3.7 Solvability of equations

For a polynomial $f\in F[X]$ , we say that $f(X)=0$ is solvable in radicals if its solutions can be obtained by the algebraic operations of addition, subtraction, multiplication, division, and the extraction of $m$ th roots, or, more precisely, if there exists a tower of fields

F=F_{0}\subset F_{1}\subset F_{2}\subset\cdots\subset F_{m}

such that

(a)

$F_{i}=F_{i-1}[\alpha_{i}]$ , $\alpha_{i}^{m_{i}}\in F_{i-1}$ ;
(b)

$F_{m}$ contains a splitting field for $f.$

Theorem 3.27 (Galois, 1832)

Let $F$ be a field of characteristic zero, and let $f\in F[X]$ . The equation $f(X)=0$ is solvable in radicals if and only if the Galois group of $f$ is solvable.

We’ll prove this later (LABEL:ag23). Also we’ll exhibit polynomials $f(X)\in\mathbb{Q}[X]$ with Galois group $S_{n}$ , which are therefore not solvable when $n\geq 5$ by GT, 4.37.

Remark 3.28

When $F$ has characteristic $p$ , the theorem fails for two reasons,

(a)

$f$ need not be separable, and so not have a Galois group;
(b)

$X^{p}-X-a=0$ need not be solvable in radicals even though it is separable with abelian Galois group (cf. Exercise 2-2).

If the definition of solvable is changed to allow extensions defined by polynomials of the type in (b) in the chain, then the theorem holds for fields $F$ of characteristic $p\neq 0$ and separable $f\in F[X]$ .

Notes

Much of what has been written about Galois is unreliable — see Tony Rothman, “Genius and Biographers: The Fictionalization of Evariste Galois,” Amer. Math. Mon. 89, 84 (1982). For a careful explanation of Galois’s “Premier Mémoire”, see Edwards, Harold M., Galois for 21st-century readers. Notices A.M.S. 59 (2012), no. 7, 912–923.