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Let $F$ be a field and $p(x)\in F[x]$ a separable polynomial, denote $K$ as the splitting field of $p$ and assume that $K/F$ is Galois with a solvable Galois group.

I don't understand if this imply of any formula (in radicals) for the roots of $p$ (however, I do understand how a formula would imply that $p$ is solvable by roots).

Is there some kind of a way to obtain the roots of a solvable polynomial ?

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up vote 7 down vote accepted

Lagrange and Vandermonde (and others) knew how to treat various cases before "Galois theory", by "Lagrange resolvents". Some of my algebra notes show how this idea recovers the formula for the solution of cubics:

By the late 19th century such ideas were well known, and in many cases of solvable Galois groups (though people were not quite able to say things so simply) this device produce solutions in radicals.

Expressing roots of unity in radicals is another example where Lagrange resolvents gives expressions in radicals. This quickly becomes computation-heavy, however. A slightly more sophisticated implementation of "Lagrange resolvents" that does bigger cyclotomic examples before getting bogged down is worked out at In that case, some additional ideas of algebraic number theory are used.

Van der Waarden's "Algebra" (based on notes from E. Noether) was the place I saw Lagrange resolvents, and it was a revelation. Indeed, many treatments of "Galois theory" give no inkling of how to do any computations.

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