# why can't quintics be solved by radicals and the relevance of permutations of roots of polynomials

I am seeking to learn about the motivation in the development of group theory. It has been a few years since algebra, and we got as far as rings and fields. I am aware that there were several motivations for the development of the theory of groups; the studies of different geometries, the groups that arose out of the study of modular arithmetic, but i am particularly interested in the motivation from the quest to disprove the solvability of the quintic equation, generally, by radicals.

I understand after the solutions for the lower order polynomial functions were discovered it was believed that there would be no like solution for that of quintic equations. I have read of the work Lagrange did that led Abel to develop his proof. Particularly the link on the history of the development of group theory provides some of the history.

My specific question relates to the discussion of the role the permutations of the roots the 5th order polynomials played. as follows from the above link

Lagrange assumes the roots of a given cubic equation are x', x'' and x'''. Then, taking 1, w, w2 as the cube roots of unity, he examines the expression

$$R = x' + wx'' + w2x'''$$

and notes that it takes just two different values under the six permutations of the roots x', x'', x'''

If R is a polynomial, and the roots are as denoted above, what are these cube roots of unity? What is the interpretation of the permutation of the roots, and their relevance to a solution by radicals?

I have read the article on the Abel-ruffini theorem, which provides a proof which is based on Galois theory. What i would like is more of an explanation in terms of the mathematics the problem was understood in during the time it was posed. i.e. Not the modern algebraic formulation it is introduced in today. I have not studied Galois theory, and considering my own limitations, perhaps the context it was understood in at the time would be more accessible to my understanding.

Also, i did try to reason through Intuition behind looking at permutations of the roots in Galois theory, and the part that was difficult was the discussion of elementary symmetric functions.

UPDATE

A paper entitled Galois theory for beginers was submitted in an answer below. I would like to include an excerpt from that paper that is more direct at part of my difficulty.

the set of elements obtainable from $a_0,\ldots,a_{n-1}$ by +, -, $\times$, $\div$ is the field $\mathbb{Q}(a_0,\ldots,a_{n-1})$. If we denote the roots of $(\ast)$ by $x_1,\ldots,x_n$, so that $$(x-x_1)\cdots(x-x_n)=x^n=a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ then $a_0,\ldots,a_{n-1}$ are polynomial functions of $x_1,\ldots,x_n$ called the elementary symmetric functions: $$a_0=(-1)^nx_1x_2\ldots x_n,\ldots,a_{n-1}=-(x_1+x_2+\cdots+x_n).$$

the link to the paper is provided in a comment below the answer. what i understand that the field $\mathbb{Q}$ is formed from the coefficients of the nth order monic polynomial, $a_0,\ldots,a_{n-1}$, closed under the operations of addition and multiplication.

But in what sense are these coefficients a polynomial of the roots of the polynomial? Also, what is the meaning of the equation $a_0=(-1)^nx_1x_2\ldots x_n,\ldots,a_{n-1}=-(x_1+x_2+\cdots+x_n)$?

Is this equation some how related to the identities relating the sum and multiplies of roots of an nth order polynomial to the quotients of the coefficients, $\alpha+\beta=\frac{-b}{a}$ and $\alpha \beta= \frac{c}{a}$ for the second order polynomial with roots $\alpha,\beta$ and coefficients a,b,c?

• I then suggest you read up a bit on symmetric functions! – Mariano Suárez-Álvarez Apr 8 '16 at 6:10
• @MarianoSuárez-Alvarez yes but my question was long enough so I will save that for a different thread. Perhaps I should have asked that first though – user74091 Apr 8 '16 at 6:13
• perhaps you would like to read Galois' original text as well.... Joking aside, the modern treatment through Galois theory is an enormous simplification over any of the original treatments. Tackling, say, Abel's work directly is going to be extremely challenging, while there are several excellent books that go from nothing to the Galois correspondence in under 150 pages, complete with all the linear algebra and group theory prerequisites, and applications not only to the insolubility of the quintic, but (why not, it's so easy once you have Galois) other famous results. – Ittay Weiss Apr 8 '16 at 7:23
• I'd be interested in such a text I your familiar with some titles – user74091 Apr 8 '16 at 7:26
• Also, I'm not opposed to original works, in fact I find them in my mind often more intuitive then the way subjects are presented in a classroom. In the case of Galois' work, I had the impression his work was challenging to the greatest minds of his time. Don't think I was the intended audience of that work ;) – user74091 Apr 8 '16 at 7:29

## 1 Answer

I think there is a very short but expository article on the topic you are considering. The proof, that a general quintic can not be solved by radicals, do not requires too much theory of field extensions, and other things, as usually seen in books. This is in fact stated in the article below (I hope you may have access to it).

• math.jhu.edu/~smahanta/Teaching/Spring10/Stillwell.pdf – user74091 Apr 8 '16 at 7:49
• ^this is a link a PDF of the above. – user74091 Apr 8 '16 at 7:49
• Thanks for putting the link: I was initially thinking to put only link, but, (1) it may not be freely accessible some places and (2) the author highlights (underlined sentence) that he want to see the un-solvability with very less prerequisite; many undergraduates (as I saw) learn this theory in abstract sense, but become unable to express this un-solvability shortly or quickly. – p Groups Apr 8 '16 at 7:57