Cyclotomic extensions with split Galois group


Consider the set of all Galois extensions $E/\Q(\zeta_n)$ of a given cyclotomic field $\Q(\zeta_n)$ such that $$\Gal(E/\Q) \simeq\Gal(E/\Q(\zeta_n)) \rtimes \Gal(\Q(\zeta_n)/\Q).$$
In other words, such that there is a homomorphism $$\Gal(E/\Q) \leftarrow \Gal(\Q(\zeta_n)/\Q)$$ inverting the natural quotient map $$\Gal(E/\Q) \to \frac{\Gal(E/\Q) }{\Gal(E/\Q(\zeta_n))}\simeq \Gal(\Q(\zeta_n)/\Q).$$

Are they classified? Is there a "largest" one? What can be said about them (or about their cohomology) in general? Are there any prominent examples of such extensions arising "in nature"?

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If n is 1 (or 2) then classifying such extensions means classifying all Galois extensions of Q. If n is 3 or 4 (or 6) then every Galois extension of Q containing the nth cyclotomic field is split by complex conjugation. –  Tom Goodwillie Nov 25 '10 at 4:47
Tom, this is a good point. So I guess this means that if E is such an extension and if s generates (Z/p), then s^{(p-1}/2} needs to map to complex conjugation in E. These sort of facts are what I was after. –  Jon Yard Nov 25 '10 at 6:12
Let $K/k$ be a finite Galois extension of number fields, and let $K'$ be the Hilbert class field of $K$. Let $e$ be the lcm of the local ramification indices for $K/k$. Wyman proves that if $e=[K\colon k]$ then the group extension $$1\rightarrow\text{Gal}(K'/K)\rightarrow\text{Gal}(K'/k)\rightarrow\text{Gal}(K‌​/k)\rightarrow 1$$ splits. In particular, this sequence splits if $k={\bf Q}$ and $K/{\bf Q}$ is cyclic. On the other hand, he shows by a counterexample that a Galois non-cyclic extension of ${\bf Q}$ need not lead to a split extension, contrary to an assertion of C. S. Herz. (MR0337916 –  Chandan Singh Dalawat Dec 12 '10 at 5:23


Abbreviate $K=\Q(\zeta_n)$. Note first that a galoisian extension $E$ of $K$ need not be galoisian over $\Q$, so I'm assuming that you are considering only those $E$ which are. We then have an exact sequenece $$1\to\Gal(E|K)\to\Gal(E|\Q)\to\Gal(K|\Q)\to1$$ in which the last group is $(\Z/n\Z)^\times$, of order $\varphi(n)$. A sufficient condition for the sequence to split is : the degree $[E:K]$ is prime to $\varphi(n)$ (Schur-Zassenhaus). I don't think there is a classification of all such extensions.

Note finally that this answer does not depend on the fact that $K$ is the cyclotomic field of level $n$, or even the fact that the base field is $\Q$. It applies to any galoisian tower $E|K|F$: the associated short exact sequence $$1\to\Gal(E|K)\to\Gal(E|F)\to\Gal(K|F)\to1$$ splits if the degrees $[E:K]$, $[K:F]$ are mutually prime.

Addendum (at Alex Bartel's suggestion): Let's return to the case $F=\Q$, $K=\Q(\zeta_n)$, $\Delta=\Gal(K|\Q)$, and suppose that $n$ is a prime $p$, for simplicity. Kummer theory tells us that abelian extensions $E|K$ of exponent dividing $p$ correspond bijectively to subgroups $D\subset K^\times/K^{\times p}$ under $E=K(\root p\of D)$; such an $E$ is galoisian over $\Q$ if and only if the subgroup $D$ is $\Delta$-stable. When such is the case, we get examples of the kind of extensions envisaged in the question, with "split Galois group". I guess the group $\Gal(E|\Q)$ will be commutative if and only if the $\Delta$-action on the $\F_p$-space $D$ is via the "mod $p$" cyclotomic character, namely the canonical isomorphism $\Delta\to\F_p^\times$.

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Since the poster asked about prominent examples arising in nature, it might be worth mentioning Kummer extensions. They are covered by your criterion, but maybe they are sufficiently prominent to receive an extra mention. –  Alex B. Nov 25 '10 at 5:28
Alex, good point about Kummer extensions, and thanks for expanding on this Chandan. –  Jon Yard Nov 25 '10 at 8:45

The splitting of the Galois group of Hilbert class fields of an extension field is discussed in the following articles

• B. Wyman, Hilbert class fields and group extensions, Scripta math. 29 (1973), 141–149
• R. Gold, Hilbert class fields and split extensions, Ill. J. Math. 21 (1977), 66–69
• R. Bond, On the splitting of the Hilbert class field, J. Number Theory 42 (1992), 349–360
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The basic tool for understanding this problem is the theorem of Weil-Shafarevich, which gives the cohomological invariant describing the group extension. A good source for this result are the notes by Artin-Tate. –  Franz Lemmermeyer Nov 26 '10 at 12:21
Check this thread, it is amusing and you are involved:  tea.mathoverflow.net/discussion/802/a-reputation-table-bug  –  Will Jagy Dec 1 '10 at 19:48
Thanks Franz. After hours of searching, I cannot find the Wyman article (or really much at all from Scripta mathematica) online. Does anyone know where to find an electronic copy? –  Jon Yard Dec 4 '10 at 19:30