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Very Quick Sketch of Galois Theory

(I'm just writing this to help myself remember. YMMV.)

Question: we can write down a general formula for the roots of a quadratic, cubic, or quartic polynomial in terms of the coefficients. Why can't we do it for a quintic polynomial?

Explanation: Extend $\mathbb{Q}$ by taking roots and you will introduce certain symmetries (automorphisms which leave $\mathbb{Q}$ fixed). All the automorphisms introduced this way are cyclic in structure.

The groups $S_2, S_3, S_4$ all have the property of decomposing in a certain nice way: they have a chain of normal subgroups which leave behind a cyclic group when quotiented out. But $S_{\geq 5}$ does not have this property (not "solvable")*.

The theorem: a polynomial $p$ is solvable by radicals $\Leftrightarrow$ $\text{Gal}(p)$ is solvable. It's true because of the isomorphism between field extensions and Galois subgroups. If we wanted to build a tower of field extensions (by taking roots) which culminated in a field that could hold the roots of the quintic, we would have to be able to add cyclic groups the whole way. But we can't reach $S_{\geq 5}$ this way, because no chain.

* The obstruction is that $A_5$, the only candidate to start the chain off with, doesn't have any normal subgroups ("simple").


  • This lovely diagram of the isomorphism for $p = x^4 - 2$.
  • Implication which I somehow never realized until right now: there exist algebraic numbers, i.e. the roots of some $p \in \mathbb{Q}[x]$, which we have no "formula for". That is, no agreed-upon way to write them other than just exhibiting $p$. (You might object that this is a matter of convention and no different than the situation with $\sqrt{2}$, and you might have a point.) At any rate, these are called non-solvable numbers.