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When trying to answer the question of whether a given equation can be solved with radicals, historically people have paid lots of attention to permutations that preserve all algebraic relations between the roots of the polynomial.

The idea finally led to the solution of the problem by Galois, but the idea of looking at such permutations pre-dates him.

Is there any nice simple explanation of why such permutations are related with solve-ability of polynomial equations with radicals? (An explanation that doesn't directly involve the Galois group).

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I have a vague recollection of people having studied resolvents before Galois. It just happens that in the quadratic, cubic and quartic cases the resolvent is of a lower degree. This fails with quintics. – Jyrki Lahtonen Aug 9 '13 at 20:08
@JyrkiLahtonen, that'd be Lagrange. – lhf Aug 12 '13 at 1:26
up vote 3 down vote accepted

Solving polynomials is all about breaking symmetry. In order to describe the solutions to $x^2 - 2 = 0$, we need to break the symmetry between $\sqrt{2}$ and $-\sqrt{2}$, for example by introducing the notion of positivity and saying that one root is positive and one root is negative. In order to describe the solutions to $x^2 + 1 = 0$, we need to break the symmetry between $i$ and $-i$, for example by introducing the complex plane and the notion of counterclockwise rotation.

If the work of solving polynomials consists of breaking symmetries, then this suggests that the more symmetry there is to break, the harder the polynomial will be to solve. For example, a polynomial that factors into linear factors over $\mathbb{Q}$ has no symmetries. A polynomial that factors into quadratic factors has symmetries involving exchanging the roots of each quadratic. And so forth.

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Thanks for this nice reply, but is there any nice way to see why one needs to break these symmetries in order to solve the polynomial. (This may be obvious but I wouldn't know how to explain it in a simple way). – Timotej Aug 10 '13 at 12:01

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