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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?

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    $\begingroup$ Is "Galois Theory" a particular textbook? If so, please edit your question to include the details of the book (e.g., its author(s)). $\endgroup$
    – Shaun
    Apr 19, 2020 at 1:00
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    $\begingroup$ And what does this have to do with "climax"? (An end need not be a "climax.") $\endgroup$ Apr 19, 2020 at 1:14
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    $\begingroup$ 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. $\endgroup$ Apr 19, 2020 at 1:28
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    $\begingroup$ 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. $\endgroup$ Apr 19, 2020 at 2:05
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    $\begingroup$ 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) $\endgroup$ Apr 19, 2020 at 2:07

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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 $F$.

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  • $\begingroup$ 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. $\endgroup$
    – Partey5
    Apr 19, 2020 at 11:27
  • $\begingroup$ @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 $x_1,\ldots,x_n$. $\endgroup$ Apr 19, 2020 at 16:42
  • $\begingroup$ @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. $\endgroup$ Apr 20, 2020 at 0:40

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