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  "body": "\n## Statement\n\nLet $G$ be a finite [[solvable group|0P3KZK0FTV]].\nThen there is a [[Galois extension|0DS2F7GK05]] $L/\\Q$ whose [[Galois group|0XVR72M51M]] $\\Gal(L/\\Q)$ is isomorphic to $G$.\n\n## Remarks\n\nThe proof is long and uses class field theory and embedding problems over number fields.\nIt builds the extension in steps, and each step solves an embedding problem.\nGiven a normal subgroup $N$ of $G$ and an extension with group $G/N$, one must find a larger extension with group $G$ that contains it.\nExample: the symmetric group $S_4$ is solvable, so it is a Galois group over $\\Q$; the splitting field of $x^4 - x - 1$ is one such extension.\nThe theorem answers the [[inverse Galois problem|YJS6X8JN7D]] for solvable groups only; nonsolvable groups need other methods.\n\n## Explanations\n\n### L1 human\n\nThe inverse Galois problem asks which finite groups arise as symmetry groups of rational polynomials.\nShafarevich's theorem says that all solvable groups do.\nThese are the groups that can be built from commutative pieces, the same groups that correspond to equations solvable by radicals.\n\n### L2 human\n\nAbelian groups are easy to realise inside cyclotomic fields, but solvable groups need many successive extensions.\nThe difficulty lies in the embedding problem: an extension with a given quotient group need not extend to one with the larger group.\nThe theorem holds over every number field, not only over $\\Q$.\n",
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