I am asking my question here, since it's been voted up a fair bit on math.SE, but without answers, so it may be harder than I assumed it was.

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 $K$, i.e. $\mu_F=\mu_K$, followed by an extension $L/F$ which is of the form $L=F(\{\omega_i\})$, where the $\omega_i$ are roots of unity? If not, are there reasonable hypotheses (e.g. separable, finite) on $L/K$ that would make this true?

My motivation was simply that the other order, i.e. breaking an arbitrary extension $L/K$ into one where $F=K(\{\omega_i\})$ for some roots of unity $\omega_i$, followed by $L/F$ where $\mu_L=\mu_K$, is obvious - specifically, set $F=K(\mu_L)$.

Now, my first (naive) approach was to try to construct the "maximum" intermediate field which does not add roots of unity by taking the compositum of all such intermediate extensions. However, this doesn't exist even for number fields, e.g. setting $K=\mathbb{Q}$, $L=\mathbb{Q}(\zeta_3,\sqrt[3]{2})$, $E_1=\mathbb{Q}(\sqrt[3]{2})$, and $E_2=\mathbb{Q}(\zeta_3\sqrt[3]{2})$, we have $\mu_{E_1}=\mu_{E_2}=\mu_K$, but $\mu_{E_1E_2}=\mu_{L}\supsetneq\mu_K$.

Note that $K=\mathbb{Q}$ and $L=\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ isn't a counterexample to the actual problem - for example, $F=E_1$ works, because $\mu_{E_1}=\mu_{\mathbb{Q}}$, and $L=E_1(\zeta_3)$.

So, to prove the claim / construct a counterexample, it seems to me that we want to look at intermediate fields $E$ which are maximal among those such that $\mu_E=\mu_K$, and determine whether or not there always exists at least one such $E$ such that $L=E(\text{some roots of unity})$.

Here is Arturo Magidin's comment on the original question:

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 $G=\text{Gal}(L/K)$. If we can break up the extension as you mention, then $L$ is Galois over $F$, and $\text{Gal}(L/F)=\text{Gal}(M/K)=G/H$. So $G$ would necessarily have normal subgroup $H$ and a subgroup isomorphic to $G/H.$