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Kronecker's theorem and Abel's theorem are stated as follow.

Kronecker's theorem: Assume $f(x) \in \mathbb{Q}[x]$ is irreducible over $\mathbb{Q}$ and $\deg{f(x)=p}$ where $p$ is an odd prime number. If $f(x)$ is solvable by radical, then either all roots of $f(x)$ are real number or only one of the roots is real number.

Abel's theorem: Given $n\geq 5$, general polynomial of degree $n$ is not solvable by radical.

General polynomial in Abel's theorem is defined as follow.

Definition. Given n indeterminate $s_1,\cdots,s_n$. $f(x)=x^n+\sum_{i=1}^{n} (-1)^i s_ix^{n-i}$ is called a general polynomial of degree n. A general polynomial of degree 5 is $f(x)=x^5-s_1x^4+s_2x^3-s_3x^2+s_4x-s_5$.

I want to deduce Abel's theorem from Kronecker's theorem, but don't know how to.

Kronecker's theorem can give examples of insolvable polynomials, e.g. $f(x)=x^5-4x+2$.

If a root of general polynomial is like: $x=\frac{\cdots}{s_4 - 2\cdot s_5}$, then this formula cannot be applied to $f(x)=x^5-4x+2$. Because $s_4=-4, s_5=-2$ and $s_4 - 2\cdot s_5 = 0$ which cannot be denominator.

Roots of cubic polynomial $f(x)=x^3+px+q$ is $$ x = \zeta^i \sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}} + \zeta^j \sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}. $$ where $\zeta$ is a primitive cube root. This formula can be applied to any number $p,q$.

There is another formula for cubic polynomial: $$ x = \zeta^i \sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}} -\frac{p/3}{ \zeta^i \sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}}$$ This formula requires $-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}\neq 0$.

Hilbert Matthew Alexander
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  • The key result (which you mention as Kronecker's) is proved by Galois: an irreducible polynomial of prime degree with rational coefficients is solvable by radicals if and only if all the roots can be expressed as rational functions of any two of its roots. But Abel's theorem does not deal with polynomials with rational coefficients, but rather the polynomial has symbolic coefficients. – Paramanand Singh Nov 16 '22 at 01:14
  • Abel's theorem fits more nicely with the idea of Galois groups and solvability by radicals. The general polynomial has Galois group $S_n$ which is not solvable if $n\geq 5$. – Paramanand Singh Nov 16 '22 at 01:17
  • Thanks for comment. Kronecker's theorem I mentioned is from a book named "100 Great Problems of Elementary Mathematics: Their History and Solution". Proof in the book uses the fact that radical extension can be viewed as linear space over base field to judge wheter roots are real or complex. Abel's and Kronecker's concern different things, i.e. rational coefficient polynomial and symbolic coefficient polynomial. Hence there are no straight connection between this two theorems unless there is a general formula? @ParamanandSingh – Hilbert Matthew Alexander Nov 16 '22 at 02:14
  • I don't see any direct connection between the two theorems unless we bring Galois groups into picture. Kronecker's theorem is a corollary of result proved by Galois. Galois proved his result using Galois groups. – Paramanand Singh Nov 16 '22 at 04:05
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    Assume that $p$ is a prime $>3$. Using the idea from Cam McLeman's old answer we can easily find sequences of irreducible polynomials of degree $p$ with exactly three real zeros. By Kronecker's theorem such a polynomial is not solvable by radicals. It seems possible to get your desired formulation of Abel's theorem from that sequence (perturbation of the coefficients becomes possible, and that could do it). – Jyrki Lahtonen Nov 16 '22 at 10:35
  • It seems similar to the idea of Arnold's topological proof of Abel theorem.
    If it is necessary to use such technique, then the two theorems I mentioned are actually independent. @JyrkiLahtonen     – Hilbert Matthew Alexander Nov 16 '22 at 11:54
  • I cannot make a precise comment @HilbertMatthewAlexander because we don't have a precise statement of Abel's theorem at hand. A key difference seems to be that you want Abel's theorem to apply to generic polynomials with coefficients drawn from a generic field. Algebra textbooks often use coefficients of a quintic that are algebraically independent transcendental elements over some base field, where as Kroneckers's theorem discusses polynomials over $\Bbb{Q}$ in particular. The missing link may be around specializations, i.e. assigning values to those transcendentals. – Jyrki Lahtonen Nov 16 '22 at 12:04
  • (cont'd) That is a far less common aspect of all this. I, for one, am largely ignorant about the basic results about such specializations. I'm just happy with the textbook versions of the results.

    Esoteric fields with suitable transcendentals is a must here. For otherwise the fundamental theorem of algebra can be seen as saying that every polynomial from $\Bbb{C}[x]$ is solvable because the splitting field is a trivial extension and hence has a solvable Galois group - clearly not what Abel wanted to say!

    I should study the topic myself before I can say anything definite. Sorry.

     – Jyrki Lahtonen Nov 16 '22 at 12:09
  • No need to apologize. Overall I just want to assign value. Rational fraction of symbols only requires the denominator is not zero polynomial. But fraction of number requires the denominator is not 0. Maybe it's better to just use Galois theory which is very clear. @JyrkiLahtonen – Hilbert Matthew Alexander Nov 16 '22 at 12:33
  • @JyrkiLahtonen: the version of Abel's theorem which I am familiar with deals with the generic polynomial $x^n+a_1x^{n-1}+\dots+a_n$ over $\mathbb{C} (a_1,\dots,a_n)$. Abel's own proof requires existence of roots of unity in base field of coefficients. I don't know if the result holds otherwise. – Paramanand Singh Nov 16 '22 at 15:35
  • Thanks @ParamanandSingh. I'm afraid I have never looked at Abel's original version. The Galois theoretic version (dealing with number fields) does not require roots of unity. Or rather (see Jacobson) leaving the roots of unity out or including them does not change the solvability of the Galois group of the splitting field. – Jyrki Lahtonen Nov 16 '22 at 17:06
  • I found a paper which complained that there are several different statement of Abel's theorem at page 5 remark 8. According to the paper, I have to use this version of Abel's theorem:"There exists insolvable quintic equation.". @JyrkiLahtonen – Hilbert Matthew Alexander Nov 18 '22 at 01:57

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