2
$\begingroup$

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,

$\endgroup$
6
$\begingroup$

Given a finite field extension $F \subseteq K$. A $\textbf{root tower}$ of $K$ over $F$ is a finite sequence of extensions

$$F = K_1 \subseteq K_2 \subseteq \dots \subseteq K_n = K$$

Such that for every $i$ where $1 \le i \le n - 1$, there exists a prime $p_i$ and an element $q_i \in K_{i+1}$ such that $q_i^{p_i} \in K_i$ and $q_i \notin K_i$.

If $f(x) \in F[x]$ where $char F = 0$, we say that the equation $f(x) = 0$ is solvable by radicals, if there exists a finite extension of the splitting field of $f(x)$ that has a root tower over $F$.

$\endgroup$
  • 1
    $\begingroup$ IMHO this is a bit too much for a school paper. :) $\endgroup$ – Michael Mar 21 '14 at 3:32
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @lhf Yes, I forgot that. It is important. $\endgroup$ – Stephen Montgomery-Smith Mar 21 '14 at 1:47
  • $\begingroup$ And it should be stressed that you can't also use conditionals ("if ..."), since this would allow you to obtain, for example, the Jordan form of the companion matrix... $\endgroup$ – Jay Jun 10 '14 at 11:03
1
$\begingroup$

You may enjoy reading these books:

and this expository paper:

See also these answers: 1, 2.

$\endgroup$
1
$\begingroup$

This is the standard dogma about solvability by radicals:

A polynomial is solvable by radicals if every root of the polynomial can be generated from rational numbers using the operations $+,-,\times,\div$, and taking $n$th roots.

For instance, this is a root that can be expressed in terms of radicals: $$ \sqrt[5]{1+\sqrt[4]{\frac{27}{8} + \sqrt 2}} - \sqrt[5]{1+\sqrt[4]{\frac{27}{8} - \sqrt 2}}. $$

I say "dogma" because the definition obfuscates a subtle but important point -- I didn't catch it until I took a class where we proved the Abel-Ruffini theorem. When I say "taking $n$th roots" I mean "taking $n$th roots of a complex number". In general, a complex number has $n$ different roots, and they are trigonometric in nature. For example, the number $1$ has seven $7$th roots, which look like this: $$ 1,\cos\Big(1\cdot\frac{2\pi}{7}\Big)+i\sin\Big(1\cdot\frac{2\pi}{7}\Big),\dots, \cos\Big(6\cdot\frac{2\pi}{7}\Big)+i\sin\Big(6\cdot\frac{2\pi}{7}\Big) $$ In other words, those seven numbers are the roots of the polynomial equation $x^7=1$. According to the standard dogma, those numbers above are "expressible as radicals": indeed, each is $\sqrt[7]{1}$. But this seems wrong according to our intuition since the roots contain trigonometric constants, not radical expressions per se.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.