I am writing a school paper about the Abel-Ruffini theorem. My teacher wants me to explain what it means when a polynomial equation is solvable by radicals. My best guess is that it means that the answer contains a root,
[Math] What does it mean for a polynomial to be solvable by radicals
polynomialsradicals
Related Solutions
As you might already know, solutions to the quintic can be expressed in terms of either ${}_4 F_3$ hypergeometric functions or Jacobi theta functions. See King or Prasolov/Solovyev for details.
For polynomials of higher degree, there is also a general formula for the roots, due to Umemura. The formulae involve the multidimensional generalization of the Jacobi theta functions (the Riemann theta function), and are a bit unwieldy; see Umemura's paper if you want more details. See also this preprint for a solution of the reduced polynomial equation $x^n-x-\alpha=0$ in terms of hypergeometric functions.
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.
Best Answer
It means that the answer can be computed using only the operations $+$, $-$, $\times$, $\div$, and $n$th roots, in a finite number of steps, using the coefficients of the equation.