How many units can a commutative ring have?

0.07 MB·

by  Sunil K. Chebolu

How many units can a commutative ring
have?
SunilK. Chebolu and Keir Lockridge
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Abstract.La´szlo´ Fuchsposedthefollowingproblemin1960,whichremainsopen:classify
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the abelian groups occurring as the group of all units in a commutative ring. In this note,
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we provide an elementary solution to a simpler, related problem: find all cardinal numbers
a
occurringasthecardinalityofthegroupofallunitsinacommutativering.Asaby-product,
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weobtainasolutiontoFuchs’problemfortheclassoffiniteabelianp-groupswhenpisan
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oddprime.

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C 1. INTRODUCTION It is well known that a positive integer k is the order of the
A multiplicative group of a finite field if and only if k is one less than a prime power.
. The correspondingfact for finite commutative rings, however,is not as well known.
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Our motivation for studying this—and the question raised in the title—stems from a
t
a problem posedby La´szlo´ Fuchs in 1960:characterizethe abelian groups that are the
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groupofunitsinacommutativering(see[8]).ThoughFuchs’problemremainsopen,it
[ hasbeensolvedforvariousspecializedclassesofgroups,wheretheringisnotassumed

1 tobecommutative.Examplesincludecyclicgroups([10]),alternating,symmetricand
v finitesimplegroups([4,5]),indecomposableabeliangroups([1]),anddihedralgroups
1 ([2]).InthisnoteweconsideramuchweakerversionofFuchs’problem,determining
4 only the possiblecardinalnumbers|R×|, whereR is a commutativering with group
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ofunitsR×.
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0 InapreviousnoteintheMONTHLY([6]),DitorshowedthatafinitegroupGofodd
. orderis thegroupofunitsofaringifandonlyifGisisomorphictoadirectproduct
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ofcyclicgroupsG ,where|G | = 2ni −1forsomepositiveintegern .Thisimplies
0 i i i
7 that an odd positive integer is the number of units in a ring if and only if it is of the
1 form t (2ni −1)forsomepositiveintegersn ,...,n .AsDitormentionedinhis
Qi=1 1 t
v: paper, this theorem may be derived from the work of Eldridge in [7] in conjunction
i with the Feit-Thompson theorem which says that every finite group of odd order is
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solvable. However, the purpose of his note was to give an elementary proof of this
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resultusingclassicalstructuretheory.Specifically,Ditor’sproofusesthefollowingkey
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ingredients:Mashke’stheorem,whichclassifies(forfinitegroups)thegroupalgebras
overafieldthataresemisimplerings;theArtin-Wedderburntheorem,whichdescribes
the structure of semisimple rings; and Wedderburn’s little theorem, which states that
everyfinitedivisionringisafield.
Inthis note wegiveanotherelementaryproofofDitor’s theoremforcommutative
rings.Wealsoextendthetheoremtoevennumbersandinfinitecardinals,providinga
completeanswertothequestionposedinthetitle;seeTheorem8.Ourapproachalso
givesanelementarysolutionto Fuchs’problemforfinite abelianp-groupswhenp is
anoddprime;seeCorollary4.
2. FINITECARDINALS Webeginwithtwolemmas.Foraprimep,letF denote
p
thefieldofpelements.Recallthatanyringhomomorphismφ: A −→ B mapsunits
tounits,soφinducesagrouphomomorphismφ×: A× −→ B×.
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Lemma1. Letφ: A −→ Bbeahomomorphismofcommutativerings.Iftheinduced
grouphomomorphismφ×: A× −→ B× issurjective,thenthereisaquotientA′ ofA
suchthat(A′)× ∼= B×.
Proof. By the first isomorphism theorem for rings, Imφ is isomorphic to a quotient
of A. It therefore suffices to prove that (Imφ)× = B×. Since ring homomorphisms
mapunitstounitsandφ×issurjective,wehaveB× ⊆(Imφ)×.Thereverseinclusion
(Imφ)× ⊆ B× holdssince everyunit in the subringImφmustalso be a unitin the
ambientringB.
Lemma2. LetV andW befinitefieldsofcharacteristic2.
1. ThetensorproductV ⊗F2 W is isomorphicasa ringto afinite directproduct
offinitefieldsofcharacteristic2.
2. AsF2-vectorspaces,dim(V ⊗F2 W)= (dimV)(dimW).
Proof. Toprove(1),letKandLbefinitefieldsofcharacteristic2.Bytheprimitiveel-
ementtheorem,wehaveL ∼=F [x]/(f(x)),wheref(x)isanirreduciblepolynomial
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inF [x].Thisimpliesthat
2
K[x]
K ⊗F2 L ∼= (f(x)).
Theirreduciblefactorsoff(x)inK[x]aredistinctsincetheextensionL/F is sep-
2
arable.Nowletf(x) = t f (x)bethefactorizationoff(x)intoits distinctirre-
Qi=1 i
duciblefactorsinK[x].Wethenhavethefollowingseriesofringisomorphisms:
t
K[x] K[x] K[x]
K ⊗F2 L ∼= (f(x)) ∼= ( t f (x)) ∼= Y (f (x)),
Qi=1 i i=1 i
wherethelastisomorphismfollowsfromtheChineseremaindertheoremforthering
K[x]. Since eachfactorK[x]/(f (x)) is afinite field ofcharacteristic2, we seethat
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K ⊗F2 Lisisomorphicasaringtoadirectproductoffinitefieldsofcharacteristic2,
asdesired.
For (2), simply note that if {v ,...,v } is a basis for V and {w ,...,w } is a
1 k 1 l
basisforW,then{vi⊗wj|1 ≤ i ≤ k,1 ≤ j ≤ l}isabasisforV ⊗F2 W.
Wemaynowclassifythefiniteabeliangroupsofoddorderthatappearasthegroup
ofunitsinacommutativering.
Proposition3. LetGbeafiniteabeliangroupofoddorder.ThegroupGisisomorphic
tothegroupofunitsinacommutativeringifandonlyifGisisomorphictothegroup
of units in a finite direct product of finite fields of characteristic 2. In particular, an
oddpositiveintegerk is thenumberofunitsinacommutativeringifandonlyifk is
oftheform t (2ni −1)forsomepositiveintegersn ,...,n .
Qi=1 1 t
Proof. The‘if’directionofthesecondstatementfollowsfrom thefactthat,forrings
AandB,(A×B)× ∼= A××B×.
For the converse, since the trivial group is the group of units of F , let G be a
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nontrivial finite abelian group of odd order and let R be a commutative ring with
group of units G. Since G has odd order, the unit −1 in R must have order 1. This
impliesthatRhascharacteristic2.
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Nowlet
G∼= Cpα11 ×···×Cpαkk
denoteadecompositionofGasadirectproductofcyclicgroupsofprimepowerorder
(theprimesp arenotnecessarilydistinct).Letg denoteageneratoroftheithfactor.
i i
DefinearingS by
F [x ,...,x ]
2 1 k
S = .
(xpα11 −1,...,xpαkk −1)
1 k
Since R is a commutative ring of characteristic 2, there is a natural ring homomor-
phism S −→ R sending x to g for all i. Since the g ’s together generate G, this
i i i
mapinducesasurjectionS× −→ R×,andhencebyLemma1thereisaquotientofS
whosegroupofunitsisisomorphictoG.
Sinceanyquotientofafinitedirectproductoffieldsisagainafinitedirectproduct
of fields (of possibly fewer factors), the proof will be complete if we can show that
S isisomorphicasaringtoafinitedirectproductoffieldsofcharacteristic2. Tosee
this,observethatthemap
F [x ]/(xpα11 −1)×···×F [x ]/(xpαkk −1) −→ S
2 1 1 2 1 k
sendingak-tupletotheproductofitsentriesissurjectiveandF -linearineachfactor;
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bytheuniversalpropertyofthetensorproduct,itinducesasurjectiveringhomomor-
phism
F2[x1]/(xp1α11 −1)⊗F2 ···⊗F2 F2[x1]/(xpkαkk −1) −→ S. (†)
Thedimensionofthesourceof(†)asanF -vectorspaceis pα1···pαk byLemma2
2 1 k
(2);thisisalsothedimensionofthetarget(countmonomialsinthepolynomialringS).
Consequently,themap(†)isanisomorphismofrings.Theirreduciblefactorsofeach
polynomialxpαii −1aredistinctsincethispolynomialhasnorootsincommonwithits
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derivative(p isodd).ThereforebytheChineseremaindertheorem,eachtensorfactor
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is a finite direct product of finite fields of characteristic 2. Since the tensor product
distributesoverfinite directproducts,wemayuseLemma2(1)toconcludethatS is
ringisomorphictoafinitedirectproductoffinitefieldsofcharacteristic2.
For any odd prime p, we now characterize the finite abelian p-groups that are re-
alizable as the group of units of a commutativering. Recall that a finite p-group is a
finite group whose order is a power of p. An elementary abelian finite p-group is a
finitegroupthatisisomorphictoafinitedirectproductofcyclicgroupsoforderp.
Corollary4. Letpbeanoddprime.Afiniteabelianp-groupGisthegroupofunits
of a commutative ring if and only if G is an elementary abelian p-group and p is a
Mersenneprime.
Proof. The ‘if’ direction follows from the fact if p = 2n −1 is a Mersenne prime,
then
(F ×···×F )× ∼= C ×···×C .
p+1 p+1 p p
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Fortheotherdirection,letpbeanoddprimeandletGbeafiniteabelianp-group.If
Gisthegroupofunitsofcommutativering,thenbyProposition3,G∼= T× whereT
isafinitedirectproductoffinitefieldsofcharacteristic2.Consequently,
G ∼= C2n1−1×···×C2nt−1.
Since each factor must be a p-group, for each i we have 2ni −1 = pzi for some
positiveintegerz .Weclaimthatz = 1foralli.Thisfollowsfrom[3,2.3],butsince
i i
theargumentisshortweincludeithereforconvenience.
Assume to the contrary that z > 1 for some i. Consider the equation pzi +1 =
i
2ni.Sincep > 1,wehaven ≥ 2andhencepzi ≡ −1 mod 4.Thismeansp ≡ −1
i
mod 4andz isodd.Sincez > 1,wehaveanontrivialfactorization
i i
2ni =pzi +1= (p+1)(pzi−1−pzi−2+···−p+1),
so pzi−1−pzi−2+···−p+1 mustbe even.On the otherhand,since z andp are
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bothodd,workingmodulo2weobtain
0 ≡ pzi−1−pzi−2+···−p+1 ≡ z ≡ 1 mod 2,
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a contradiction. Hence z = 1 for all i, so p is Mersenne and G an is elementary
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abelianp-group.
TheabovecorollarydoesnotholdfortheMersenneprimep = 2;forexample,C =
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F×.Asfarasweknow,Fuchs’problemforfiniteabelian2-groupsremainsopen.
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We next provide a simple example demonstrating that every even number is the
numberofunitsinacommutativering.
Proposition5. Everyevennumberisthenumberofunitsinacommutativering.
Proof. Letmbeapositiveintegerandconsiderthecommutativering
Z[x]
R = .
2m (x2,mx)
Every element in this ring can be uniquely represented by an element of the form
a+bx, where a is an arbitrary integerand 0 ≤ b ≤ m−1. We will now show that
a+bxisaunitinthisringifandonlyifaiseither1or−1;thisimpliestheringhas
exactly2munits.(Infact,itcanbeshownthatR× ∼= C ×C .)
2m 2 m
If a+bx is a unit in R , there there exits an element a′ +b′x such that (a+
2m
bx)(a′ +b′x) = 1inR .Sincex2 = 0inR ,wemusthaveaa′ = 1inZ;i.e.,a
2m 2m
is1or−1.Conversely,ifais1or−1,weseethat(a+bx)(a−bx) = 1inR .
2m
3. INFINITECARDINALS Propositions3and5solveourproblemforfinitecardi-
nals.Forinfinitecardinals,wewillprovethefollowingproposition.
Proposition6. Everyinfinitecardinalisthenumberofunitsinacommutativering.
OurproofreliesmainlyontheCantor-Bernsteintheorem:
Theorem 7 (Cantor-Bernstein). Let A and B be any two sets. If there exist injec-
tive mappings f: A −→ B and g: B −→ A, then there exits a bijective mapping
h: A −→ B.Inotherwords,if|A| ≤ |B|and|B|≤ |A|,then|A| = |B|.
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We also use otherstandard facts from set theory which may be found in [9]. For ex-
ample,wemakefrequentuseofthe factthatwheneverαandβ areinfinite cardinals
withα ≤ β,thenαβ ≤ β.Recallthatℵ denotesthecardinalityofthesetofnatural
0
numbers.
ProofofProposition6. LetλbeaninfinitecardinalandletS beasetwhosecardinal-
ityisλ.ConsiderF (S),thefieldofrationalfunctionsintheelementsofS.Weclaim
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that |F (S)×| = λ. By the Cantor-Bernstein theorem,it suffices to provethat |S| ≤
2
|F (S)×|and|F (S)×|≤ |S|.SinceS ⊆ F (S)×,itisclearthat|S| ≤ |F (S)×|.
2 2 2 2
Forthereverseinequality,firstobservethatifAisafiniteset,then|F [A]| = ℵ .
2 0
This follows by induction on the size of A, because F [x] is countable and R[x] is
2
countablewheneverRiscountable.Wenowhavethefollowing:
|F (S)×| ≤ |F (S)|
2 2
≤ |F [S]×F [S]| (everyrationalfunctionisaratiooftwopolynomials)
2 2
= |F [S]|2
2
= |F [S]|
2
≤ X |F2[A]|
A⊂S,1≤|A|<ℵ0
= X ℵ0
A⊂S,1≤|A|<ℵ0
∞ ∞
≤ X|S|iℵ0 = X|S|ℵ0 = |S|ℵ20 = |S|ℵ0 = |S|.
i=1 i=1
CombiningPropositions3,5,and6,weobtainourmainresult.
Theorem 8. Let λ be a cardinal number. There exists a commutative ring R with
|R×| = λifandonlyifλisequalto
1. anoddnumberoftheform t (2ni −1)forsomepositiveintegersn ,...,n ,
Qi=1 1 t
2. anevennumber,or
3. aninfinitecardinalnumber.
ACKNOWLEDGMENTS. WewouldliketothankGeorgeSeelingerforsimplifyingourpresentationofa
ringwithanevennumberofunits.Wealsowouldliketothanktheanonymousrefereesfortheircomments.
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DepartmentofMathematics,IllinoisStateUniversity,Normal,IL61790,USA
[email protected]
DepartmentofMathematics,GettysburgCollege,Gettysburg,PA17325,USA
[email protected]
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