Trace map and regularity of finite extensions of a DVR

Contents lists available atScienceDirect

Journal

of

Number

Theory

www.elsevier.com/locate/jnt

Trace

map

and

regularity

of

finite

extensions

of a DVR

Fabio Tonini

Freie Universität Berlin, FB Mathematik und Informatik, Arnimallee 3, Zimmer 112A, 14195 Berlin, Germany

a r t i c l e i n f o a b s t r a c t

Article history: Received10May2016

Receivedinrevisedform20August 2016

Accepted22August2016 Availableonline8October2016

CommunicatedbyD.Goss Keywords: Regularity Tracemap Ramificationindex Tame

Weinterprettheregularity of a finite andflat extensionof a discrete valuation ring in terms of thetrace map of the extension.

©2016ElsevierInc.Allrightsreserved.

1. Introduction

LetR bearingand A beanR-algebra whichisprojective andfinitely generated as

R-module. We denote by trA/R: A −→ R the trace map and by trA/R: A −→ A∨ =

HomR(A,R) the map a −→ trA/R(a· −). It is a well known result of commutative

algebra that the étaleness of the extension A/R isentirely encoded in the trace map trA/R: A/R is étale if and only if the map trA/R: A −→ A∨ is an isomorphism (see

[Gro71,Proposition4.10]).Inthiscase,ifR isaDVR(discretevaluationring)itfollows

E-mail address:tonini@zedat.fu-berlin.de. http://dx.doi.org/10.1016/j.jnt.2016.08.019 0022-314X/©2016ElsevierInc.Allrightsreserved.

thatA is regular(thatisaproductofDedekinddomains), whiletheconverseisclearly nottruebecauseextensions ofDedekinddomainsareoftenramified.

InthispaperweshowhowtoreadtheregularityofA intermsofthetracemaptrA/R.

In order to express our result we need some notations and definitions. Let us assume from nowonthattheringR isaDVRwithresiduefieldkR.Wefirstextendthenotion

of tameextensionsand ramificationindex:givenamaximalidealp ofA weset

e(p, A/R) = dimkR(Ap⊗RkR)

[k(p) : kR]

wherek(p)= A/p,andwecallittheramificationindex ofp intheextensionA/R.Notice thate(p,A/R)∈ N (seeLemma 2.3).WesaythatA/R is tame (overthemaximal ideal of R) if the ramification indexes of allmaximal ideals of A are coprime with char kR.

Those definitionsagree withtheusualoneswhenA is aDedekinddomain.Wealso set

QA/R= Coker(A  trA/R

−→ A∨_{), f}A/R_{= l(Q} A/R)

where l denotesthelengthfunction.AlternativelyfA/R _{canbe seenasthevaluationof}

the discriminant section det trA/R. We also denote by |Spec(A⊗RkR)| the numberof

primes ofA⊗RkR:thisnumbercanalsobecomputedas

| Spec(A ⊗RkR)| =



p maximal ideals of A

[Fp: kR]

where Fp denotesthemaximal separableextension ofkR insidek(p)= A/p (see

Corol-lary 2.4). Finally we will say thatA/R has separable residuefields (over themaximal idealofR)ifforallmaximalidealsp ofA thefiniteextensionk(p)/kRisseparable.The

theorem wearegoingtoproveisthefollowing:

Maintheorem. LetR beaDVR andA beafiniteandflat R-algebra.Thenwe havethe inequality

fA/R≥ rk A − | Spec(A ⊗RkR)|

and thefollowingconditionsare equivalent:

1) theequalityholdsin theinequalityabove;

2) A is regularandA/R istame withseparableresiduefields;

3) A/R is tame with separable residue fields and the R-module QA/R is defined over

Theimplication 2) =⇒ 1) is classical,becauseit follows directlyfrom the relation betweenthedifferentandramificationindexes(see[Ser79,III,§6,Proposition13]). No-ticethattheétalenessofA/R isequivalenttoanyofthefollowingconditions:fA/R= 0, rk A=|Spec(A⊗RkR)|, QA/R= 0.

TheaboveresulthasalreadybeenprovedinmyPh.D. thesis[Ton13,Theorem4.4.4]

underthe assumption thatafinite solvable groupG acts on A in away that AG _{= R}

and using a completely different strategy: using induction and finding a filtration of

R-algebrasofR⊆ A startingfromafiltrationofnormalsubgroupsofG.Maintheorem

isanessentialingredientintheproof of[Ton15,Theorem C]whichgeneralizes [Ton13, Theorem4.4.7].

Thepaperis organizedas follows. Inthefirstsectionwe discussthenotionof tame-nessandramification index andrecall somebasicfactsof commutativealgebraand,in particular,aboutthetracemap.InthesecondsectionweprovetheMaintheorem.

1.1. Notation

Allringsandalgebras inthis paperarecommutativewith unity.

Given a ring R and aprime q we denote by k(q) the residue fieldof Rq. If A is a

finiteR-algebra wesaythatA/R hasseparable residuefieldsover aprimeq of R if for allprimesp ofA lyingoverq thefiniteextensionoffieldsk(p)/k(q) isseparable.Wesay thatA/R hasseparableresiduefields ifithasthisproperty overallprimesofR.

Thefollowingconditionsareequivalent(see[Gro64,Proposition1.4.7]):A/R isfinite, flat and finitely presented; A is finitely generated and projective as R-module; A is

finitely presented as R-module and for all primes q of R the Rq-module Aq is free. In

this situation there is a well-defined trace functionHomR(A,A) −→ R which extends

the usual trace of matrices and commutes with arbitrary extensions of scalars. The trace map of A/R, denoted by trA/R: A −→ R (or simply trA), is the composition

A−→ HomR(A,A)−→ R,wherethefirstmapisinducedbythemultiplicationinA.

If R isa local ring we denote by mR its maximal ideal and by kR its residue field.

A DVRisadiscretevaluation ring.

Ifk isafieldwe denoteby k analgebraic closureof k andbyks aseparable closure ofk.

2. Preliminariesontamenessandtrace map

We fix a ring R and a finite, flat and finitely presented R-algebra A over R, that is an R-algebra A which is finitely presented as R-module and such that Aq is a free

Rq-moduleoffiniterankforallprimeq ofR.

Definition2.1.Given aprimeidealp ofA lying abovetheprimeq ofR weset

e(p, A/R) = dimk(q)(Ap⊗Rqk(q))

and we call it the ramification index of p in the extension A/R. We say that A/R is tame in p∈ Spec A ife(p,A/R) is coprimewithchar k(q),is tame overq∈ Spec R ifit is tameover allprimesof A overq and,finally, wesaythatitistame ifitis tameover allprimes ofR (orinallprimesofA).

Remark2.2.Thenumbere(p,A/R) isalwaysanaturalnumberasshownbelow.Moreover ifR andA areDedekinddomains,thenotionoframificationindexagreeswiththeusual one.

Lemma 2.3. Let p be a prime of A lying over the prime q of R and denote by F the maximal separable extension of k(q) inside k(p). Then e(p,A/R) is a natural number and

Ap⊗Rqk(q) B1× · · · × Bswith Bi local, dimk(q)Bi= e(p, A/R)[k(p) : F ]

and s = [F : k(q)]

Proof. WecanassumethatR = k(q)= k isafieldandthatA islocal.SetalsoL= k(p) and write

L⊗kk C1× · · · × Cr, A⊗kk B1× · · · × Bs

for thedecompositions intolocal rings. Inparticular wehavethatr = s because A⊗k

k−→ L⊗kk issurjectivewithnilpotentkernel.Moreoverthismapsplitsasaproductof

surjectivemapsBi−→ Ci.DenotebyJitheirkernelsandbyφ : A⊗kk−→ B1×· · ·×Bs

thepreviousisomorphismofrings.Forallt∈ N (includingt= 0)wehave

φ((mA⊗kk)t) = J1t× · · · × Jst

Inparticular forallt∈ N wehave  Jt 1 J₁t+1  × · · · ×  Jt s Jst+1  (mA⊗kk)t (mA⊗kk)t+1 mtA mt+1_A ⊗kk (L ⊗kk)f (t) C1f (t)× · · · × Csf (t)

as L⊗kk-modules,where f (t)= dimL(mtA/m t+1

A ).Wecanconcludethat

 Jt i J_it+1  Cf (t) i for all i, t Inparticular dim_kBi=  t∈N dim_k  Jt i J_it+1  = (dim_kCi)  t∈N f (t)

becauseJi isnilpotent.Similarlywehave dimkA =  t∈N dimk  mt A mt+1_A  = [L : k] t∈N f (t)

Inparticular [L: k]| dimkA, so thattheramification index isanaturalnumber.Bya

directcomputation we see thateverything follows if we show thatdim_kCi = [L : F ].

Since F/k is separable we know that F⊗k k k [F :k]

. Each factor corresponds to an embeddingσ : F−→ k such thatσ_|k= idk and

L⊗kk L ⊗F (F ⊗kk)



σ∈Homk(F,k)

L⊗F,σk

This isexactlythe decompositioninto local rings because,sinceL/F is purely insepa-rable,alltheringsL⊗F,σk arelocal.Thisends theproof. 2

Corollary2.4. Letq beaprimeofR. Then

| Spec(A ⊗Rk(q))| =



p primes of A over q

[Fp: k(q)]

whereFp denotesthemaximal separableextensionof k(q) insidek(p).

Proof. Itfollows from Lemma 2.3 using the factthat A⊗Rk(q) is the product of the

Ap⊗Rqk(q) forp running throughallprimes ofA overq. 2

Definition2.5.Letp beaprimeofA lyingovertheprimeq ofR.Wedenotebyh(p,A/R)

thecommonlengthofthelocalizations ofAp⊗Rqk(q),thatis,following notationfrom

Lemma 2.3,h(p,A/R)= dimkBi= e(p,A/R)[k(p): F ].

Lemma2.6.Letp beaprimeofA,RbeanR-algebraandpbeaprimeofA= A⊗RR

overtheprimep. Then

h(p, A/R) = h(p, A/R)

Proof. WecanassumethatR = k andR = karefieldsandthatA islocal.Moreoverby definitionofthefunctionh(−) wecanalsoassumethatk andk arealgebraicallyclosed. Inthis caseh(p,A/k)= dimkA and, sinceA = A⊗k k is again local,h(p,A/k)=

dimkA = dimkA. 2

Corollary2.7.Letp beaprimeof A.ThenA/R is tamein p andk(p)/k(q) isseparable ifand onlyif thenumberh(p,A/R) iscoprimewith char k(q).

Proof. We canassumethatR = k = k(q) isafieldand A islocal with residuefieldL.

LetalsoF bethemaximalseparableextensionofk insideL.ThankstoLemma 2.3,the last condition inthe statementis thatthe numbere(mA,A/k)[L : F ] is coprimewith

char k. SinceL/F ispurely inseparable,sothat[L: F ] iseither1 orapowerofchar k, this isthesameas A/k beingtame andL= F ,thatisL/k is separable. 2

Remark2.8.Letq beaprimeofR,R be anR-algebraandq beaprimeofR overR.

ByLemma 2.6andCorollary 2.7itfollowsthatA/R istame overq andA⊗Rk(q) has

separableresiduefieldsifandonlyifthesameistruefortheextension(A⊗RR)/Rwith

respecttotheprimeq.Ontheotherhandtamenessalonedoesnotsatisfythesamebase changeproperty, and,inparticular,thefunctione(−) doesnotsatisfyLemma 2.6. The counterexampleisafinitepurelyinseparableextensionL/k:wehavethate(mk,L/k)=

1,so thatL/k istame,whilee(m_k,L⊗kk/k)= [L: k] becauseL⊗kk islocal,so that

L⊗kk/k is nottame.

Lemma 2.9. Assume that R = k is a field, that A is local andset π : A−→ kA for the

projection. Then

trA/k= e(mA, A/k) trkA/k◦π

Proof. Set P = mA,L = kA andlet x1,i,. . . ,xri,i ∈ P

i _{be elements whose projections}

form anL-basisofPi_/Pi+1_{.Weset x}

1,0 = 1.Letalsoy1,. . . ,ys∈ A beelementswhose

projectionsform ak-basisofL, wheres= dimkL. Itiseasytoseethatthecollection

{xα,iyβ}1≤α≤ri,1≤β≤s

isak-basisofPi_/Pi+1 _{foralli≥ 0.Byaninverseinductionstartingfromthenilpotent}

index ofP ,italsofollowsthat

Bn={xα,iyβ}1≤α≤ri,1≤β≤s,i≥n

is ak-basisofPn _{foralln≥ 0.Inparticular, whenn= 0 we getak-basisofA= P}0_.

We are going to compute the trace map trA over the basis B0. Consider an index

i> 0.Forallpossible α,β,γ,δ,j wehavethat

z = (xα,iyβ)(xγ,jyδ)∈ Pi+j ⊆ Pj+1

Thus z isa linearcombination of vectors inBj+1, which doesnot contain (xγ,jyδ). It

follows that trA(xα,iyβ) = 0 for all i > 0, that is trA(P ) = 0 which agrees with the

formula inthestatement.ItremainstocomputetrA(yβ).Write

yβyδ =



q

Itfollowsthat

trL/k(π(yβ)) =



δ

bβ,δ,δ

Let’smultiplynowyβ with anelement xα,iyδ,obtaining

z = yβ(xα,iyδ) = xα,iuβ,δ+



q

bβ,δ,qxα,iyq

Ifi= 0,sothatα = 1 andx1,0= 1,thecoefficientofz withrespecttoyδ isbβ,δ,δbecause

uβ,δ ∈ P .Ifi> 0 thenthecoefficientofz withrespectto(xα,iyδ) isagainbβ,δ,δ because

xα,iuβ,δ ∈ Pi+1 and thuscan be writtenusing only the vectorsin Bi+1. Inconclusion

wehavethat trA(yβ) =  i,α,δ bβ,δ,δ = (  δ bβ,δ,δ)(  α,i 1) = trL(π(yβ))C with C = (  i,α 1)

ThustrA= C trL◦π and,finally,

C = ( i,α 1) = i≥0 dimL( Pi Pi+1) = dimkA [L : k] = e(mA, A/k) 2 Corollary2.10.Assumethat R andA arelocal. Then

1) trA/R(mA)⊆ mR;

2) ifkA= kR andrk A∈ R∗ thenKer trA/R ⊆ mA.

Proof. Since trA/R⊗RkR = tr(A⊗RkR)/kR point 1) follows from Lemma 2.9 because

tr(A⊗RkR)/kR(mA⊗RkR) = 0. Assume now the hypothesis of 2) and let x∈ Ker trA. If

x∈ m/ A thereexistsλ∈ R∗ suchthaty = x− λ∈ mA,so that

trA(x) = 0 = rk Aλ + trA(y)∈ R∗+ mR= R∗

whichisimpossible. 2

Lemma2.11.If R andA arelocal then

trA/R: A−→ R is surjective ⇐⇒ h(mA, A/R) and char kR are coprime

Proof. ByNakayama’slemmatrA/R: A−→ R issurjectiveifandonlyiftrA/R⊗RkR=

trA⊗kR/kR: A ⊗ kR −→ kR is so. Thus we can assume that R = k is a field. By

Lemma 2.9 trA/k is surjective ifand only iftrkA/k is surjective, i.e. kA/k is separable,

3. RegularityoffiniteextensionsofDVR

WefixaDVRR andafiniteandflatR-algebraA,sothatA isfreeoffiniterankrk A as R-module.Wewill usethefollowing notation

 trA/R: A−→ A∨, a−→ trA/R(a· −) QA/R= Coker(A  trA/R −→ A∨_{), f}A/R_{= l(Q} A/R)

where l denotes the length function. For simplicity we will replace A/R with A if no confusion canarise.

Remark3.1.BystandardargumentswehavethatfA_{coincides withthevaluationover}

R ofdet( trA).Moreoverthefollowingconditionsareequivalent:

1) A isgenericallyétaleoverR;

2) fA<∞;

3) trA: A−→ A∨ isinjective.

In particular we see that all three conditions in Main theorem implies that A/R is

genericallyétale.This alsomeansthatA/R istame withseparableresiduefields ifand onlyifA⊗RkR/kR, orA⊗RkR/kR,isso.

Ifβ ={x0,. . . ,xn} isanR-basisofA thenthematrixoftrA: A−→ A∨withrespect

to β and its dualis given byT = (trA(xixj))i,j. In particularwe canconclude that,if

trA: A−→ R is notsurjective, thenfA≥ rk A,becauseallentriesofT belongstomR.

IftrA(1)= rk A∈ R∗wehaveadecompositionA= R1⊕Ker trA.Inparticularifx0= 1

and x1,. . . ,xn∈ Ker trA (suchabasisalwaysexistslocally)thematrixT hastheform



rk A 0

0 N



and thereforeQA Coker N.

Remark3.2.LetRsbethestrictHenselizationofR andsetAs= A⊗RRs.Inparticular

Rs _{is adiscretevaluation ring with residuefieldthe separableclosure k}s

R of kR. Using

thattheextensionR−→ Rs_{isfaithfullyflatand unramified,itiseasytoseethat}

fA/R= fAs/Rs, | Spec(A ⊗RkR)| = | Spec(As⊗Rsk_Rs)|

thatQA/R isdefinedoverkR ifand onlyifQAs_/Rs isdefinedoverk_Rs and thatA/R is

tame withseparableresiduefields ifandonlyifAs_/Rs_isso.

Since Rs _{is Henselian, we have a decomposition A}s _{= A}

1× · · · × Aq where the Aj

|Spec(A⊗RkR)|.SincetrAs_/Rs: As−→ (As)∨ isthedirectsumofthetr_Aj/Rs: A_j −→ (Aj)∨ wehave QAs_/Rs j QAj/Rs and f As_/Rs = j fAj/Rs

Finally we have thatthe following three conditions are equivalent: A is regular; As is regular;Aj is regularforallj = 1,. . . ,q.

Proof. (of Main theorem) By Remark 3.1 and Remark 3.2 we can assume that R is

strictly Henselian, that A is local, so that |Spec(A⊗R kR)| = 1, and that A/R is

generically étale. Moreover in this case the following three conditions are equivalent byCorollary 2.7and Lemma 2.11:A/R is tame withseparable residuefields;kA= kR

andrk A∈ R∗; trA/R: A−→ R issurjective.

Inequality and 1) ⇐⇒ 3). By Remark 3.1 we can assume that trA: A −→ R is

surjective, so thatkA = kR,rk A∈ R∗. ByCorollary 2.10 wealso haveKer trA ⊆ mA

and trA(mA)⊆ mR. Using Remark 3.1 and itsnotation, we see thatfA = vR(det N ).

Ifi,j≥ 1 then xixj ∈ mA and thustrA(xixj)∈ mR, thatisallentriesofN belongsto

mR.Ifπ∈ mR isanuniformizerofR wecanwriteN = πN whereN isamatrixwith

entriesinR.Inparticular

det N = πrk A−1det N

Applying thevaluation of R we get fA = vR(det N ) = rk A− 1+ vR(det N) and the

desiredinequality.

IfQA Coker(N) is definedover kR we obtainasurjective map krk AR −1 −→ QA⊗

kR QAandthereforethatfA= l(QA)≤ rk A− 1.Conversely,iffA= rk A− 1,which

meansthatN: Rrk A−1 _{−→ R}rk A−1 _{isanisomorphism,we haveQ}

A Coker(πN),so

thatQA kRrk A−1 isdefinedoverkR asrequired.

2) =⇒ 1) Lett∈ A be ageneratorof themaximalideal.ThekR-algebraA⊗ kR is

local,with residuefieldkR and its maximalideal isgenerated by theprojection t oft.

It is easyto conclude thatA⊗ kR = kR[X](XN) where N = rk A andX corresponds

to t.ByNakayama’slemmaitfollowsthat1,t,. . . ,tN−1isanR-basisofA andthusthat

A R[Y ]/(YN_{− g(Y )) whereY correspondstot,deg g < N andallcoefficientsofg are}

inmR.SincemA=t,mRA,theconditionthatA isregulartellsusthatvR(g(0))= 1.

We are going to compute the valuation of the determinant of trA: A −→ A∨ writing

thismapintermsofthebasis1,t,. . . ,tN−1,thatisthevaluationofthedeterminantof thematrix(trA(ti+j))0≤i,j<N (seeRemark 3.1).Setqs= trA(ts).ByCorollary 2.10ora

directcomputationwe knowthatqS ∈ mR fors> 0.Inparticular, sincevR(g(0))= 1,

wealsohavevR(qN)= 1.MoreoveritfollowsbyinductionthatvR(qs)> 1 ifs> N .Set

SN forthegroupofpermutationsoftheset{0,1,. . . ,N− 1}. Wehave

det((trA(ti+j))0≤i,j<N) =



σ∈SN

WeclaimthatvR(zσ)≥ N forallσ∈ Snbutthepermutationσ(0)= 0 andσ(i)= N−i

fori= 0,forwhichvR(zσ)= N−1.Thiswillconcludetheproof.Letσ∈ SN.Ifσ(0)= 0,

then allthe N -factors of zσ are in mR, which implies thatvR(zσ)≥ N. Thus assume

thatσ(0)= 0.Sinceq0= trA(1)= rk A∈ R∗ weseethatzσ is,uptoq0,theproductof

N−1 elementsofmR.Ifoneofthosefactorsisoftheformqi+σ(i)withi+ σ(i)> N then

vR(zσ)≥ N becausevR(qs)≥ 2 ifs> N . Thustheonlycaseleft iswhenσ(i)≤ N − i

for all 0< i < N and σ(0) = 0. But, arguing by induction,this σ is uniqueand it is given byσ(0)= 0 andσ(i)= N− i. Inthis casewe have

zσ= rk A(tr(tN))N−1

and thereforevR(zσ)= N− 1 asrequired.

1) =⇒ 2) Since 1) ⇐⇒ 3), we already know that A/R is tame with separable residue fields.We haveto provethatA is regular.Set k(R) forthe fraction fieldof R.

SinceA⊗ k(R) isetaleoverk(R),itisaproductoffieldsL1,. . . ,Lswhichareseparable

extensions of k(R).Let B be the integral closure of R insideA⊗ k(R). Wehave that

A⊆ B andthatB = B1× · · · × BswhereBi istheintegralclosureofR insideLi.Since

R isstrictly HenselianandtheBi aredomains, itfollowsthattheyarelocal.Moreover

sinceR isaDVRwecanalsoconcludethattheBi areDVR.Wearegoingtoprovethat

s= 1 andA= B.Noticethatrk B = rk A anddenote byj : A−→ B theinclusion.By computing trA andtrB over A⊗ k(R) B ⊗ k(R) wecanconcludethat(trB)|A= trA.

Inparticular weobtainacommutativediagram offreeR-modules

A A∨ B B∨  trA j  trB j∨

Noticethatdet j = det j∨.Thustakingdeterminantsandthenvaluationsweobtainthe expression fA= 2vR(det j) + fB = 2vR(det j) + s  i=1 fBi

Sincek isseparablyclosedandthankstoCorollary 2.4wehavethat|Spec(Bi⊗Rk)|= 1.

InparticularfBi≥ rk B

i− 1 bytheinequalityinthestatement.SincefA= rk A− 1 we

get

s≥ 1 + 2vR(det j)

WearegoingtoprovethatvR(det j)≥ s− 1. Thiswillendtheproofbecauseitimplies

vR(det j) = l(B/A)≥ dimkR(B/(A + mAB))

Denote by e1,. . . ,es ∈ B the idempotents corresponding to the decomposition B =

B1×· · ·×Bs.Wewillprovethate2,. . . ,esarekR-linearlyindependentinB/(A+mAB),

thatisweprovethatif

x = x1e1+· · · + xses∈ A + mAB where x1= 0 and xi∈ R

thenallxiareinthemaximalidealmR.Noticethat,sincekA= kR,1 andmAgenerates

A as R-module. In particular A + mAB is generated by 1 and mAB as R-module.

Moreoversince themaps A−→ B −→ Bi map mA insidemBi it follows thatmAB ⊆

mB1× · · · × mBs.Thusx canbewrittenas

x = α + c1e1+· · · + cseswith α∈ R and ci∈ mBi

insideB.In particular0= x1= α + c1 inB1,whichimplies thatα∈ mB1∩ R = mR. Thusifi> 0 wehave

xi= α + ci ∈ mBi∩ R = mR

asrequired. 2

Weshowviasomeexampleshowtheconditionsoftamenessandseparabilityofresidue fieldsinMaintheorem cannotbeomitted.

Example3.3.LetR =Z2be theringof 2-adicnumbersandconsider theR-algebra

A = R[X]

(X2_{− c)} with c∈ R

WehavethattrA(X)= 0 andtrA(X2)= 2c,sothatthematrixoftrA: A−→ A∨ is

M =  2 0 0 2c  InparticularfA_{= v}

R(det M )≥ 2> rk A− |Spec(A⊗RkR)| andQA isdefinedoverF2

ifandonlyifc∈ R∗.

Assume c = 2t+ d2 _{with t,d ∈ R, so that A/m}

RA F2[Y ]/(Y2), A is local with

maximalideal(mR,X− d),hasseparableresiduefieldsanditisnottame.Weseethat

QA isdefinedoverF2 ifand onlyifd∈ R∗,whileA isregularifandonlyift∈ R∗.

Assumec∈ R∗andthatc isnotasquareinF2. InthiscaseA islocalwithmaximal

idealmRA,A/R istame,itsresiduefieldisnotseparable,itisregularandQAisdefined

Acknowledgments

IwouldliketothankAngeloVistoliandDajanoTossiciforalltheusefulconversations we hadandallthesuggestionsI received.

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