Impossibility of the Quintic Formula

Statement

There is no general formula using radicals (addition, subtraction, multiplication, division, and $n$-th roots) that solves all polynomial equations of degree $\geq 5$.

The Chain of Ideas

1. Field extensions: Given a polynomial $f(x)$ over $\mathbb{Q}$, adjoin its roots to get the splitting field $K$.
2. Galois group: The group $\text{Gal}(K/\mathbb{Q})$ permutes the roots of $f$. The Fundamental Theorem of Galois TheoryFundamental Theorem of Galois Theory$$\{\text{intermediate fields}\} \longleftrightarrow \{\text{subgroups of } \text{Gal}(K/F)\}$$Bijection between intermediate fields and subgroups of the Galois groupRead more → turns subgroups of this group into intermediate fields.
3. Solvable groups: A polynomial is solvable by radicals if and only if its Galois group is a solvable group.
4. $S_5$ is not solvable: The symmetric group $S_5$ contains $A_5$, which is simple and non-abelian.
5. Conclusion: The general quintic has Galois group $S_5$, hence is not solvable by radicals.

Concrete Example

The polynomial $f(x) = x^5 - 4x + 2$ has Galois group $S_5$ over $\mathbb{Q}$ — it is irreducible by Eisenstein's criterion at $p = 2$, and one can verify that the discriminant is not a perfect square.

Connections

This result is a direct consequence of the Fundamental Theorem of Galois TheoryFundamental Theorem of Galois Theory$$\{\text{intermediate fields}\} \longleftrightarrow \{\text{subgroups of } \text{Gal}(K/F)\}$$Bijection between intermediate fields and subgroups of the Galois groupRead more →. The question of Constructible NumbersConstructible Numbers$$\text{Regular } n\text{-gon constructible} \iff n = 2^a \cdot p_1 \cdot p_2 \cdots p_k$$Which regular n-gons can be constructed with compass and straightedge?Read more → is also resolved using similar group-theoretic techniques.