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,
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1Possible duplicate of http://math.stackexchange.com/questions/116814/what-does-insolvability-of-the-quintic-mean-exactly. – lhf Mar 21 '14 at 01:48
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Relevant: How is the general solution for algebraic equations of degree five formulated? – MJD Mar 21 '14 at 02:01
4 Answers
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.
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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... – Jay Jun 10 '14 at 11:03
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 $F$.
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Why assume that the splitting field embeds into a radical tower, can't we say the splitting field is a radical tower? Is there a counterexample? – Espace' etale Jul 13 '20 at 11:50
You may enjoy reading these books:
Abel's Proof by Peter Pesic
Galois Theory for Beginners: A Historical Perspective by Jörg Bewersdorff (MAA review)
and this expository paper:
- Abel and the Insolvability of the Quintic by Jim Brown
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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.
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Regarding your concluding comments, look up the phrase real radicals, which is the name for a certain proper subfield of the real algebraic numbers. – Dave L. Renfro Mar 21 '14 at 15:14
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See also rational angles with sines expressible with radicals and When $\cos{n^{\circ}}$ can be expressed in real radicals?, and the references/web-pages given in the several comments I left in various places at each of these. – Dave L. Renfro Mar 21 '14 at 17:41