Factoring a field extension into one which adds no roots of unity, followed by one which adds only roots of unity

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 $G/H.$

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As a variation of Peter and Chandan's answers, ${\mathbb Q}(\sqrt[4]{-3})/{\mathbb Q}$ is also a counterexample. (Or ${\mathbb Q}(i,\sqrt[4]{-3})/{\mathbb Q}(i)$ if you prefer a Galois extension.) –  Tim Dokchitser Dec 20 '10 at 14:14

The answer is no. The idea is to consider a tamely ramified extension of a local field. The remaining paragraphs provide details of a proof.

Take K=ℚp. Let M be a degree d unramified extension of K. Let L be an extension of M obtained by adjoining a n-th root of $pa$ for some $a$ of norm 1 in M that is not in ℚp. We will eventually take n=pd-1.

Now suppose that such a field F exists. Then F must be a totally ramified extension of ℚp with L a totally unramified extension of F. So F has degree n over ℚp. Let x be a uniformiser of F. Then $x=yb$ where $y^n=pa$ and $b$ is of norm 1. Consider the minimal polynomial of x, namely $\sum_{i=0}^n a_i x^i=0$ with $a_i$ integral, WLOG $a_n=1$. Examining valuations, $a_0=pc$ for some integral c, and $(x^n+a_0)/p=0 \pmod p$, ie $ab^n+c$ has zero image in the residue field.

Now to get a contracition, set n=pd-1. Thus we get $a+c$ has zero image in the finite field GF(pd) but $c \in\mathbb Q_p$ and $a \notin \mathbb Q_p$.

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Thanks for your answer! I've only just started learning about local fields, so I'm afraid I don't follow some parts of your argument. Here are my main questions: Why must $F/\mathbb{Q}_p$ be totally ramified and $L/F$ be unramified? Why is $(x^n+a_0)/p\equiv 0\bmod p$ (I think I remember that the minimal polynomial of a uniformizer must be Eisenstein, but that doesn't appear to suffice for this claim)? Why isn't the residue field of $F$ just $\mathbb{F}_p$, regardless of what we set $n$ to be, because we assumed $F$ is totally ramified and hence $f(y/p)=[\bar{F}:\mathbb{F}_p]=1$? –  Zev Chonoles Dec 20 '10 at 3:13
Feel free to just point me to a book or something for any standard facts - it's more likely than not that I haven't seen them yet. –  Zev Chonoles Dec 20 '10 at 3:13
@Zev: I'm freely using the fact that n is coprime to p in claiming F totally ramified and L/F unramified - adjoining roots of unity of order prime to p are precisely all unramified extensions. For your 2nd Q, the congruence should be modulo a uniformiser of F (my typo), to see it consider minimal polynomial modulo uniformiser^1,2,...,n+1 successively. Residue field of F is of order p, calculations are taking place inside L, which has larger residue field. As per the comment in Chandan's answer, we could also give this example in positive characteristic. –  Peter McNamara Dec 20 '10 at 6:43
Thanks for the helpful explanation - I think that pretty much covers it :) –  Zev Chonoles Dec 20 '10 at 20:12


Let $K$ be a finite extension of $\Q_p$, let $l$ be a prime $\neq p$, and suppose that $K$ does not contain a primitive $l$-th root $\zeta$ of $1$, so that the extension $K'=K(\zeta)$ of $K$ has degree $>1$. Note that $K'|K$ is unramified, of group $\Gamma$ (say) which is naturally a subgroup of $\F_l^\times$.

Cyclic extensions $L|K'$ of degree $l$ correspond to $\F_l$-lines $D$ in $K'^\times/K'^{\times l}$ under $L=K'(\root l\of D)$. Such an $L$ is of the form $L=F(\zeta)$ for some degree-$l$ extension $F$ of $K$ if and only if the line $D$ is $\Gamma$-stable.

Now, as a $\Gamma$-module, the $\F_l$-space $K'^\times/K'^{\times l}$ is isomorphic to $\mu_l\oplus\mathbf{F}_l$, so it certainly contains lines which are not $\Gamma$-stable.

In other words, there are degree-$l$ cyclic extensions $L$ of $K'=K(\zeta)$ which are not of the form $L=F(\zeta)$ for a degree-$l$ extension $F$ of $K$.

The simplest instance of this phenomenon occurs for $K=\Q_2$, $l=3$: the extension $L=\Q_2(\zeta,\root3\of{2\zeta})$ of $\Q_2$ is not of the form $F(\zeta)$ for any cubic extension $F$ of $\Q_2$ basically because otherwise $L|\Q_2$ would be galoisian, which it is not.

Note finally that in this entire discussion the finite extension $K$ of $\Q_p$ can be replaced by $k((T))$ for any finite extension $k$ of $\F_p$.

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