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  "body": "\n## Statement\n\nLet $K$ be a field of [[characteristic|1ZB13N78XW]] $0$, and let $f \\in K[x]$ be a nonconstant polynomial.\nThen $f$ is [[solvable by radicals|J7YS0DA3HQ]] over $K$ if and only if the [[Galois group of $f$|0JZY3R908H]] over $K$ is a [[solvable group|0P3KZK0FTV]].\n\n## Remarks\n\nFor one direction, put a radical extension containing the roots of $f$ inside a Galois radical extension that first adjoins enough roots of unity.\nEach later step then adjoins an $n$-th root in the presence of the $n$-th roots of unity, which gives an abelian extension.\nSo the big Galois group is solvable, and so is its quotient, the Galois group of $f$.\nFor the other direction, a solvable group has a chain of subgroups with cyclic quotients of prime order.\nAfter adjoining roots of unity, each cyclic step of prime degree $q$ is generated by a $q$-th root, found with a [[Lagrange resolvent|2GDFFH7Y7T]].\nExample: $x^5 - 4x + 2$ over $\\Q$ has Galois group $S_5$, which is not solvable, so its roots cannot be written with radicals.\nThe characteristic hypothesis matters: in characteristic $p$, the [[Artin–Schreier|7YNMNRGH3T]] extensions are cyclic of degree $p$ but in general not radical.\n\n## Explanations\n\n### L1 human\n\nThe theorem answers an old question: when can the roots of a polynomial be written with $+$, $-$, $\\times$, $\\div$ and $n$-th roots?\nGalois' answer is to look at the group of permutations of the roots that respect all algebraic relations.\nThe roots have such a formula exactly when this group can be broken down into commutative pieces.\n\n### L2 human\n\nThe theorem reduces a question about formulas to a question about finite groups.\nSince $S_n$ is solvable for $n \\le 4$, every polynomial of degree at most $4$ is solvable by radicals, which matches the [[cubic formula|FB6J5WAT5V]] and the [[quartic formula|4CDFMKKZ01]].\nSince [[$S_n$ is not solvable for $n \\ge 5$|0J53NE674D]], the [[Abel–Ruffini theorem|163A2WQQJX]] follows once a polynomial with Galois group $S_n$ is found.\nThe proof uses the [[fundamental theorem of Galois theory|0T1Q8D0FW8]] together with [[Kummer theory|XZQENYV15Y]] for the cyclic steps.\n",
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