p -GROUPS AS p -CLASS TOWER GROUPS

. — We give a streamlined and eﬀective proof of Ozaki’s theorem that any ﬁnite p -group Γ is the Galois group of the p -Hilbert class ﬁeld tower of some number ﬁeld F. Our work is inspired by Ozaki’s and applies in broader circumstances. While his theorem is in the totally complex setting, we obtain the result in any mixed signature setting for which there exists a number ﬁeld k 0 with class number prime to p . We construct F { k 0 by a sequence of Z { p -extensions ramiﬁed only at ﬁnite tame primes and also give explicit bounds on r F : k 0 s and the number of ramiﬁed primes of F { k 0 in terms of #Γ.


Introduction
For a number field k, define L p (k) to be the compositum of all finite unramified Galois p-extensions of k.The extension L p (k)/k is called the p-Hilbert class field tower of k, and its Galois group Gal(L p (k)/k) is its p-class tower group.Ozaki [2011] proved that every finite p-group occurs as Gal(L p (F)/F) for some totally complex number field F. His strategy is as follows.
As finite p-groups are solvable, it is natural to proceed by induction.After establishing the base case (realizing ‫/ޚ‬ p as a p-class tower group), it remains to show that given any short exact sequence of finite p-groups where G := Gal(L p (k)/k), one can realize G ′ as Gal(L p (k ′ )/k ′ ) for some number field k ′ .Ozaki constructs such a k ′ /k via a sequence of carefully chosen ‫/ޚ‬p-extensions.
In this paper, we provide a streamlined and effective proof of Ozaki's theorem.Some differences between our work and Ozaki's are: 772 Farshid Hajir, Christian Maire and Ravi Ramakrishna • Our result is effective and we are able to obtain explicit upper bounds on [F : k 0 ] and the number of ramified primes in F/k 0 , all of which are tame and finite.
• Moreover, we bypass some of the most delicate and involved arguments of [Ozaki 2011].
We prove: Theorem.Let be a finite p-group and k 0 a number field with (# Cl k 0 , p) = 1.There exist infinitely many number fields F/k 0 such that Gal(L p (F)/F) ≃ and • if µ p ̸ ⊂ k 0 then F/k 0 is of degree at most p 2 • # and is ramified at at most 2 + 2 log p (# ) finite tame primes, • if µ p ⊂ k 0 then F/k 0 is of degree at most p • (# ) 2 and is ramified at at most 1 + 3 log p (# ) finite tame primes.
Remark.If our starting field k 0 has infinite p-Hilbert class field tower, there is no hope of solving the problem with a finite extension of k 0 .If on the other hand the tower is finite, one can simply pass to the number field L p (k 0 ), which has the same signature ratio as k 0 , and use that as the starting point to realize .
As any (topologically) countably generated pro-p group is the inverse limit of finite p-groups, Ozaki shows any such is the Galois group of the maximal unramified p-extension of some infinite extension of ‫.ޑ‬The corresponding corollary of our theorem is: Corollary.Any (topologically) countably generated pro-p group , including p-adic analytic , can be realized as Gal(L p (F)/F) for a totally real tamely ramified infinite extension F/‫.ޑ‬We now give details about the structure of our proof and the difference between our methods and Ozaki's, though we were very much inspired by Ozaki's beautiful theorem and techniques.
We start the base case of the inductive process with any number field k 0 , of any signature, whose class number is prime to p. Referring to the group extension (1) with G being trivial, one has to find an extension k ′ /k 0 such that k ′ has p-class group tower exactly ‫/ޚ‬p, which is equivalent to the p-class group being ‫/ޚ‬p.This is a standard argument and is part of Proposition 2.15.
The base case being done, we proceed to the inductive step (with our base field relabeled k).There are two cases, depending on whether (1) splits or not.For the sake of brevity, we only outline the nonsplit case in this introduction; the split case is handled similarly.For a set of places of k, we say that an extension k ′ /k is exactly ramified at S if it is ramified at all the places in S and nowhere else.
We need to find a suitable tame prime v 1 of k such that: • v 1 splits completely in L p (k)/k.
• There is no ‫/ޚ‬p-extension of k exactly ramified at v 1 .
• The maximal p-extension L p (k) {v 1 } /L p (k) exactly ramified at the primes of L p (k) above v 1 is of degree p and solves the embedding problem (1).
Arranging this and its split analog are the main technical difficulties.One then chooses a second prime v 2 that also solves the embedding problem as above and remains prime in L p (k) {v 1 } /L p (k).The existence of v 1 and v 2 will follow from Chebotarev's theorem.The compositum of these two solutions, after a ‫/ޚ‬p-base change k ′ /k ramified at both v 1 and v 2 (which exists!), gives the unramified solution to the embedding problem (1) which we show is L p (k ′ ).This is done in the proof of Theorem 3.3.
Our ability to choose primes v i as above depends upon the existence of Minkowski units in the tower L p (k)/k, namely on the condition that -torsion module and λ is a large enough integer.In some situations, Minkowski units are rare; see Section 5.3 of [Hajir et al. 2021].By contrast, both for Ozaki's proof (implicitly) and ours (explicitly), much of the work involves seeking fields for which they exist in abundance.
If µ p ⊂ k, we may not be able to make our choices of v i as above to both split completely in L p (k)/k and solve the nonsplit embedding problem (1).In this case we need to perform an extra base change k/k to shift the obstruction to the embedding problem so that we can proceed as above.The base change k/k must preserve the tower, that is L p ( k) = L p (k) k.Theorem 3.2 provides such a k.
Finally we check that the condition "λ is large enough" persists, that is there are enough Minkowski units to keep the induction going.Proposition 2.14 guarantees this.To sum up, the key ingredients of the proof of the above theorem and corollary are Theorems 3.2 and 3.3 and Proposition 2.14.
We now explain in some detail Ozaki's approach and our simplifications: • Using a result of Horie [1987], Ozaki starts with a quadratic imaginary field with class number prime to p in which p is inert.He then chooses a suitable layer k in the cyclotomic ‫ޚ‬ p -extension as the starting point of his induction.Assuming the problem solved for G in (1) and relabelling k as his base field, he proceeds inductively with the goal to find a k ′ ⊃ k whose p-Hilbert class field tower has Galois group G ′ .For the induction to go forward, Ozaki needs r 2 (k) ≥ B p (k) (implicit in this inequality is the existence of enough Minkowski units) where B p (k) is a certain explicit quantity depending on k, G and the p-part of the class group of K := L p (k)(µ p ).This involves delicate estimates in Section 4 of [Ozaki 2011].We replace r 2 (k) ≥ B p (k) with f (k) ≥ 2h 1 (G) + 3 where h i (G) := dim H i (G, ‫/ޚ‬p) and f (k), which is a lower bound for the number of Minkowski units in L p (k)/k, depends only on h 1 (G), h 2 (G) and the signature of k.We neither consider K nor invoke the estimates of Section 4 of [loc.cit.].
• Ozaki [2011, Section 6] proves his base change Proposition 1, namely he shows there exists a ramified ‫/ޚ‬p-extension k/k such that Gal(L p ( k)/ k) ≃ Gal(L p (k)/k).He uses this repeatedly when solving each embedding problem (1).Several tame primes are ramified in k/k and he also needs that K and K k have the same p-class group.This makes the proof significantly more involved.Theorem 3.2 of this paper, our version of his Proposition 1, has only one tame prime of ramification and K plays no role.We only invoke Theorem 3.2 when µ p ⊂ k.In particular, for p odd, our Corollary above makes no use of Theorem 3.2.
• To solve the embedding problem (1), Ozaki base changes several times (to a field relabeled k) and then uses a wildly ramified ‫/ޚ‬p-extension L/L p (k) to solve (1).After more base changes this is switched to a solution ramified at one tame prime.He then proceeds as in the description of this work using two such solutions and a base change that absorbs the ramification at both tame primes to find a k ′ such that Gal(L p (k ′ )/k ′ ) = G ′ .We go directly to this last step and require at most two ‫/ޚ‬p-base changes to solve the embedding problem.This allows us to quantify explicitly both the degree and number of ramified primes of F/k 0 .
Notations.Let p be a prime number: • L is a number field, O L its ring of integers, O × L its units and Cl L and Cl L [ p ∞ ] are, respectively, the class group of L and its p-Sylow subgroup.
• For a finite set S of primes of L, set In particular, one has the exact sequence • The superscript ∧ indicates the Kummer dual of an object Z defined over a number field L, though we never work with the Gal(L(µ p )/L) action on Z ∧ .
• L S is the maximal pro-p-extension of L unramified outside S, G S := Gal(L S /L) and L p (L) := L ∅ , the maximal unramified pro-p-extension of L, as it will ease notation at various points.
• Gov(L):=L(µ p )( p V L,∅ ), the governing field of L. The span of The following may be helpful in orienting the reader: • We frequently use finite tame primes with desired splitting properties in number field extensions.We always use Chebotarev's theorem for the existence of such primes.
• Our ‫/ޚ‬p-extensions L ′ /L of number fields are only ramified at (one or two) finite tame primes so • Note that k 0 is our given base field, whereas k is a field used in the inductive process with p-class tower group G from (1).Our task is to construct k ′ with p-class tower group G ′ .Finally, k/k is an extension having p-class tower group G, the same as for k.
2. Tools for the proof 2A.‫ކ‬ p [G]-modules and Minkowski units.Let G be a finite group, a p-group in our situation.We record a few basic facts about finitely generated ‫ކ‬ p [G]-modules M; see [Curtis and Reiner 1962, Section 62].
and where λ depends only on M.
Proof.As free modules are clearly projective, Theorem 62.3 of [Curtis and Reiner 1962]  1) .Apply the same argument to M (1) and iterate until, at the λ-th stage there are no copies of ‫ކ‬ p [G] in M (λ) .Thus for every m 0 ∈ M (λ) we have and thus m 0 has nontrivial annihilator.The result is established.
} be a system of minimal generators of G.By Nakayama's lemma and the fact that , I G can be generated, as G-(right or left)-module, by the elements 2B. Extensions ramified at a tame set of primes.We recall a standard formula on the number of ‫/ޚ‬pextensions of a number field with given tame ramification; see Section 11.3 of [Koch 2002] for a proof.
Recall that for a field L, Proposition 2.5.Let L be a number field, p a prime number and X a set of tame primes of L prime to p. (2) Our v ∈ X are always finite and have norm congruent to 1 mod p so δ(L v ) = 1.
Fact 2.6.Let S be a set of tame primes of L as above.For each v ∈ S let When µ p ̸ ⊂ L, Fr v is only well-defined up to nonzero scalar multiplication.
Proof.In (2), as we vary X from ∅ to S, we are adding v∈S δ(L v ) = #S to the right side, but also subtracting dim(V L,∅ /L × p ) − dim(V L,X /L × p ) from the right side.This last quantity is #S − d. □ Fact 2.7.Let L be a number field such that (# Cl L , p) = 1.Let L ′ /L be a ‫/ޚ‬ p-extension exactly ramified at S = {v 1 , . . ., v r } where the v i are finite and tame.Then (# Cl L ′ , p) = 1 if and only if L ′ /L is the unique ‫/ޚ‬ p-extension of L unramified outside S. In particular, that is the case when |S| = 1.
Proof.Indeed, (# Cl L ′ , p) ̸ = 1 if and only if there exists an unramified ‫/ޚ‬ p-extension H/L ′ such that H/L is Galois (use the fact the action of a p-group on a p-group always has fixed points).Observe that H/L cannot be cyclic of degree p 2 as all inertial elements of Gal(H/L) have order p and they would thus fix an unramified extension of L, a contradiction.So Gal(H/L) ≃ ‫/ޚ‬ p × ‫/ޚ‬ p, and L has at least two disjoints ‫/ޚ‬ p-extension unramified outside S, also a contradiction.
Let λ L be the number of Minkowski units in L p (L)/L.
This result is Theorem 2.9 of [Hajir et al. 2021], but we sketch the proof for the sake of keeping this paper self-contained.
Proof.Set G = Gal(L p (L)/L).We consider two "norm maps" induced by the norm map on units When µ p ⊂ L, note that the "difference" between the images of N G and To conclude, we use the well-known equality (see [Roquette 1967, Lemma 9]) . □ 2C. Solving the ramified embedding problem with one tame prime.We start with our nonsplit exact sequence 1 given by the element 0 We assume that G = Gal(L p (k)/k).
Set S = {v} where v is a finite tame prime of k.We first show the existence of a lift of G to G ′ in some k S /k for certain v of k.We call this solving the embedding problem (3) in k S .
Recall that X 2 k,S → B k,S by Fact 2.8.Here , ‫/ޚ‬p) be the inflation map.We have the commutative diagram: By Hoeschmann's criteria (see [Neukirch et al. 2008, Chapter 3, Section 5]), the embedding problem has a solution in k S if and only if Inf S (ε) = 0.As L p (k)/k is unramified, Inf S (ε) ∈ X 2 k,S and as g(Inf S (ε)) = f S (h(ε)) ∈ B k,S , the embedding problem has a solution if and only if h(ε) ∈ Ker( f S ).
Set Gov S (k) := k(µ p )( p V k,S ).In the governing extensions k(µ p ) ⊂ Gov S (k) ⊂ Gov(k), one sees that the kernel of the map f S : B k,∅ ↠ B k,S is exactly the (unramified) decomposition group D v of the prime v.As noted in Fact 2.6, if w 1 , w 2 | v are two primes of k(µ p ), their Frobenius elements in Gal(Gov(k)/k(µ p )) differ by a nonzero scalar multiple.
We have proved: Lemma 2.10.The embedding problem (3) has a solution in k S /k if and only if h(ε) ∈ D v .Thus it has a solution in k S /k if we choose the prime v such that ⟨Fr v ⟩ = ⟨h(ε)⟩ in M(k), that is the lines spanned by these elements in M(k) are equal.This is always possible by Chebotarev's theorem.
2D. Cohomological facts implying the persistence of Minkowski units.Our main aim in this paper is to show that given a short exact sequence of finite p-groups where G = Gal(L p (k)/k), there exists a finite tamely ramified extension k ′ /k with G ′ = Gal(L p (k ′ )/k ′ ).To solve this embedding problem using Theorem 3.3, the tower L p (k)/k must have 2h 1 (G) Minkowski units.Proposition 2.14 below shows that if we start with enough Minkowski units, after a base change that realizes G ′ , we will be able to continue the induction.Proposition 2.13, which is only needed in the case when µ p ⊂ k, shows that given at least h 1 (G) Minkowski units, we can perform a base change that preserves the tower and the number of Minkowski units increases.Proposition 2.11 is a basic group theory result bounding h 1 (G ′ ) and h 2 (G ′ ) in terms of h 1 (G) and h 2 (G).Furuta proves a similar result in Lemma 2 of [Furuta 1972]. Set Proof.The h 1 result is clear.For the h 2 statement we have the long exact sequence (see for instance [Dekimpe et al. 2012]) In the nonsplit case we have Definition 2.12.For a number field L set G = Gal(L p (L)/L).Define f as follows: Fact 2.9 implies f (L) is a lower bound on the number of Minkowski units of L p (L)/L.
Proposition 2.13.Let k/k be a ‫/ޚ‬p-extension ramified at finite tame primes such that Proof.This follows immediately as we have the same group G for k and k, Let f (k) be as in Definition 2.12.Then Proof.We do the case µ p ̸ ⊂ k first.We need to prove and by Proposition 2.11 we have This holds for all p.When µ p ⊂ k.We need to prove Again using Proposition 2.11 and that r i (k ′ ) = p • r i (k) it suffices to show which holds for all p. □ Proposition 2.15 below provides the base case of the induction.
Proposition 2.15.Recall (# Cl k 0 , p) = 1.There exists a tamely ramified extension k ′ /k 0 such that Proof.Since L p (k 0 ) = k 0 , we see G = {e}.Choose a tame prime v of k whose Frobenius is trivial in the governing Galois group M(k).By Fact 2.6 there is a unique ‫/ޚ‬p-extension k 1 /k 0 unramified outside v.That (# Cl k 1 , p) = 1 follows from Fact 2.7.Repeat this process with k 1 to get a field k 2 with (# Cl k 2 , p) = 1.
We do one more base change to find a field k ′ with class group ‫/ޚ‬p.This is proved more generally as part of Theorem 3.3, but we include a short proof here.
Choose v 1 a finite tame prime of k 2 with trivial Frobenius in M(k 2 ) so that by Fact 2.6 there exists a unique D 1 /k 2 ramified at v 1 .As D 1 ∩ Gov(k 2 ) = k 2 , we may choose v 2 a finite tame prime of k 2 with trivial Frobenius in Gov(k 2 ) such that v 2 remains prime in D 1 /k 2 .Again by Fact 2.6 there exists a unique D 2 /k 2 ramified at v 2 .
Let D/k 2 be any of the p − 1 "diagonal" ‫/ޚ‬p-extensions of k 2 between D 1 and D 2 so D 1 D 2 /D is everywhere unramified.We claim D 1 D 2 = L p (D).Indeed, by Fact 2.7 applied to D 1 /k 2 we see Depending on p and the signature of k 0 one can decrease the number of base changes, but this analysis complicates the statement of the main theorem without significant gain.

Solving the embedding problem
Having established the base case of our induction, we now prove Theorem 3.3.
be exact and let k be a number field with Gal(L p (k)/k) = G and f (k) ≥ 2h 1 (G) + 3. Then there exists a number field Theorem 3.2 below is only necessary for the key inductive step, Theorem 3.3, when µ p ⊂ k.Set K := L p (k)(µ p ).We only consider finite tame primes v of k that split completely in K/k.When µ p ̸ ⊂ k, our Frobenius elements in governing fields (or their subfields) are only defined up to scalar multiples.We write ⟨Fr v ⟩ Gov(k)/k(µ p ) for the well-defined line spanned by Frobenius at v in Gal(Gov(k)/k(µ p )).When the Frobenius is trivial there is no ambiguity so we write We have no knowledge of N and must work with the free part to control things over L p (k).We then use Proposition 3.1 to control things over k.
Proposition 3.1.Let F ⊂ Gov(L p (k)) be the field fixed by I G • M(L p (k)).For v of k splitting completely in K and w | v in K, the lines ⟨Fr w ⟩ F/K do not depend on w so we may write ⟨Fr v ⟩ F/K .Then by computing the dimension of h 1 (G)  i=1 Ann(x i ).This intersection is the annihilator of I G which by Proposition 2.2 is just ‫ކ‬ p T G , establishing our dimension result.By Fact 2.6 there is a unique extension over L p (k) ramified at v and thus it must be L and the extension is nonsplit, k ′ can be realized as a compositum of two successive ‫/ޚ‬p-extensions and L p (k ′ )/k ′ has at least 2h 1 (G ′ ) + 3 Minkowski units.
Proof.Recall that our finite tame primes split completely in K/k.We first treat the split case.This is independent of whether or not µ p ⊂ k.
Split case.Choose tame degree one primes w 1 and w 2 of Gov(k)K such that . This is possible as the tuple lies in I G • M(L p (k)) and Gov(k)K ⊂ F. As Fr w 1 projects to 0 in Gal(F/K), we see for v 1 of k below w 1 that ⟨Fr v 1 ⟩ F/K = 0 so by Proposition 3.1 ⟨Fr v 1 ⟩ Gov(k)/k(µ p ) = 0.By Fact 2.6 applied to k there is one ‫/ޚ‬p-extension D 1 /k ramified at v 1 .Fact 2.6 also gives (see the proof of Theorem 3.2 as well) a unique ‫/ޚ‬p-extension of L p (k) ramified at v 1 , namely D 1 L p (k)/L p (k).
Fact 2.6 and our choices of the Frobenius elements of v 1 and v 2 imply using that the span of the Frobenius elements above them in Gal(Gov(L p (k))/ Gov(k)K) ⊂ M(L p (k)) has dimension 2#G − 2 and Fact 2.6.(With only h 1 (G) Minkowski units, we would again have had In this case the span of the Frobenius elements above and note L/D is unramified as D/k has absorbed all ramification at {v 1 , v 2 }.We will solve the problem by showing Since (# Cl L p (k) , p) = 1 and our choice of v 1 is such that Fact 2.7 applied to J/L p (k) implies (# Cl J , p) = 1.We now prove that there exists a unique ‫/ޚ‬ p-extension over J unramified outside v 2 , namely L. Set = Gal(J/L p (k)), J p,el {v 2 } to be the maximal elementary p-abelian extension of J inside J {v 2 } , and = Gal(J p,el {v 2 } /J).Then acts on and trivially on Gal(L/J).We claim this is the only ‫/ޚ‬p-extension of J in J p,el {v 2 } /J on which acts trivially: If not, there exists another ‫/ޚ‬ p-extension H/J unramified outside v 2 and Galois over L p (k). Hence Gal(H/L p (k)) has order p 2 and is abelian.The extension H/L p (k) cannot be cyclic because all inertia elements have order p and would then fix an everywhere unramified extension of L p (k), a contradiction.Suppose now that Gal(H/L p (k)) ≃ ‫/ޚ‬ p × ‫/ޚ‬ p, with The final possibility is that there exists a ‫/ޚ‬ p-extension E 0 /J unramified outside v 2 , different from L/J and not fixed by ; let S 0 be the set of ramification of E 0 /J.As primes above v 2 in L p (k) are inert in J/L p (k), (S 0 ) = S 0 : then takes E 0 to another ‫/ޚ‬ p-extension E 1 /J exactly ramified at S 0 and such that E 1 ̸ = E 0 .The compositum E 1 E 0 /J contains a ‫/ޚ‬ p-extension E ′ 0 /J exactly ramified at a set S ′ 0 ⊊ S 0 .Observe that E ′ 0 ̸ = L since L/J is totally ramified at every prime above v 2 .Continuing the process, we obtain an unramified ‫/ޚ‬ p-extension H/J, which is impossible since (# Cl J , p) = 1.Thus L/J is the unique ‫/ޚ‬ p-extension unramified outside v 2 .Fact 2.7 applied to L/J implies (# Cl L , p) = 1.
We have solved the split embedding problem with k ′ = D and Gal(L p (k ′ )/k ′ ) = G × ‫/ޚ‬p.It required one base change ramified at two tame finite primes.Proposition 2.14 implies f (k ′ ) ≥ 2h 1 (G ′ ) + 3 so the induction can proceed.
For the nonsplit case we treat µ p ̸ ⊂ k and µ p ⊂ k separately.Theorem 3.2 is only used in the nonsplit case when µ p ⊂ k.
The nonsplit case, µ p ̸ ⊂ k.By Lemma 2.10 we may use one tame prime v of k to find a ramified solution to the embedding problem.As and by Fact 2.6 there is no ‫/ޚ‬p-extension of L p (k) ramified at the w | v, contradicting that we are solving an embedding problem with v.If n 0 ∈ I G • N , then the projection of Fr w to Gal(F/K) is trivial so Proposition 3.1 implies ⟨Fr v ⟩ Gov(k)/k(µ p ) = 0 and the embedding problem we are solving is split, also a contradiction.
As the free components of Fr w , Fr w 1 and Fr w 2 are all in I λ k G , their projections to Gal(F/K) depend only on n 0 and Proposition 3.1 implies Thus there is no extension of k ramified at either v 1 or v 2 , but, by Fact 2.6, there is a ‫/ޚ‬p-extension of k ramified at {v That D 1 D 2 has trivial p-class group follows exactly as it did in the split case and we may set k We have solved the embedding problem in the nonsplit case when µ p ̸ ⊂ k.We performed one base change ramified at two tame finite primes and Proposition 2.14 implies f (k ′ ) ≥ 2h 1 (G ′ ) + 3 so the induction can proceed.
The nonsplit case, µ p ⊂ k.We can no longer assume L p (k) ∩ Gov(k) = k.
Let 0 ̸ = ε ∈ X 2 k,∅ be the obstruction to our embedding problem G ′ ↠ G. Using Lemma 2.10, let v of k be a tame prime annihilating ε.The difficulty is that in the diagram below we may have L p (k)∩Gov(k) ⊋ k and that Fr v , which is necessarily nonzero in M(k), may also be nonzero in Gal((L p (k) ∩ Gov(k))/k).This prevents us from also choosing v to split completely in L p (k)/k and as we need in Gov(L p (k))/L p (k) to ensure there is only one extension of L p (k) ramified at the primes of L p (k) above v.If we could choose v to annihilate ε such that Fr v = 0 ∈ Gal(L p (k)/k), we would be able to proceed as in the µ p ̸ ⊂ k case.We get around this by a base change.
By Kummer theory and the definition of governing fields, Gal(Gov(L)/L(µ p )) is an elementary pabelian group.Let k/k be a tamely ramified ‫/ޚ‬p-extension as given by Theorem 3.2 so Gal(L p ( k)/ k) = G.By Proposition 2.13 we have λ k ≥ 2h 1 (G) + 3: As Gov(k) ∩ k = k, we may choose a prime v to solve the embedding problem for k whose Frobenius is nontrivial in Gal( k/k), that is v remains prime in k/k.As observed above, L p ( k) ∩ Gov( k)/ k is a (‫/ޚ‬p) r -extension for some r and, as Gal(L p (k)/k) = Gal(L p ( k)/ k) = G, it is the base change of such a subextension of L p (k)/k from k so L p ( k) ∩ Gov( k)/k is a (‫/ޚ‬p) r +1 -extension.Since v remains prime in k/k and residue field extensions are cyclic, it splits completely in L p ( k) ∩ Gov( k)/ k.As the embedding problem is solvable over k by allowing ramification at v, it is also solvable over k by allowing ramification at the unique prime of k above v.Thus ε ∈ X 2 k,∅ → B k,∅ = M( k) actually lies in Gal(Gov( k)/(L p ( k) ∩ Gov( k))).The base change shifted the obstruction to outside of our p-Hilbert class field tower!The rest of the proof is identical to the µ p ̸ ⊂ k case. □ We now prove the main theorem of the introduction.
Proof.We have verified the base case of the induction in Proposition 2.15 and the inductive step with Theorem 3.3.It remains to count degrees and ramified primes.Proposition 2.15 involved three ‫/ޚ‬p-base changes, the first two ramified at one tame prime and the last at two tame primes.The inductive steps breaks into cases as follows: • µ p ̸ ⊂ k 0 : At each of the log p (# ) − 1 inductive stages we need one base change ramified at two primes for a total of 3 + (log p (# ) − 1) base changes ramified at 4 + 2(log p (# ) − 1) primes.
• µ p ⊂ k 0 : At each of the log p (# ) − 1 inductive stages we need at most two base changes and at most three ramified tame primes so in total there are at most 3 + 2(log p (# ) − 1) base changes ramified at most 4 + 3(log p (# ) − 1) primes.□ stable under the action of the p-group G and thus has a fixed point.But it is easy to see the only fixed points of ‫ކ‬ p [G] are multiples of T G so T G ∈ Ann G (n) as desired.□ Definition 2.4.We say the tower L p (k)/k with Galois group G has λ Minkowski units if, as