© Hindawi Publishing Corp.
TWISTED FORMS OF FINITE ÉTALE EXTENSIONS
AND SEPARABLE POLYNOMIALS
FRANK DEMEYER
(Received 25 April 2000)
Abstract.Examples of twisted forms of finite étale extensions and separable polynomials are calculated using Mayer-Vietoris sequences for non-abelian cohomology.
2000 Mathematics Subject Classification. 13B05, 13B25, 13B40.
1. Introduction. LetSbe a finite étale extension of a commutative Noetherian ring R(a finitely generated projective separable extension ofR). A twisted form ofS(in the Zariski topology) is a finite étale extensionTofRwithRP⊗SRP⊗T asRP-algebras for each prime idealP ofR. In this caseS is locally isomorphic to T. IfQ⊂P are prime ideals ofRandRP⊗SRP⊗T, thenRQ⊗SRQ⊗Tso prime can be replaced by maximal in the definition of the twisted form. In this paper, we study the set of isomorphism classes of twisted forms of S. We especially concentrate on the case whereSR[t]/(p(t)), wherep(t)is a separable polynomial inR[t]. Throughout this paper,Rdenotes a commutative Noetherian ring.
We first observe the well-known facts that ifRis an integrally closed domain, then there are no twisted forms ofS and in general the twisted forms ofSare in bijective correspondence withH1(X,Aut()), where Aut
R()is the sheaf of automorphisms onX=Spec(R)associated to AutR(S). We check that H1(X,Aut())is unchanged modulo a nilpotent ideal.
With some hypotheses on the sheaf AutR(), whenRis a one-dimensional domain or ifRis a reduced one-dimensional ring with connected spectrum,H1(X,Aut())fits into a Mayer-Vietoris sequence which makes computations possible. These computa-tions are the point of this article. For infinitely many prime numbersp,qwe give a class of examples of integral domainsRand separable polynomialstp−q∈R[t]with the cardinality of the set of isomorphism classes of twisted forms ofSR[t]/(p(t))equal to(p−1)!. Whenp=3 these twisted formsT are isomorphicto algebrasT ⊕2
j=0Ijtj witht3=qandIis a fractional ideal withI3=R. We give an example of twisted forms that do not have this structure. We also give one-dimensional rings over which finite étale extensions may not have either a primitive element nor a normal basis but which are twisted forms of extensions which do. We also give a separable polynomial which is irreducible overRbut factors into linear factors at each localizationRP ofRand modulo each maximal ideal ofR.
with the Zariski topology andᏻXthe associated sheaf of rings onX,the sheaf ofᏻX -algebras associated toS(see [5, pages 70 and 130]). For each open setU⊂Xassociate toU the group ofR-algebra automorphisms Autᏻ(U)((U)), and if V⊂U associate the natural restriction Autᏻ(U)((U))→Autᏻ(V )((V)). We begin by recording for the readers convenience that AutR()forms a sheaf onXand give some of its properties (see also [3]).
Lemma2.1. LetSbe a finite étaleR-algebra.
(a)IfUis an open set inXand{Vi}is an open cover ofU, and ifσ∈Autᏻ(U)((U))
satisfies1⊗σ=1inAutᏻ(Vi)((Vi))for alli, thenσ=1inAutᏻ(U)((U)).
(b)IfUis an open set inXand{Vi}is an open cover ofU, and ifσi∈Autᏻ(Vi)((Vi))
with restrictions σi = σj in Autᏻ(Vi∩Vj)((Vi∩Vj)), then there is an element σ ∈
Autᏻ(U)((U))whose restriction toAutᏻ(Vi)((Vi))isσi.
Lemma 2.2. Let S,T be finite étaleR-algebras andP a prime ideal in R. Letσ :
RP⊗S→RP⊗T be anRP-algebra homomorphism. Then there is an open setU inX
withP∈U and anᏻ(U)-algebra homomorphismτ:ᏻ(U)⊗S →ᏻ(U)⊗T such that
1⊗τ=σ∈AlgRP(RP⊗S,RP⊗T ). Ifσ is an isomorphism thenUcan be chosen soτis
an isomorphism.
IfS,T are finite étaleR-algebras andRP⊗SRP⊗T for all prime idealsP ofRwe say thatSandTare locally isomorphic or thatTis a twisted form ofS. IfTis a twisted form ofS thenLemma 2.2implies there is an open coverᐁ= {Ui}ofXand isomor-phismsσi:ᏻ(Ui)⊗S→ᏻ(Ui)⊗T for alli. Define an elementa∈Z1(ᐁ,Aut())by assigning to the index pairi,jthe automorphisma(i,j)=σ−1
i σj∈Autᏻ(Ui∩Uj)(ᏻ(Ui∩
Uj)⊗S). Passing to the limit over all covers ofX gives an injection from the set of isomorphism classes of twisted forms ofStoH1(X,Aut()). By descent, (see [7, 2.2, page 110] or [8, page 19]), this assignment is onto so H1(X,Aut())classifies the twisted forms ofS. In the next result we point out that, as with the Brauer group, there are no nontrivial twisted forms in the geometrically irreducible case.
Proposition2.3. IfRis an integrally closed domain andSis a finite étaleR-algebra, thenH1(X,Aut())= {1}.
Proof. We can writeS =S1⊕ ··· ⊕Sk, where eachSihas a connected spectrum. By [6, Theorem 4.3] or [8, Proposition 3.19, page 28], eachSiis an integrally closed domain. LetKbe the quotient field ofR. ThenK⊗S= ⊕k
i=1K⊗Si, where eachK⊗Siis a finite-dimensional separable field extension ofK, andSiis the integral closure ofR inK⊗Si. Letσ∈AutK(K⊗S), then since the image of an integral element is integral,
σ|S∈AutR(S)andσ=1⊗σ|S. Therefore, the natural map AutR(S)→AutK(K⊗S)is a bijection which implies that the sheaf Aut()is constant. Therefore,H1(X,Aut())= {1}.
Corollary2.4. LetRbe an integrally closed domain andS,Tfinite étaleR-algebras.
IfRp⊗SRP⊗T asRP-algebras for eachP∈Spec(R), thenST asR-algebras.
(a)IfRhas a connected spectrum thenρis a one-to-one map.
(b)IfIis nilpotent thenρis a bijection map.
Proof. (a) Assume first thatS is connected and Galois over R. Then AutR(S) = Galois group ofSoverR= the Galois group ofS/ISoverR/I⊂AutR/I(S/IS)soρis a one-to-one map in this case. IfSis connected but not necessarily Galois, imbedSin a connected Galois extensionNofR(see [2, Theorem 3.2.9]). EveryR-automorphism of Sextends to an automorphism ofN, and any two such extensions differ by an element ofH= {σ∈AutR(N)|σ|S=1}(see [2, Chapter 3]). By flatness,S/ISis a subalgebra ofN/IN and His the subgroup of the Galois group ofN/IN over R/IfixingS/IS. Ifτ,σ ∈AutR(S)with extensions ¯τ,σ¯ toNand with the natural image ofτ =σ in AutR/I(S/IS), then ¯τ−1σ¯∈Hsoτ=σ onSandρis a one-to-one map in this case.
IfR is connected thenS=Se1⊕ ··· ⊕Sm witheiej=eiδi,j, Sei connected for all
i,j. Let σ ∈AutR(S) and assume σ induces the identity automorphism on S/IS. Thenσ (ei)=ei for all isinceei+IS≠ej+IS for anyi≠j. Thereforeσ induces
(σ1,...,σm)∈ ×iAutR(Sei). Sinceσ is the identity on each of these summands, by the previous paragraphσ is the identity onS andρis a one-to-one map.
(b) We can writeR=R1⊕ ··· ⊕Rk, where eachRihas a connected spectrum. Then there are the corresponding decompositions S =S1⊕ ··· ⊕Sk and I =I1⊕ ··· ⊕Ik withIj a nilpotent ideal ofRj (in particular, noIj=Rj and AutR(S)= ×jAutRj(Sj), AutR/I(S/IS)= ×jAutR/Ij(Rj⊗S/Ij(Rj⊗S))). Thus we can assumeRis connected.
IfIis nilpotent, thenISis nilpotent and idempotents can be lifted modulo a nilpo-tent ideal, soR/Ihas a connected spectrum and ifS is connected thenS/ISis con-nected. AssumeSis connected and Galois. Then AutR(S)= the Galois group ofSover
R = the Galois groupS/IS over R/I= AutR/I(S/IS)soρis bijective in this case. If
S/Ris Galois, thenS=Se1⊕···⊕Semwitheiej=eiδi,j,Seiconnected,SeiSejfor alli,j, and eachSei Galois over R with Galois group of ordern=rank(Sei). Thus
|AutR(S)| =m!nm. Since idempotents can be lifted moduloI, we get the same count for|AutR/I(S/IS)|, so by part (a),ρis onto in this case.
IfS/Ris finite étale, there is a Galois extensionNofRcontainingSconstructed in the following way. WriteS=Se1⊕···⊕Semas above and letLbe a connected Galois extension ofRcontaining all theSei. LetN=Le1⊕ ··· ⊕Lem. Let ¯τ∈AutR/I(S/IS). Then one can extend ¯τ to ¯γ∈AutR/I(N/IN)which corresponds, by the paragraph above, toγ∈AutR(N). Then(γ(S)+IS)/I=S/Isoγ(S)⊂S. Thereforeγ|S∈AutR(S) andρ(γ|S)=τ¯. Thusρis a bijection in every case.
Corollary2.6. LetX=Spec(R),Ithe nil radical ofR andXred=Spec(R/I). IfS
is a finite étaleR-algebra, thenH1(X,Aut())andH1(Xred,Aut(/Ᏽ))are bijective
with one another.
Proof. ByLemma 2.5,X=Xredand Aut()=Aut(/Ᏽ).
Example2.7. (a) LetRdenote the real numbers andCthe complex numbers. Let R=R⊕RandS=C⊕C. Letσ be complex conjugation. Then(1,1)=(1,σ )on the first summand but(1,1)≠(1,σ )so the map AutR(S)→AutR/I(S/IS)is not always a one-to-one map.
S=R[t]/(p(t)). Sincep(t)is irreducible, AutR(S)=C3(the cyclic group of order 3) butR/(x)=CandS/(x)S=C⊕C⊕ChasC-automorphism groupS3(the symmetric group on three letters). It is not always the case that AutR(S)→AutR/I(S/IS)is onto.
Mayer-Vietoris I. LetRbe a one-dimensional integral domain with module finite integral closure ¯R and conductorc= {x∈R|R¯·x⊂R}. LetS be a finite étaleR -algebra. Assume the following.
(a) IfPis a maximal ideal inRcontainingcthen there is only one maximal idealQ in ¯Rlying overP.
(b) If P is a maximal ideal in R containing c then the natural map AutR(S)→ AutR/P(S/PS)is surjective.
(c) ¯R⊗SR¯(n), wheren=rank R(S).
Then there is an exact sequence of pointed sets
1 →AutR(S) α→AutR¯(R¯⊗S)×AutR/cR/c⊗S β
→AutR/c¯ R/c¯ ⊗S γ→H1X,Aut() →1.
(2.1)
Sketch of the proof. We define explicitly the maps in the sequence. Checking exactness at each term is then a straightforward computation.
The map α is given as α(σ )= (1⊗σ ,1⊗σ ). The mapβ is given as β(τ,ρ)= (1⊗τ)(1⊗ρ)−1.
Let P1,...,Pk be the maximal ideals of R lying over c. Using hypothesis (a), let
Q1,...,Qkbe the maximal ideals in ¯Rlying overcwithQi∩R=Pi(1≤i≤k). Then
c= ∩iPifi = ∩Qeii so R/c = ⊕R/Pfi and ¯R/c= ⊕R/Q¯ ei. Moreover, AutR/c(S/cS)=
×iAutR/Pfi(R/Pifi⊗S)and AutR/c¯ (R/c¯ ⊗S)= ×iAutR/Q¯ ei(R/Q¯ ei⊗S). If(...,σ¯i,...)∈ AutR/c¯ (R/c¯ ⊗S) then by hypothesis (c) and Lemma 2.5, AutR¯(R¯⊗ S) = Sn = AutR/Q¯ eii (R/Q¯
ei
i ⊗S)so there existsσi∈AutR¯(R¯⊗S)withσia lift of ¯σi. Letᐁ= {Uj} be a cover ofX=Spec(R)withPi∈Ujif and only ifi=j. Assign toUj the identity automorphism if noPiis inUj. SinceUi∩Ujdoes not contain any points lying over
c, Autᏻ(Ui∩Uj)((Ui∩Uj))=Snand therefore contains the elementa(i,j)=σi−1σj. It is now easy to checka∈Z1(ᐁ,Aut())is a 1-cocycle and a different choice of cover gives an equivalent cocycle modulo coboundaries, soγ is defined byγ(...,σ¯i,...)=
|a| ∈H1(X,Aut()).
Mayer-Vietoris II. Let R be a reduced ring with X = Spec(R) connected. Let I1,...,Iqbe the set of minimal prime ideals ofRand ¯R= ⊕qj=1R/Ij. Assume dimR/Ij=1 for allj. IdentifyRwith its natural image in ¯Rand letc= {x∈R|R¯·x⊂R}be the conductor. LetY =Spec(R)¯ . LetS be a finite étaleR-algebra.
(a) Assume for each maximal idealQin ¯Rlying overcthe natural map AutR(S)→ AutR/Q¯ (R/Q⊗¯ S)is a surjection.
Then there is an exact sequence of pointed sets
1 →AutR(S) α→AutR¯(R¯⊗S)×AutR/cR/c⊗S β
→AutR/c¯ R/c¯ ⊗S γ→H1X,Aut() δ→H1Y ,Aut(R¯⊗).
Sketch of the proof. As in Mayer-Vietoris I, we give the maps explicitly, then checking exactness is a straightforward computation. The mapαis defined asα(σ )= (1⊗σ ,1⊗σ ). The mapβisβ(τ,ρ)=(1⊗τ)(1⊗ρ)−1.
LetP1,...,Pmbe the maximal ideals inRlying overcandQi,jthe maximal ideals in ¯R lying overcwhere the projection ofQi,jonR/Ijis proper. Writec=∩km=1Pkfk=∩i,jQei,ji,j. Then AutR¯(R⊗S)¯ = ×qj=1AutR/Ij(R/Ij⊗S), AutR/c(R/c⊗S)=×mk=1AutR/Pfk
k (R/P fk k ⊗S), and AutR/c¯ (R/c¯ ⊗S)= ×i,jAutR/Q¯ ei,j
i,j ( ¯
R/Qei,ji,j ⊗S). Let(...,σ¯i,j,...)∈AutR/c¯ (R/c¯ ⊗S)=
×i,jAutR/Q¯ ei,j i,j (
¯
R/Qei,ji,j ⊗S). By hypothesis (a) there isσi,j∈AutR/Ii(R/Ii⊗S)which re-duces to ¯σi,j. For a fixedi,{σi,j}determines an elementσi∈AutR¯(R¯⊗S)where we let σibe the identity in AutR/Ik(R/Ik⊗S)ifkis not anyj. Letᐁ= {Ui}be an open cover of
X=Spec(R)wherePi∈Ujif and only ifi=j. Letγ(...,σ¯i,j,...)= |a| ∈H1(X,Aut()) wherea∈Z1(ᐁ,Aut()) is given bya(i,k)=σ−1
i σk∈Autᏻ(Ui∩Uk)(ᏻ(Ui∩Uk)⊗S). Note,σ−1
i σkis defined sinceUi∩Ukcontains noPjsoᏻ(Ui∩Uj)⊗S=ᏻ(Ui∩Uj)⊗R¯R⊗S¯ . It is clear that the definition ofγ is independent of the choice of cover and our as-signment gives a well-defined map.
Letᐁ= {Ui}be an open cover ofX constructed as above and let π:Y →X be given by restriction. LetVi=π−1(Ui)soᐂ= {Vi}is an open cover ofY. Givena∈
Z1(ᐁ,Aut()), letδ(a)∈Z1(ᐂ,Aut(R¯⊗))byδ(a)(i,j)=a(i,j). This assignment is well defined sinceᏻX(Ui∩Uj)=ᏻY(Vi∩Vj).
Note2.8. If eachR/Ijin Mayer-Vietoris II is integrally closed,H1(Y ,(R¯⊗))= {1} byProposition 2.3. This is the case in all the following examples.
Example2.9. LetQdenote the rational numbers, letp,qbe prime integers, and let ωbe a primitive complexpth root of 1. LetF=Q(ω)andR=F[x,y]/(xp−qyp(y−1)p). Ifqis irreducible inZ[ω]then by Eisenstein’s criterionxp−qyp(y−1)pis irreducible inF[x,y]soRis a one-dimensional Noetherian integral domain. Note thatqis irre-ducible inZ[ω]wheneverpq−1,p≠q. LetS=R[t]/(tp−q). ThenSis a finite étale
R-algebra which is connected sincetp−qis irreducible inF[t]andR/(x,y)=F. Iden-tifyx,ywith their images inR. The integral closure ¯RofRisR(x/y(y−1))and since (x/y(y−1))p=q,tp−q=Πp−1
i=0(t−ωi(x/(y(y−1))))∈R[t]¯ so ¯R⊗SR¯(p). The maximal ideals lying overcinRare(x,y)and(x,y−1)and the only maximal ideal in
¯
Rlying over(x,y)is(y), the only maximal ideal in ¯Rlying over(x,y−1)is(y−1). SinceR/(x,y)F, andR/(x,y−1)F, andtp−qis irreducible in F[t],R satis-fies the hypothesis of Mayer-Vietoris I. But AutR(S)=Cp, the cyclic group of orderp. AutR¯(R¯⊗S)=Sp, the symmetricgroup onp-letters. AutR/c(R/c⊗S)=Cp×Cpand AutR/c¯ (R/c¯ ⊗S)=Sp×Spso for this example the Mayer-Vietoris I sequence becomes
1→Cp →Sp×Cp×CP →Sp×Sp →H1X,Aut() →1. (2.3)
LetK= {(τρ−1,τσ−1)|τ∈S
Example2.10. (a) LetR be a Noetherian domain with quotient fieldK, assumeR contains a primitiventh root of 1 andIis a fractionalR-ideal inKwithIn=R. Assume
p(t)=tn−a∈R[t]is a separable polynomial, andS=R[t]/(p(t)),L=K[t]/(p(t)). Then the subsetT= ⊕n−1
j=0IjtjofLis a twisted form ofS.
(b) LetRbe a reduced Noetherian ring with minimal prime idealsI1,...,Iqand as-sume the dimension of eachR/Ijis one. LetK= ⊕R/Ij. AssumeRcontains a primitive
nth root of 1 andI is a finitely generatedR-submodule ofK with In=R. Assume
p(t)=tn−a∈R[t]is a separable polynomial andS=R[t]/(p(t)),L=K[t]/(p(t)). Then the subsetT= ⊕n−1
j=0IjtjofLis a twisted form ofS.
Note2.11. In Example 2.9ifp=3 andq=2 then the nontrivial twisted formT is constructed as inExample 2.10where the idealI =(y,y−1x), as one can check by showing the associated cocycle inH1(X,Aut())is not a coboundary. NoticeT is free as anR-module. LetᏲ(S) be the set of isomorphism classes of twisted forms ofS which are free asR-modules and assumeS is free as anR-module. Then there is an exact sequence of pointed sets 1→Ᏺ(S)→H1(X,Aut())→H1(X,Gl()). The types of examples given inExample 2.10all lie inᏲ(S), but we give inExample 2.12a twisted form ofSwhose image inH1(X,Gl())is not the identity.
Example 2.12. Let R = R[x,y]/(y −1)(y−x2) and ¯R = R[x,y]/(y−1)⊕ R[x,y]/(y−x2), whereRis the set of the real numbers. ThenR={(p(x,y),q(x,y))∈
¯
R|p(1,1)=q(1,1); p(−1,1)=q(−1,1)}. LetI= {(p(x,y),q(x,y))∈R¯|p(1,1)= q(1,1); p(−1,1)= −q(−1,1)}. ThenIis anR-submodule of ¯R,I2=R, andR
P⊗IRP for all prime idealsPofR. Letp(t)=t2+1, letS=R[t]/(t2+1), and takeT=R⊕It. ThenT is a twisted form ofS andT is not a freeR-module by cancellation [9] soT is a nontrivial twisted form ofS. Notice thatT is a Galois extension ofRwith Galois group of order two induced by complex conjugation but sinceT is not free,T does not have either a normal basis or a primitive element.
Example2.13. ConsiderTas constructed inExample 2.12and letw2+1∈T [w]. Thenq(w)=w2+1 is irreducible in T [w]but for each prime idealQof T,q(w) is reducible inTQ[w]. This gives an example of an irreducible separable polynomial over a connected commutative ring which factors into linear factors over the local-ization at every prime ideal or modulo each maximal ideal. Ifr (w)=w2−1 then S1=R[w]/(q(w)) is not isomorphictoS2=R[w]/(r (w)) sinceS2R⊕Rbut S1 andS2are locally isomorphic. This is an example of two separable polynomials that are locally isomorphicbut not isomorphic(in the sense of [4]).
Example2.14. Let R=Q[x,y]/(y−1)(y−x2)as in Example 2.12. Let p(t)= t3−3t+1 and S =R[t]/(p(t)). Using the Mayer-Vietoris sequence for the Picard group, [1] or [5], one can check the torsion part of the Picard group isC2. But Mayer-Vietoris II givesH1(X,Aut())=C3so ifT is a nontrivial twisted form ofS, thenT is not isomorphictoR⊕It⊕It2for any fractional ideal ofRwithI3=R.
ideal ofRbutp(t)is irreducible inR[t]. Hypothesis (a) of Mayer-Vietoris II fails to hold for this example.
Acknowledgement. This article was written while the author was a visitor at the Eidgenossische Technische Hochschule (ETH) in Zurich, Switzerland. I would like to thank Max Knus and Beno Eckmann of the ETH for several stimulating conversations.
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Frank DeMeyer: Department of Mathematics, Colorado State University, Fort Collins, CO80523, USA
