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At the end of Galois Theory we have the theorem:

“Over a field of characteristic zero, a polynomial is solvable by radicals if and only if its Galois group is solvable”

I don’t understand how this connects to being able to find a general formula for the roots of a polynomial of degree $n$ in terms of its coefficients.

I see that for some polynomials of degree $5$, its Galois group is not solvable and so the polynomial cannot be solvable by radicals i.e. its roots are not radical expressions and therefore there must be no general formula for the radical roots of a Quintic equation.

However I don’t understand the converse. If a Galois group of a polynomial $f\in K[x]\setminus\{K\}$, is solvable then the polynomial is solvable by radicals i.e. its roots must be radical expressions of elements in the coefficient field $K[x]$. Good. However the polynomial being solvable by radicals doesn’t imply its roots are radical expressions of precisely the coefficients of $f$.

How then do we know Galois group being solvable means there’s a general radical formula for the roots in terms of the polynomials coefficients?

Partey5
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    Is "Galois Theory" a particular textbook? If so, please [edit] your question to include the details of the book (e.g., its author(s)). – Shaun Apr 19 '20 at 01:00
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    And what does this have to do with "climax"? (An end need not be a "climax.") – David G. Stork Apr 19 '20 at 01:14
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    The point of the theorem is that the roots are found in an extension that is obtained by radical extensions if and only if the Galois group has certain group-theoretic properties, properties that are then given the name “solvability” (for the group). That does not mean, by itself, that there is a “general radical formula”; but of course if there is at least one polynomial whose roots cannot be so expressed, then there can be no such “general formula”. A priori, it could be that every polynomial of degree $n$ is solvable by radicals without a general formula. – Arturo Magidin Apr 19 '20 at 01:28
  • I meant Galois Theory as a subject. This theorem is the whole idea of Galois and it took a lot of work to reach this point so climax of Galois theory seemed appropriate. – Partey5 Apr 19 '20 at 01:35
  • @Arturo_Magidin Is that because only radicals are the roots of general algebraic Equations? – Partey5 Apr 19 '20 at 01:44
  • I understand what solvability means but what does a galois group being isomorphic to insolvable group $S_5$ really say about the roots that means they can't be obtained by radicals. What was the big idea of Galois which made him think the Galois group resembling $S_5$ would mean there wouldn't a formula to get the roots? – Partey5 Apr 19 '20 at 01:49
  • Like I don't think Galois just stumbled upon this. He must have noticed something about the structure of $S_5$ which means you can't obtain the roots from a formula. I dont understand the thinking of why not being solvable would imply this (obviously this is proven but Galois must have had an intuition of why this must be the case). – Partey5 Apr 19 '20 at 01:54
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    The definition of “solvable by radicals” is that the roots can be expressed in terms of sum, products, differences, quotients, and radicals of the coefficients and such combinations. The second, third, and fourth degree general equations show that those can be solved that way in general. Some equations of higher degree can be solved that way. The point of the Theorem is to relate the solvability of the equation to a group which can be studied abstractly. As soon as you get one equation with group $S_5$, that equation cannot be solved by radicals, hence no general formula. – Arturo Magidin Apr 19 '20 at 02:05
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    As to what Galois was thinking, there are excellent books that go over the original papers and the antecedents, such as a recent translation (I think by Peter Neumann, but I could be wrong). Abel had already noticed that there was something different about the degree 5 (and higher) equations. He wasn’t just idling in the wind, he was working in a context. The objective wasn’t to end up with $S_5$, it was to study to roots and how they relate to the coefficients, and how their permutations relate to the coefficients (cont) – Arturo Magidin Apr 19 '20 at 02:07
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    See for example this. But now you are mixing two different things. The question seems to be about what the theorem says; now you seem to be asking “how could he have come up with the theorem?” That’s a very different question. – Arturo Magidin Apr 19 '20 at 02:08
  • And so if polynomial can't be solved by radicals then its roots can't all be expressed as radicals involving the ground field, say $\mathbb {Q}$ and therefore certainly not as radicals involving specifically the coefficients of the polynomial? – Partey5 Apr 19 '20 at 02:09
  • Sorry for mixing up the ideas. I think you have answered my written question I.e. the theorem does not say there is/ isn't a formula for the roots but we can conclude from the theorem that there isn't for degree 5 or higher a formula by looking at certain galois groups of quintics. However for <5 there is no implication of a formula since these groups are solvable. However as we know such formulas exist for < 5 we find there is only a general formula for <5. I keep this thinking of mine here as a reference for me to read again in future if I forget this. Thanks. – Partey5 Apr 19 '20 at 02:15
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    You are clearly confused about what “solvable by radicals” means. Perhaps you should start there instead of at the end of Galois Theory. Also, note that group theory as an abstract subject did not even exist when Galois died: Cayley would not introduce the abstract definition of a group until 1854, and Galois died in 1832. – Arturo Magidin Apr 19 '20 at 02:18
  • I have the definition: a polynomial over $K$ is solvable by radicals if there a Galois extension $L:K$ of radicals which contains its splitting field. – Partey5 Apr 19 '20 at 02:21
  • Which I like to think this means its roots must be radical expressions involving elements of K – Partey5 Apr 19 '20 at 02:22

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If there were a general formula for solving a degree $n$ polynomial by radicals, then certainly every degree $n$ polynomial would have solvable Galois group: the formula would apply, so the polynomial would be solvable, hence the Galois group would be solvable.

By contrapositive, if there were at least one degree $n$ polynomial that does not have solvable Galois group, then there cannot be a general formula for solving polynomials of degree $n$ by radicals.

One can show, explicitly, that there are polynomials of degree $5$ that have non-solvable Galois group (isomorphic to $A_5$ or to $S_5$). Thus, there can be no general formula for solving quintics by radicals.

In principle, it could be that each polynomial is solvable by radicals but there is no general formula; at least, we would not be able to discard that possibility just from the theorem you quote.

But in fact it turns out this is not the case, which can be checked by looking at the “general degree $n$ equation”, which means working with the polynomial $$F=(x-x_1)\cdots(x-x_n)$$ in the field $K(x_1,\ldots,x_n)$ of rational functions in $x_1,\ldots,x_n$ with coefficients in $K$. This polynomial has Galois group $S_n$, and so is not solvable by radicals for $n\geq 5$. This means that you cannot have a general formula for degree $n\geq 5$, as that formula would imply the solvability of $F$.

Arturo Magidin
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  • How can you decide a Galois group for a general equation without knowing its coefficients? Some quintics are isomorphic to $A_5$ and others $S_5$ like you mention. – Partey5 Apr 19 '20 at 11:27
  • @Anteater23: The “general equation” is the polynomial $(x-x_1)\cdots(x-x_n)$ with coefficients in $K(x_1,\ldots,x_n)$; you do know the coefficients, they are the elementary symmetric functions on $x_1,\ldots,x_n$. – Arturo Magidin Apr 19 '20 at 16:42
  • @AderinsolaJoshua: This is not related to my answer; it is in bad taste for you to clutter the comment threat in an attempt to publicize your question. – Arturo Magidin Apr 20 '20 at 00:40