What is Kronecker's theorem/lemma for root solutions?

[swampwiz]

I am going through this article, and it mentions a Kronecker made a discovery about odd prime degree polynomials that makes Abel's Theorem on the impossibility of the quintic easier to prove:

https://hubpages.com/education/Abel...oduction-to-the-Sublime-Beauty-of-Mathematics

Obviously, Kronecker has produced a lot of mathematical work, and I find it hard to find his proof on this.

 

[fresh_42]

If we had more than one zero in a field extension, then there would be a field in between with those zeros and no others of degree greater than two. But the total degree of the splitting field is prime and the degree of the smaller field would be a divisor of that prime. Thus there can only be one zero or already all.

 

[swampwiz]

If we had more than one zero in a field extension, then there would be a field in between with those zeros and no others of degree greater than two. But the total degree of the splitting field is prime and the degree of the smaller field would be a divisor of that prime. Thus there can only be one zero or already all.

Hi, do you have link that discusses the proof for this?

 

[fresh_42]

I hope it's in here: https://www.jmilne.org/math/CourseNotes/FT.pdf but any book on Galois theory should do. The formula for $K \subseteq L \subseteq M$ is $[M:L]\cdot [L:K]=[M:K]\,.$

 

[sysprog]

The textbook, https://danboak.files.wordpress.com/2017/08/afirstcourseinabstractalgebra.pdf, by Marlow Anderson and Todd Feil, has a proof of Kronecker’s Theorem, and uses and discusses it extensively.

Summary for Chapter 48 Solving Polynomials by Radicals:

In this chapter we prove that if a polynomial over a subfield of the complex numbers can be solved by radicals, then the Galois group of its splitting field over the base field is necessarily a solvable group. We can then easily exhibit a fifth degree polynomial over the rational numbers that cannot be solved by radicals.​