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Can we always break an arbitrary field extension $L/K$ into an extension $F/K$ in which the only roots of unity of $F$ are those in $K$, followed by an extension $L/F$ which is of the form $L=F(\{\omega_i\})$ where the $\omega_i$ are roots of unity? What if we restrict to $L/K$ separable, or finite?

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@Zev: not a proof/counterexample, but an observation: suppose $L$ is Galois over $K$; we can let $M$ be the extension of $K$ obtained by adding all roots of unity in $L$; this is Galois over $K$, so corresponds to a normal subgroup $H$ of $G=\mathrm{Gal}(L/K)$. If we can break up the extension as you mention, then $L$ is Galois over $F$, and $\mathrm{Gal}(L/F)=\mathrm{Gal}(M/K) = G/H$. So $G$ would necessarily have normal subgroup $H$ and a subgroup isomorphic to $G/H$. –  Arturo Magidin Nov 7 '10 at 1:37
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