How many units can a commutative ring
have?

arXiv:1701.02341v1 [] 9 Jan 2017

Sunil K. Chebolu and Keir Lockridge
Abstract. László Fuchs posed the following problem in 1960, which remains open: classify
the abelian groups occurring as the group of all units in a commutative ring. In this note,
we provide an elementary solution to a simpler, related problem: find all cardinal numbers
occurring as the cardinality of the group of all units in a commutative ring. As a by-product,
we obtain a solution to Fuchs’ problem for the class of finite abelian p-groups when p is an
odd prime.

1. INTRODUCTION It is well known that a positive integer k is the order of the
multiplicative group of a finite field if and only if k is one less than a prime power.
The corresponding fact for finite commutative rings, however, is not as well known.
Our motivation for studying this—and the question raised in the title—stems from a
problem posed by László Fuchs in 1960: characterize the abelian groups that are the
group of units in a commutative ring (see [8]). Though Fuchs’ problem remains open, it
has been solved for various specialized classes of groups, where the ring is not assumed
to be commutative. Examples include cyclic groups ([10]), alternating, symmetric and
finite simple groups ([4, 5]), indecomposable abelian groups ([1]), and dihedral groups
([2]). In this note we consider a much weaker version of Fuchs’ problem, determining
only the possible cardinal numbers |R× |, where R is a commutative ring with group
of units R× .
In a previous note in the M ONTHLY ([6]), Ditor showed that a finite group G of odd
order is the group of units of a ring if and only if G is isomorphic to a direct product
of cyclic groups Gi , where |Gi | = 2ni − 1 for some positive integer ni . This implies
that an
positive integer is the number of units in a ring if and only if it is of the
Qodd
t
form i=1 (2ni − 1) for some positive integers n1 , . . . , nt . As Ditor mentioned in his
paper, this theorem may be derived from the work of Eldridge in [7] in conjunction
with the Feit-Thompson theorem which says that every finite group of odd order is
solvable. However, the purpose of his note was to give an elementary proof of this
result using classical structure theory. Specifically, Ditor’s proof uses the following key
ingredients: Mashke’s theorem, which classifies (for finite groups) the group algebras
over a field that are semisimple rings; the Artin-Wedderburn theorem, which describes
the structure of semisimple rings; and Wedderburn’s little theorem, which states that
every finite division ring is a field.
In this note we give another elementary proof of Ditor’s theorem for commutative
rings. We also extend the theorem to even numbers and infinite cardinals, providing a
complete answer to the question posed in the title; see Theorem 8. Our approach also
gives an elementary solution to Fuchs’ problem for finite abelian p-groups when p is
an odd prime; see Corollary 4.
2. FINITE CARDINALS We begin with two lemmas. For a prime p, let Fp denote
the field of p elements. Recall that any ring homomorphism φ : A −→ B maps units
to units, so φ induces a group homomorphism φ× : A× −→ B × .

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Lemma 1. Let φ : A −→ B be a homomorphism of commutative rings. If the induced
group homomorphism φ× : A× −→ B × is surjective, then there is a quotient A′ of A
such that (A′ )× ∼
= B×.
Proof. By the first isomorphism theorem for rings, Im φ is isomorphic to a quotient
of A. It therefore suffices to prove that (Im φ)× = B × . Since ring homomorphisms
map units to units and φ× is surjective, we have B × ⊆ (Im φ)× . The reverse inclusion
(Im φ)× ⊆ B × holds since every unit in the subring Im φ must also be a unit in the
ambient ring B .
Lemma 2. Let V and W be finite fields of characteristic 2.
1. The tensor product V ⊗F2 W is isomorphic as a ring to a finite direct product
of finite fields of characteristic 2.
2. As F2 -vector spaces, dim(V ⊗F2 W ) = (dim V )(dim W ).
Proof. To prove (1), let K and L be finite fields of characteristic 2. By the primitive element theorem, we have L ∼
= F2 [x]/(f (x)), where f (x) is an irreducible polynomial
in F2 [x]. This implies that

K ⊗ F2 L ∼
=

K[x]
.
(f (x))

The irreducible factors of f (x) in K[x] are distinct since the extension L/F2 is sepQt
arable. Now let f (x) = i=1 fi (x) be the factorization of f (x) into its distinct irreducible factors in K[x]. We then have the following series of ring isomorphisms:
Y K[x]
K[x]
K[x] ∼
∼
,
L∼
=
= Qt
=
(f (x))
( i=1 fi (x)) i=1 (fi (x))
t

K ⊗ F2

where the last isomorphism follows from the Chinese remainder theorem for the ring
K[x]. Since each factor K[x]/(fi (x)) is a finite field of characteristic 2, we see that
K ⊗F2 L is isomorphic as a ring to a direct product of finite fields of characteristic 2,
as desired.
For (2), simply note that if {v1 , . . . , vk } is a basis for V and {w1 , . . . , wl } is a
basis for W , then {vi ⊗ wj | 1 ≤ i ≤ k, 1 ≤ j ≤ l} is a basis for V ⊗F2 W .
We may now classify the finite abelian groups of odd order that appear as the group
of units in a commutative ring.
Proposition 3. Let G be a finite abelian group of odd order. The group G is isomorphic
to the group of units in a commutative ring if and only if G is isomorphic to the group
of units in a finite direct product of finite fields of characteristic 2. In particular, an
odd positiveQinteger k is the number of units in a commutative ring if and only if k is
t
of the form i=1 (2ni − 1) for some positive integers n1 , . . . , nt .
Proof. The ‘if’ direction of the second statement follows from the fact that, for rings
A and B , (A × B)× ∼
= A× × B × .
For the converse, since the trivial group is the group of units of F2 , let G be a
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
implies that R has characteristic 2.

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Now let

G∼
= Cpα1 × · · · × Cpαk
1

k

denote a decomposition of G as a direct product of cyclic groups of prime power order
(the primes pi are not necessarily distinct). Let gi denote a generator of the ith factor.
Define a ring S by

S=

F2 [x1 , . . . , xk ]

.

α
p k

α

p 1
(x11

− 1, . . . , xkk − 1)

Since R is a commutative ring of characteristic 2, there is a natural ring homomorphism S −→ R sending xi to gi for all i. Since the gi ’s together generate G, this
map induces a surjection S × −→ R× , and hence by Lemma 1 there is a quotient of S
whose group of units is isomorphic to G.
Since any quotient of a finite direct product of fields is again a finite direct product
of fields (of possibly fewer factors), the proof will be complete if we can show that
S is isomorphic as a ring to a finite direct product of fields of characteristic 2. To see
this, observe that the map
α
p k

α
p 1

F2 [x1 ]/(x11 − 1) × · · · × F2 [x1 ]/(xkk − 1) −→ S
sending a k -tuple to the product of its entries is surjective and F2 -linear in each factor;
by the universal property of the tensor product, it induces a surjective ring homomorphism
α
p k

α
p 1

F2 [x1 ]/(x11 − 1) ⊗F2 · · · ⊗F2 F2 [x1 ]/(xkk − 1) −→ S.
α

(†)

α

The dimension of the source of (†) as an F2 -vector space is p1 1 · · · pk k by Lemma 2
(2); this is also the dimension of the target (count monomials in the polynomial ring S ).
Consequently, the map (†) is an isomorphism of rings. The irreducible factors of each
α
p i

polynomial xi i − 1 are distinct since this polynomial has no roots in common with its
derivative (pi is odd). Therefore by the Chinese remainder theorem, each tensor factor
is a finite direct product of finite fields of characteristic 2. Since the tensor product
distributes over finite direct products, we may use Lemma 2 (1) to conclude that S is
ring isomorphic to a finite direct product of finite fields of characteristic 2.
For any odd prime p, we now characterize the finite abelian p-groups that are realizable as the group of units of a commutative ring. 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
finite group that is isomorphic to a finite direct product of cyclic groups of order p.
Corollary 4. Let p be an odd prime. A finite abelian p-group G is the group of units
of a commutative ring if and only if G is an elementary abelian p-group and p is a
Mersenne prime.
Proof. The ‘if’ direction follows from the fact if p = 2n − 1 is a Mersenne prime,
then

(Fp+1 × · · · × Fp+1 )× ∼
= Cp × · · · × Cp .

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For the other direction, let p be an odd prime and let G be a finite abelian p-group. If
G is the group of units of commutative ring, then by Proposition 3, G ∼
= T × where T
is a finite direct product of finite fields of characteristic 2. Consequently,

G∼
= C2n1 −1 × · · · × C2nt −1 .
Since each factor must be a p-group, for each i we have 2ni − 1 = pzi for some
positive integer zi . We claim that zi = 1 for all i. This follows from [3, 2.3], but since
the argument is short we include it here for convenience.
Assume to the contrary that zi > 1 for some i. Consider the equation pzi + 1 =
ni
2 . Since p > 1, we have ni ≥ 2 and hence pzi ≡ −1 mod 4. This means p ≡ −1
mod 4 and zi is odd. Since zi > 1, we have a nontrivial factorization

2ni = pzi + 1 = (p + 1)(pzi −1 − pzi −2 + · · · − p + 1),
so pzi −1 − pzi −2 + · · · − p + 1 must be even. On the other hand, since zi and p are
both odd, working modulo 2 we obtain

0 ≡ pzi −1 − pzi −2 + · · · − p + 1 ≡ zi ≡ 1 mod 2,
a contradiction. Hence zi = 1 for all i, so p is Mersenne and G an is elementary
abelian p-group.
The above corollary does not hold for the Mersenne prime p = 2; for example, C4 =
F×
5 . As far as we know, Fuchs’ problem for finite abelian 2-groups remains open.
We next provide a simple example demonstrating that every even number is the
number of units in a commutative ring.
Proposition 5. Every even number is the number of units in a commutative ring.
Proof. Let m be a positive integer and consider the commutative ring

R2m =

Z[x]
.
(x2 , mx)

Every element in this ring can be uniquely represented by an element of the form
a + bx, where a is an arbitrary integer and 0 ≤ b ≤ m − 1. We will now show that
a + bx is a unit in this ring if and only if a is either 1 or −1; this implies the ring has
× ∼
exactly 2m units. (In fact, it can be shown that R2m
= C2 × Cm .)
If a + bx is a unit in R2m , there there exits an element a′ + b′ x such that (a +
bx)(a′ + b′ x) = 1 in R2m . Since x2 = 0 in R2m , we must have aa′ = 1 in Z; i.e., a
is 1 or −1. Conversely, if a is 1 or −1, we see that (a + bx)(a − bx) = 1 in R2m .
3. INFINITE CARDINALS Propositions 3 and 5 solve our problem for finite cardinals. For infinite cardinals, we will prove the following proposition.
Proposition 6. Every infinite cardinal is the number of units in a commutative ring.
Our proof relies mainly on the Cantor-Bernstein theorem:
Theorem 7 (Cantor-Bernstein). Let A and B be any two sets. If there exist injective mappings f : A −→ B and g : B −→ A, then there exits a bijective mapping
h : A −→ B . In other words, if |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.

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We also use other standard facts from set theory which may be found in [9]. For example, we make frequent use of the fact that whenever α and β are infinite cardinals
with α ≤ β , then αβ ≤ β . Recall that ℵ0 denotes the cardinality of the set of natural
numbers.
Proof of Proposition 6. Let λ be an infinite cardinal and let S be a set whose cardinality is λ. Consider F2 (S), the field of rational functions in the elements of S . We claim
that |F2 (S)× | = λ. By the Cantor-Bernstein theorem, it suffices to prove that |S| ≤
|F2 (S)× | and |F2 (S)× | ≤ |S|. Since S ⊆ F2 (S)× , it is clear that |S| ≤ |F2 (S)× |.
For the reverse inequality, first observe that if A is a finite set, then |F2 [A]| = ℵ0 .
This follows by induction on the size of A, because F2 [x] is countable and R[x] is
countable whenever R is countable. We now have the following:

|F2 (S)× | ≤ |F2 (S)|
≤ |F2 [S] × F2 [S]| (every rational function is a ratio of two polynomials)
= |F2 [S]|2
= |F2 [S]|
X
≤

|F2 [A]|

A⊂S, 1≤|A|<ℵ0

X

=

ℵ0

A⊂S, 1≤|A|<ℵ0

≤

∞
X
i=1

i

|S| ℵ0 =

∞
X

|S|ℵ0 = |S|ℵ20 = |S|ℵ0 = |S|.

i=1

Combining Propositions 3, 5, and 6, we obtain our main result.
Theorem 8. Let λ be a cardinal number. There exists a commutative ring R with
|R× | = λ if and only if λ is equal to
Qt
1. an odd number of the form i=1 (2ni − 1) for some positive integers n1 , . . . , nt ,
2. an even number, or
3. an infinite cardinal number.
ACKNOWLEDGMENTS. We would like to thank George Seelinger for simplifying our presentation of a
ring with an even number of units. We also would like to thank the anonymous referees for their comments.

REFERENCES
1. S. K. Chebolu, K. Lockridge, Fuchs’ problem for indecomposable abelian groups, J. Algebra 438 (2015)
325–336.
2. ———, Fuchs’ problem for dihedral groups, J. Pure Appl. Algebra 221 no. 2 (2017) 971–982.
3. ———, Fields with indecomposable multiplicative groups, Expo. Math. 34 (2016) 237–242.
4. C. Davis, T. Occhipinti, Which finite simple groups are unit groups? J. Pure Appl. Algebra 218 no. 4
(2014) 743–744.
5. ———, Which alternating and symmetric groups are unit groups? J. Algebra Appl. 13 no. 3 (2014).
6. S. Z. Ditor, On the group of units of a ring, Amer. Math. Monthly 78 (1971) 522–523.

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7. K. E. Eldridge, On ring structures determined by groups, Proc. Amer. Math. Soc. 23 (1969) 472–477.
8. L. Fuchs, Abelian groups. International Series of Monographs on Pure and Applied Mathematics, Pergamon Press, New York-Oxford-London-Paris, 1960.
9. P. R. Halmos, Naive set theory. The University Series in Undergraduate Mathematics, D. Van Nostrand
Co., Princeton, NJ-Toronto-London-New York, 1960.
10. K. R. Pearson, J. E. Schneider, Rings with a cyclic group of units, J. Algebra 16 (1970) 243–251.
Department of Mathematics, Illinois State University, Normal, IL 61790, USA
schebol@ilstu.edu
Department of Mathematics, Gettysburg College, Gettysburg, PA 17325, USA
klockrid@gettysburg.edu

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