elementary and didn't use Galois theory (which just didn't exist yet). See math.caltech.edu/~jimlb/abel.pdf – Martin Brandenburg Dec 29 '13 at 2:28