CYCLIC ALGEBRAS, SCHUR INDICES, NORMS, AND GALOIS MODULES 0 1 0 JAŃ MINAĆˇ∗, ANDREW SCHULTZ, AND JOHN SWALLOW† 2 n To Paulo Ribenboim a J 1 Abstract. LetpbeaprimeandsupposethatK/F isacyclicex- 2 tension of degree pn with group G, with charF =p. Let J be the 6 F G-module K×/K×p ofpth-powerclasses. Inourpreviouspaper p ] we established precise conditions for J to contain an indecompos- T able direct summand of dimension not a power of p. At most one N such summand exists, and its dimension must be pi+1 for some . h 0 i . (In fact p ∈ K/Ks ∈ { } −∞ it is an easy exercise to show that N (K×) N (K×) if i j K/Ki ⊆ K/Kj ≤ directly without referring to Albert’s result.) The submodules Y are produced naturally using norms from dif- i ferent layers of the tower of field extensions. However, the remaining submodule X is more mysterious, and we consider a first problem con- cerning the classification of all F G-modules occurring as J(K/F): p Given n 1 and d an element of the set ≥ 1,p0 +1,...,pn−1 +1 , { } CYCLIC ALGEBRAS, SCHUR INDICES, NORMS, AND GALOIS MODULES 3 does there exist a cyclic extension K/F with ξ F× such that the p ∈ exceptional summand X has dimension d? (We remark that in the case of n = 1, ξ F×, the full realization p ∈ problem—realizingallpossibleisomorphismclassesfortheF G-module p J(K/F)—has been solved [MS2].) Itturnsoutthatwemayanswer thisquestionintheaffirmativeusing a construction of cyclic division algebras due to Brauer-Rowen. For a particular extension K/F with Gal(K/F) = σ Z/pnZ, the cyclic h i ≃ algebra A(K/F) = (K/F,σ,ξ ) p will beinvestigated for its connection to the module-theoretic structure of K×/K×p. For instance, we will see in section 1 that the invariant m(K/F) above can be recast as , if A(K/F) splits, m(K/F) = −∞ min s : K splits A(K/F) 1, otherwise. ( s { }− (1) In section 1 we begin by recalling basic facts about cyclic algebras, including the algebra A(K/F). We go on to describe how the Schur index of A(K/F) is bounded by m(K/F). In some important cases thesetwoinvariantscoincide; however, weshowinsection5thatwecan have a strict inequality. Thus it appears that m(K/F) is a genuinely new invariant of K/F. Further investigation of m(K/F) and its relation with the associated cyclotomic algebra A(K/F) is of considerable interest. The case when this algebra is a division algebra is considered in the last section. We also highlight in section 3 the interesting relations between the integer part function and the basic rule for cyclic algebras when passing to a cyclic extension. In section 2 we show that m(K/F) ,0 in the ∈ {−∞ } case of local fields, and we provide examples for both cases. Finally, in section 4 we show how to use just a simple observation about norms to bypass cyclic algebras and get a direct proof of Theorem 2. However, we stress the intrinsic interest of the relationship between m(K/F) and the Schur index, and we hope that this paper will promote its further investigation. 4 JAŃ MINAĆˇ, ANDREWSCHULTZ,AND JOHN SWALLOW 1. Cyclic Cyclotomic Algebras, m(K/F), Schur’s Index, and Norms First we reformulate m(K/F) in terms of cyclic algebras and then we use the construction of Brauer-Rowen of suitable cyclic algebras. We will then prove the following theorem: Theorem 2. Let n N and t ,0,1,...,n 1 . Then there ∈ ∈ {−∞ − } exists a cyclic extension K/F of degree pn with ξ F× and m(K/F) = p ∈ t. Before turning to the proof of the theorem, we recall some basic facts about cyclic algebras. If L/E is a cyclic extension of degree r > 1, with Galois group G = Gal(L/E) = τ , and b E×, then h i ∈ B = (L/E,τ,b) is a central simple algebra such that B = ujL, 0≤j 2 or a / 4F×4 if p = 2. Let K = F( p√n a). ∈ ∈ − Then we see that K/F is a cyclic extension of degree pn. (See [La, Theorems 8.1 and 9.1].) Furthermore, because N (ξ ) = ξ , we K/F pn+1 p conclude that m(K/F) = . −∞ Example 2 (A Case of m(K/F) = 0). Following [HS, 3] we provide § a simple example of a local field extension with m(K/F) = 0. Dirichlet’s theorem shows that there exists a prime q such that q = 1 + pn + kpn+1 = 1 + pn(1 + kp) for some k N 0 . For such q ∈ ∪ { } we know that ξ Q and ξ / Q (see [Ko, 3.3]). On the other pn q pn+1 q ∈ ∈ § hand by [Ko, 3.3] we know that Q (ξ ) is a totally ramified cyclic q q § extension of Q of degree q 1. Let K/Q be the cyclic extension q q − K Q (ξ ) of degree pn. Then by [HS, Proposition 4.1] we see that q q ⊆ K cannot be embedded into a cyclic extension L/Q of degree pn+1. q Hence m(K/Q ) = 0. q 3. Cocycles, Carrying, and Cyclic Algebras In section 1 in the proof of Theorem 2, we used [P, Corollary 15.1b]: Let F K E be a tower of cyclic extensions such that G = ⊆ ⊆ Gal(E/F) = σ , σ¯ = restriction of σ on K, and [E : F] = a = bq h i where b = [K : F] and q = a. Then b [(K/F,σ¯,α)] = [(E/F,σ,αq)] Br(F) ∈ 8 JAŃ MINAĆˇ, ANDREWSCHULTZ,AND JOHN SWALLOW for each α F×. ∈ In [P] the proof of this corollary uses a case-by-case calculation of cocycles. We feel that because of this, a beautiful and interesting connection between the integer parts of rational numbers and cohomology may be lost. We shall here highlight this connection. For r Q, let [r] Z denote the integer part of r Q, that is, the ∈ ∈ ∈ value [r] satisfying [r] r < [r]+1. As in [P, p. 278] the essential part ≤ ofthe proofof thisstatement is toprove that thefollowing two cocycles Ψ and Φ determine the same cohomology class in H2(G,E×): α αq 1, k +l < b Ψ (σi,σj) = α α, k +l b, ( ≥ where i = b i +k and j = b j +l; and b b (cid:2) (cid:3) (cid:2) (cid:3) 1, i+j < a Φ (σi,σj) = αq αq, i+j a. ( ≥ where 0 i,j < a. ≤ Let M = α be the subgroup of F× (and therefore also E×) gen- h i erated by α. Then we show that our cocycles Ψ and Φ already α αq determine the same cohomology class in H2(G,M) and therefore by naturality also the same class in H2(G,E×). Both classes in H2(G,M) determine the group extensions [Ψ ] and α [Φ ] of shapes αq 1 M H G 1 Ψ −→ −→ −→ −→ and 1 M H G 1. Φ −→ −→ −→ −→ Choose lifts σˆ H and σ˜ H for σ G, and let 0 i,j < a be Ψ Φ ∈ ∈ ∈ ≤ given. As above, write i = b i +k and j = b j +l, and also write b b i + j = a i+j + i+j. Notice that if k + l < b then we have both a (cid:2) (cid:3) (cid:2) (cid:3) i + j = i+j and also σîσˆj = σî+j, so b b (cid:2) b(cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) σîα[bi]σˆjα[jb] = σî+jα[i+bj]. If, on the other hand, we have k +l b, then i + j + 1 = i+j ; ≥ b b b since σîσˆj = σî+jα by our cocycle condition, this gives (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) σîα[bi]σˆjα[jb] = σîσˆjα[bi]+[jb] = σî+jα[bi]+[jb]+1 = σî+jα[i+bj]. CYCLIC ALGEBRAS, SCHUR INDICES, NORMS, AND GALOIS MODULES 9 From this calculation we learn two things: first, that the map γ : H H defined by Φ Ψ → γ(σ˜i) = σîα[bi] γ(α) = α is an isomorphism between [Φ ] and [Ψ ], thus completing our proof; αq α and, second, that the cocycle Ψ is essentially the difference i+j α b − i j . More specifically, this difference is the logarithm of Ψ with b − b (cid:2)α (cid:3) base α. (That “carrying” is a cocycle was already observed in [Is].) (cid:2) (cid:3) (cid:2) (cid:3) The referee of this article introduced us to the following elegant and conceptual alternative proof of the main result of this section. This proof also shows that this statement is a straightforward consequence of the bilinearity of the cup product. Let Γ be the Galois group of a separable closure F of F containing s E. Let χ H1(Γ,Q/Z) be the character of Γ given by χ(γ) = k/a, ∈ where the restriction of γ to E is σk. Let ∂χ be the image of χ under the connecting map ∂, ∂ : H1(Γ,Q/Z) H2(Γ,Z) → associated with 0 Z Q Q/Z 0. Viewing a given β F× as → → → → ∈ an element in H0(Γ,F×), one can then check through explicit cocycle s calculations that ∂χ β = [(E/F,σ,β)] Br(F) H2(Γ,F×). ∪ ∈ ≃ s In the case where β = αq, we therefore have ∂χ αq = [(E/F,σ,αq)]. ∪ On the other hand, the cup product’s bilinearity gives ∂χ αq = ∪ ∂(qχ) α, and since qχ is the character of Γ associated to K/F (and its ∪ generator σ¯) we therefore have ∂(qχ) α = [(K/F,σ¯,α)], as desired. ∪ 4. Bypassing Cyclic Algebras with a Norm Calculation Brauer-Rowen’s construction of the cyclic algebras above suggests replacing Q(ξ )(µ ,...,µ ) by any quotient field L of a unique factor- q 1 pn ization domain D. Let U(D) be the group of units of D. In this case we have Proposition 1. Let G be a group of order n acting on a unique fac- torization domain D as a group of n distinct automorphisms of D. Assume also that U(D) is a G-invariant subgroup of D. Then N (L) U(D) = (U(D))n. L/LG ∩ 10 JAŃ MINAĆˇ, ANDREWSCHULTZ,AND JOHN SWALLOW Proof. If u = wn U(D)n for some w U(D), then u = N (w) L/LG ∈ ∈ and therefore U(D)n N (L) U(D). L/LG ⊆ ∩ Assume now that u U(D) N (L). Then u = N (w) where L/LG L/LG ∈ ∩ p ...p 1 t w = d q ...q 1 s and d U(D) and p ...p ,q ...q are primes in D. 1 t 1 s ∈ Therefore s t g(q ) N (w) = dn g(p ). i L/LG i ! g∈Gi=1 g∈Gi=1 YY YY FromtheuniquefactorizationpropertyofD weconcludethatt = sand for each i 1,...,t there exists g G,j 1,...,t and d U(D) i ∈ { } ∈ ∈ { } ∈ such that p = d g(q ). i i j Hence N (p ) = dnN (q ) and we see that L/LG i i L/LG j t u = N (w) = dn dn U(D)n. L/LG i ∈ i=1 Y (cid:3) Using the proposition above, one can directly see, without the use of cyclotomic cyclic algebras, that the example employed in our original proof, namely K/F = Q(ξ )(µ ,...,µ )/(Q(ξ )(µ ,...,µ ))hσi, q 1 pn q 1 pn withq = pn−t, works. Infactthefollowingalternativetheoremprovides more examples. (We leave out the easy case when t = .) −∞ Theorem 3. Let ξpn−t F but ξpn−t+1 / F, where t 0,1,...,n 1 . ∈ ∈ ∈ { − } Let C act on L = F(µ ,...,µ ) by extending a faithful linear repre- pn 1 pn sentation ψ : C GL (F) on an F-vector space V L generated pn pn → ⊆ by µ ,...,µ . Let E be a fixed field of C . Then m(L/E) = t. 1 pn pn Proof. Set D = F[µ ,µ ,...,µ ]. 1 2 pn ThenD isauniquefactorizationdomain. Moreover, L = F(µ ,...,µ ) 1 pn is the quotient field of D, and C acts naturally on L via an extension pn

Cyclic algebras, Schur indices, norms, and Galois modules

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