I think that I am misunderstanding something fundamental about the technique used to decide if higher order polynomials are solvable by radicals using Galois theory. If we have a cubic it's not to bad because it's just a matter of deciding whether the discriminant is a square or not ( because if the discriminant is not square then clearly its square root will not be rational and so a subfield must exist $\Rightarrow, Gal(f)\cong S_3$ as $C_3$ has no subgroups.) But when we get to quintics I'm confused (Note: I really just want understand this method first without bringing in any additional theory that I haven't mentioned here. ) Also I may be misunderstanding multiple parts So I'll label the steps and my questions in each.

So say that we have the quintic equation $f(x)=x^5-6x+2 $. (Which is irreducible over $\Bbb Q$ Eisenstein $p=2$)

First we use Rolle's theorem to analyse the number of roots $f'(x)=5x^4-6$, so there are two real roots which implies that $f(x)$ has at most 3 real roots .Then using the intermediate value thorem we see that it has exactly three real roots, moreover the roots of the derivative are not in the intervals where $f$ has roots and so this equation is separable, and now we know that the quintic must have 3 real roots and two complex conjugate roots.

Question 1: This is all when and good because we got lucky and there happened to be exactly three real roots using the IVT but if there had only been two or one, then I don't see how this method could work as the IVT wouldn't tell you anything, you would just have to use your intuition and say well its seems like this is only going to have so many roots. Is that the best we can do or is there a method to determine that it has exactly two or one real root.

Furthemore if we know that it has three real roots then we know that it has a complex conjugate pair , I assume that if it has 1 real root we can take it that it has two complex conjugate conjugate pairs. But what if it has two real roots , does that then imply that at least one of its roots is not distinct ?

Step 2: We know that the total number of roots is 5 and so any possible Galois group must be a subgroup of $S_5$. We also know that it contains a transposition and a three cycle so it must a group which is larger than or equal to $S_3$.

Now here is where I get confused , my lecturer says that we let $E$ be a splitting field $\Bbb Q(\alpha)\subsetneq E$, where $\alpha$ is one of the roots of the polynomial, $|\Bbb Q(\alpha):\Bbb Q|=degf=5$ ( sure because f was irreducible and monic so it's the minimal polynomial of $\alpha$).

$|Gal(E/F)|=|E:\Bbb Q|\Rightarrow 5||Gal(E/F)|$. So then by Cauchy's theorem we know that an element of order 5 exists and so a 5 cycle exists and therefore the Galois group is isomorphic to $S_5$ which is not solvable.

My question is how is this valid if we choose a root and find the degree of it's extension over $\Bbb Q$ then won't it always be 5 and so wouldn't that imply that the group will always be $S_5$, which I know is not true as not all quintics are solvable by radicals . So what am I missing here ?

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    $\begingroup$ In your last paragraph, I believe you mean "not all quintics are not solvable by radicals". The method you mention here relies on the quntic being irreducible and the Galois group containing a transposition, either one of which is not always the case. $\endgroup$ – RghtHndSd Aug 11 at 13:37
  • $\begingroup$ Essentially, where there are no transposition elements, we have the Galois group $A_5$. $\endgroup$ – Don Thousand Aug 11 at 13:54
  • $\begingroup$ @DonThousand Okay sure so the final paragraph where we deduce that 5 divides the order of the group will always be true for quintics ( as there are 5 cycles in $A_5$ too ), but I still don't understand how we can determine there are no transposition elements because the Ivt only really works if it turns out we do have exactly three real roots $\endgroup$ – excalibirr Aug 11 at 14:05
  • $\begingroup$ See this $\endgroup$ – Don Thousand Aug 11 at 14:07

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