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Questions:

  • Does there exist a proof of the Abel-Ruffini theorem without using Galois theory?
  • Does there exist a proof that there exists a polynomial $P$ with $\deg P = 5$ such that the roots are not expressible in radicals? Can this be proven without Abel-Ruffini?
  • Does there exist a polynomial $P$ with $\deg P = 6$ such that the minima and maxima are not expressible in radicals? Can this be proven without Abel-Ruffini?

I originally had the following proof for the third question:

The minima and maxima of a 6th degree polynomial are the roots of its derivative, which is a fifth degree polynomial. There exists a 5th degree polynomial such that the roots are not expressible in radicals, otherwise the roots of every 5th degree polynomial would be expressible in radicals, contradicting the more general form of the Abel-Ruffini theorem proven by Galois. Therefore there exists a 6th degree polynomial whose minima and maxima are not expressible in radicals.

However, I realized only until later, that it also can be done vice versa. However, the question brought up the several other questions which I listed on top of the question. Therefore I have clarified those questions up in the answer below.

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  • $\begingroup$ Correct me if I'm wrong, suppose there's a proof that doesn't depend on Galois$\iff$ it can be proven that there exists a polynomial of degree $\ge5$ whose roots aren't expressible in radicals without Galois $\endgroup$ – Jack's wasted life Jul 28 '15 at 14:44
  • $\begingroup$ Just vor the records: you've read en.m.wikipedia.org/wiki/Quintic_function#Solvable_quintics, don't you? $\endgroup$ – Michael Hoppe Jul 28 '15 at 19:51
  • $\begingroup$ Proving something algebraic is not expressible in radicals without using Galois theory ???? I'm quite skeptical about that. $\endgroup$ – Ewan Delanoy Jul 31 '15 at 15:45
  • $\begingroup$ @EwanDelanoy (1) Where did I claim they had to be global? When talking about maxima and minima, one talks about both global and local ones. (2) Abel-Ruffini theorem has been proven without Galois theory. $\endgroup$ – wythagoras Jul 31 '15 at 15:56
  • $\begingroup$ @wythagoras (1) Sorry, my bad. I dont understand your point (2) though. It would seem from your original question that you want neither the Abel-Ruffini theorem nor Galois theory. $\endgroup$ – Ewan Delanoy Jul 31 '15 at 16:10
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Let me try to explain my understanding of what Abel and Ruffini did. I am thinking of using this as my opening lecture next time I teach Galois theory, so this is useful for me. Disclaimer: I am a mathematician, not a historian.


Here is a theorem which I can prove in the space on this page.

Theorem. Let $L= \mathbb{C}(x_1,\ldots, x_5)$ be the ring of rational functions in $5$ variables, and let $K$ be the subfield of symmetric functions. So $K$ is generated by $e_1 = x_1+x_2+\cdots+x_5$, $e_2=x_1 x_2 + x_1 x_3 + \cdots + x_4 x_5$, ... and $e_5 = x_1 x_2 \cdots x_5$. Suppose we add elements to $K$ by the operations of $+$, $-$, $\times$, $\div$ and $\sqrt[n]{}$, while staying within $L$. Then we can never reach the element $x_1 \in L$.

Proof: The group $S_5$ clearly acts on $L$ by permuting the $x_i$. Let $\sigma$ and $\tau$ be the permutations $(123)$ and $(345)$, and let $F$ be the field of functions in $L$ fixed by $\sigma$ and $\tau$. Clearly, $K \subset F$. We claim that you cannot escape $F$ by the operations $+$, $-$, $\times$, $\div$, $\sqrt[n]{}$. For the first four operations, this is obvious.

Suppose that $b \in L$ and $b^n=a \in F$. Then $\sigma(b)^n=\sigma(b^n) = \sigma(a) = a=b^n$. So $(\sigma(b)/b)^n=1$ and $\sigma(b) = \zeta b$ for some $n$-th root of unity $\zeta$. Similarly, $\tau(b) = \omega b$ for some $n$-th root of unity $\omega$.

Now, $\sigma^3=\mathrm{Id}$ so $b=\sigma^3(b)=\zeta^3 b$. Similarly, we deduce that $\omega^3=1$. Also, $(\sigma \tau)^5=\mathrm{Id}$, so $\zeta^5 \omega^5=1$, and $(\sigma^2 \tau)^5 = \mathrm{Id}$ so $\zeta^{10} \omega^5=1$. Combining all of these equations, $(\zeta^3)^2 (\zeta^5 \omega^5) (\zeta^{10} \omega^5)^{-1} = \zeta =1$ and, similarly, $\omega=1$. So $\sigma(b) = \tau(b) = b$, and $b$ is in $F$ after all.

Since $x_1 \not \in F$, we have proved the theorem $\square$.


Note that the quadratic, cubic and biquadratic formulas do stay within $L$.

For example, the cubic formula is (Double check before using!) $$x_1 = \frac{1}{3} \left( \sqrt[3]{\frac{S+\sqrt{S^2+4 T^3}}{2}} + \sqrt[3]{\frac{S-\sqrt{S^2+4 T^3}}{2}} + e_1 \right)$$ $$S = 2 e_1^3-9 e_1 e_2+27 e_3, \quad T= 3e_2-e_1^2.$$ The functions $S$ and $T$ are formed by field operations and I believe you will find that $\sqrt{S^2+4T^3} = (x_1-x_2)(x_1-x_3)(x_2-x_3)$ and $$\sqrt[3]{\frac{S+\sqrt{S^2+4T^3}}{2}} = x_1 + \frac{-1+\sqrt{3} i}{2} x_2 + \frac{-1-\sqrt{3}i}{2} x_3$$ So we have stayed within $L$ while working our way up to $x_1$.


I explain how to think about this computation in modern terms here. In brief, the group generated by $\sigma$ and $\tau$ is $A_5$, and the above computations verify that $A_5^{ab}$ is trivial.

As I understand it, both Abel and Ruffini gave correct proofs of the theorem in the box. These proofs were basically the same as the above, but longer because words like "field", "group" and "action" hadn't been invented yet. Both of them aimed to go further, and show that there was no universal formula for $x_1$ in terms of $e_1$, $e_2$, ..., $e_5$, $+$, $-$, $\times$, $\div$ and $\sqrt[n]{}$, whether or not we are required to stay in $L$. This was very confusing, as it wasn't clear what sort of algebraic object the formulas would live in once we left $L$.

As I understand it, the current consensus is that Ruffini's attempt failed, but Abel succeeded by proving the result now called the Theorem of Natural Irrationalities.

From a modern Galois theory perspective, there is no problem. Suppose we have some chain of fields $K \subset F \subset L \subset M$, and $K=K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_r = M$, where each $K_{i+1}/K_i$ is a radical extension. Without loss of generality, we may assume $M/K$ is Galois, with Galois group $G$. Then the chain of radical extensions shows $G$ is solvable. But the chain $K \subset F \subset L \subset M$ shows that $A_5 = \mathrm{Gal}(L/F)=\mathrm{Gal}(M/F)/\mathrm{Gal}(M/L)$ occurs as a composition factor for $G$, so $A_5$ must also be solvable. This contradicts that our computation that $A_5^{ab}$ is trivial.

What Galois contributed was the ability to work with arbitrary field extensions, not just polynomials/functions with various symmetry, and therefore be able to speak cleanly about the field $L$ and its symmetries. He also was the first to create tools that would let us prove a particular quintic over $\mathbb{Q}$ was not solvable by radicals.

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  • $\begingroup$ Very-very big thanks and +1 (actually +10) for greate explanation. But what can you say about initial problem (about minima/maxima)? $\endgroup$ – Michael Galuza Aug 5 '15 at 13:31
  • $\begingroup$ @David Speyer. Thank you for the great answer. However I have one question. Why $\sigma(b)=\zeta b$? $\endgroup$ – wythagoras Aug 5 '15 at 14:23
  • $\begingroup$ @wythagoras $\sigma(b)^n = \sigma(b^n) = \sigma(a) = a = b^n$. So $(\sigma(b)/b)^n=1$. Note that $\sigma(b)/b$ is a rational function of $x_1$, ..., $x_5$, so it is going to be some particular $n$-th root of unity $\zeta$. I'll see if I can edit something to help. $\endgroup$ – David E Speyer Aug 5 '15 at 15:31
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There cannot really be a simpler proof, in the sense that this result immediately implies Abel-Ruffini for 5th degree polynomials. Note that user wythagoras (comment on Aug 1st) says Abel-Ruffini has been proven without Galois theory, which I'd like to see.

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Q1. Yes. Abel's original proof is an example. Another example is the proof provided by Arnold, which makes use of group theory and topology.

Q2. This implies the Abel-Ruffini theorem since if there exists a polynomial $P$ with $\deg P = 5$ such that the roots are not expressible in radicals there is certainly no general formula that gives the roots. Note that Abel-Ruffini doesn't imply this. In fact the result is by Galois.

Q3. If there exists a polynomial $P$ with $\deg P = 6$ such that the minima and maxima are not expressible in radicals, then the roots of the derivative, a fifth degree polynomial, are not expressible in radicals. By the result in Q2, we conclude this also implies the Abel-Ruffini theorem.

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