Table Of ContentLAMBDA-RINGS
ZEHUI CHEN
Abstract. Wegiveabriefintroductiontothetheoryofλ-ringsaccessibletoundergraduate
students. We begin by introducing the concepts of tensor product, exterior power, and
graded vector spaces. Then we give the definition of λ-rings and discuss several examples.
Finally,weexplainhowdirectsums,tensorproducts,andexteriorpowersof(graded)vector
spaces induce λ-ring structures on the Grothendieck groups of these categories.
Contents
1. Introduction 1
2. Tensor Product 2
3. Exterior Power 4
4. Graded Vector Spaces 6
5. λ-Rings 8
6. Direct Sum, Tensor Product, Exterior Power and λ-Ring Structure 12
References 14
1. Introduction
In algebra, a λ-ring is a commutative ring equipped with a λ-operation. The study of
λ-rings is mainly motivated by the behaviour of tensor products and exterior powers over
vector spaces. The categories of vector spaces and graded vector spaces, with direct sum,
tensor product, and exterior power, induce the addition, multiplication, and λ-operation in
λ-rings of integers Z and Laurent polynomials.
The purpose of this paper is to give a brief introduction to the subject accessible to under-
graduate students with basic knowledge of linear algebra and group theory. In Section 2, we
provide the definition of the tensor product and show its basic properties, i.e. the universal
property, its dimension, and a kind of distributivity. In Section 3, we define the exterior
power of vector spaces and prove the theorem of dimensions of exterior powers. After that,
weexplaingradedvectorspacesandgradeddimensionoftensorproductsandexteriorpowers
in Section 4. Then, in Section 5, the axioms of λ-rings are given as well as some simple exam-
ples of λ-rings, including the integers Z, the ring of power series, and the ring of symmetric
functions. Finally, we explain how the operations of direct sum, tensor product, and exterior
power of (graded) vector spaces, induce λ-ring structures on the integers and on the ring
of Laurent polynomials, when we identify these rings with the Grothendieck groups of the
categories of finite-dimensional vector spaces and finite-dimensional graded vector spaces.
2 ZEHUI CHEN
Acknowledgements. This project was completed as part of an Undergraduate Research
Project at the University of Ottawa. I would like to thank Professor Alistair Savage at the
University of Ottawa for his advice and support.
2. Tensor Product
Definition 2.1. (Bilinear map). Let U, V, and W be three vector spaces over the same
field F. Then a bilinear map b: U ×V → W is a mapping such that
(a) b(u +u ,v) = b(u ,v)+b(u ,v) for all u ,u ∈ U, v ∈ V
1 2 1 2 1 2
(b) b(u,v +v ) = b(u,v )+b(u,v ) for all u ∈ U, v ,v ∈ V
1 2 1 2 1 2
(c) b(αu,v) = αb(u,v) = b(u,αv) for all u ∈ U, v ∈ V, α ∈ F
In other words, b(·,v) is linear from U to W for all v ∈ V, and b(u,·) is linear from V to W
for all u ∈ U.
Example 2.2. Let M be an m×n complex matrix, U be the vector space Cm, and V be
the vector space Cn. Define the map b: U ×V → C by b(u,v) = uTMv for all u ∈ U and
v ∈ V where uT is the transpose of vector u. Then, the function b is a bilinear map.
Definition 2.3. (Tensor product). Let U,V be vector spaces. The tensor product of U and
V, denoted U ⊗V, is a vector space together with a bilinear map ⊗: U ×V → U ⊗V such
that for any bilinear map b: U ×V → W, there exists a unique linear map (cid:96): U ⊗V → W
such that b = (cid:96)◦⊗, that is, such that the following diagram commutes:
⊗
U ×V U ⊗V
(cid:96)
b
W
Definition 2.4. (Free vector space). Let S be a set and F be a field. The free vector space
F(S) generated by S is defined as
F(S) = {f: S → F | f(s) = 0 for all s ∈ S \S(cid:48) where S(cid:48) is some finite subset of S}
where the operations on F(S) are defined as pointwise addition and scalar multiplication.
Definition 2.5. (Quotient space). Let V be a vector space and W ⊆ V be a subspace of
V. Define ∼ to be the equivalence relation such that v ∼ v if and only if v −v ∈ W for
1 2 1 2
all v ,v ∈ V. Then, the quotient space V/W is defined as V/ ∼ and the elements in V/W
1 2
are equivalence classes [v] for v ∈ V.
Definition 2.6. (Tensor product). Let U and V be vector spaces over field F. Define
W ⊆ F(U ×V) to be
W = {(u +u ,v)−(u ,v)−(u ,v) | u ,u ∈ U, v ∈ V}
1 2 1 2 1 2
∪{(u,v +v )−(u,v )−(u,v ) | u ∈ U, v ,v ∈ V}
1 2 1 2 1 2
∪{α(u,v)−(αu,v) | u ∈ U, v ∈ V, α ∈ F}
∪{α(u,v)−(u,αv) | u ∈ U, v ∈ V, α ∈ F}.
LAMBDA-RINGS 3
Then, thetensor product ofU andV, denotedU⊗V, isdefinedasU⊗V = F(U×V)/W. For
u ∈ U and v ∈ V, the tensor product of u and v, denoted u⊗v, is defined as u⊗v = [(u,v)].
It follows from the definition of the quotient space that the tensor product satisfies the
following properties:
(a) (u +u )⊗v = u ⊗v +u ⊗v for all u ,u ∈ U and v ∈ V.
1 2 1 2 1 2
(b) u⊗(v +v ) = u⊗v +u⊗v for all u ∈ U and v ,v ∈ V.
1 2 1 2 1 2
(c) αu⊗v = α(u⊗v) = u⊗αv for all u ∈ U,v ∈ V, and α ∈ F
Remark 2.7. One can show that the tensor product as defined in Definition 2.6 satisfies
the universal property of Definition 2.3.
Theorem 2.8. Let U and V be two finite-dimensional vector spaces over field F. Then,
dim(U ⊗V) = dimU dimV.
Proof. Let dimU = m and dimV = n. Then, U has a basis S = {u ,u ,...,u } and V
1 1 2 m
has a basis S = {v ,v ,...,v }. Now, define S = {u ⊗v : u ∈ S , v ∈ S }. The set S is
2 1 2 n i j i 1 j 2
a subset of U ⊗V with mn elements, so it is enough to prove that S is a basis of U ⊗V.
For any element that can be written as u⊗v ∈ U ⊗V, let u = α u +α u +...+α u
1 1 2 2 m m
and v = β v +β v +...+β v for α ,β ∈ F. Then, it becomes
1 1 2 2 n n i j
m n
(cid:88)(cid:88)
u⊗v = (α u +α u +...+α u )⊗(β v +β v +...+β v ) = α β (u ⊗v ).
1 1 2 2 m m 1 1 2 2 n n i j i j
i=1 j=1
Then, any arbitrary element x = u⊗v +u(cid:48) ⊗v(cid:48) +... can be written as
m n
(cid:88)(cid:88)
x = c (u ⊗v ).
ij i j
i=1 j=1
It follows x ∈ Span(S), so U ⊗V ⊆ Span(S). Obviously, S ⊆ U ⊗V, so Span(S) ⊆ U ⊗V.
Thus, we have U ⊗V = Span(S).
Let ψ ,...,ψ be the dual basis to S and let φ ,...,φ be the dual basis to S . Then
1 m 1 1 n 2
define the maps b (u,v) = ψ (u)φ (v) for all 1 ≤ i ≤ m and 1 ≤ j ≤ n. When v is fixed,
i,j i j
φ (v) becomes a fixed scalar c ∈ F, then b (u,v) = cψ (u). The dual basis ψ (u) is a linear
j i,j i i
transformation, so b (·,v) is linear from U to F for all v ∈ V. Similarly, b (u,·) is linear
i,j i,j
from V to F for all u ∈ U. Hence, b (u,v) = ψ (u)φ (v) are bilinear maps. By the universal
i,j i j
property, these induce maps on the tensor product, i.e. for all 1 ≤ k ≤ m,1 ≤ l ≤ n, there
exist (cid:96) such that (cid:96) (u⊗v) = b (u,v).
k,l k,l k,l
Now, fix 1 ≤ k ≤ m,1 ≤ l ≤ n, suppose
m n
(cid:88)(cid:88)
c u ⊗v = 0 for some c ∈ F,1 ≤ i ≤ m,1 ≤ j ≤ n.
i,j i j i,j
i=1 j=1
Then, (cid:96) (u⊗v) = (cid:96) (0) = 0, which means
k,l k,l
m n
(cid:88)(cid:88)
c b (u ,v ) = c = 0 for 1 ≤ i ≤ m,1 ≤ j ≤ n.
i,j k,l i j k,l
i=1 j=1
Thus, c = 0 for all 1 ≤ k ≤ m,1 ≤ l ≤ n. Consequently, S is linearly independent.
k,l
Therefore, S is a basis of U⊗V. So, we have dim(U⊗V) = |S| = mn = dimU dimV. (cid:3)
4 ZEHUI CHEN
Proposition 2.9 (Distributive law of tensor product over direct sum). Let U,V,W be finite-
∼
dimensional vector spaces, then U ⊗(V ⊕W) = (U ⊗V)⊕(U ⊗W).
Proof. Let U be an l-dimensional vector space, V be an m-dimensional vector space and W
be an n-dimensional vector space. Then, by Theorem 2.8, the dimension of U ⊗(V ⊕W) is
dim(U ⊗(V ⊕W)) = dim(U)dim(V ⊕W) = dim(U)(dim(V)+dim(W))
= dim(U)dim(V)+dim(U)dim(W).
On the other hand, we have
dim((U ⊗V)⊕(U ⊗W)) = dim(U ⊗V)+dim(U ⊗W) = dim(U)dim(V)+dim(U)dim(W).
Thus, U⊗(V ⊕W) and (U⊗V)⊕(U⊗W) have the same dimension, so they are isomorphic.
(cid:3)
3. Exterior Power
If we generalize the idea of tensor product in Definition 2.3, we can replace the vector
spaces U,V with vector spaces V ,V ,...V and change the bilinear map to multilinear map.
1 2 k
Then, we have the tensor product of k vector spaces, V ⊗ V ⊗ ... ⊗ V . Let V be some
1 2 k
vector space over field F, define
T0(V) = F,
T1(V) = V, and
Tk(V) = V⊗k = V ⊗V ⊗...V for k ≥ 2.
(cid:124) (cid:123)(cid:122) (cid:125)
k
Definition 3.1. (Exterior power). Let Ak be the subspace of Tk(V) spanned by vectors in
S = {v ⊗v ⊗...⊗v | v ,v ,...v ∈ V, v = v for some i (cid:54)= j}.
1 2 k 1 2 k i j
Then, the k-th exterior power of V is the quotient space (cid:86)k(V) = Tk(V)/Ak. It follows that
each element in (cid:86)k(V) is of the form [u ⊗u ⊗...⊗u +v ⊗v ⊗...⊗v +...], denoted
1 2 k 1 2 k
u ∧u ∧···∧u +v ∧v ∧···∧v +... for some u ,u ,...,u ,v ,v ,...,v ,... ∈ V.
1 2 k 1 2 k 1 2 k 1 2 k
Definition 3.2. (Alternatingmultilinearmap)Letf: Vk → W beamultilinearmap. Then,
f is alternating if f(v ,...,v ) = 0 whenever v = v for some 1 ≤ i ≤ n − 1. Then, all
1 k i i+1
alternating multilinear maps have the property f(v ,...,v ) = 0 whenever v = v for some
1 k i j
i (cid:54)= j.
Remark 3.3. It follows from the definition that the exterior power satisfies a universal
property. Precisely, for any alternating multilinear map m: Vk → W, there exists a unique
linear map (cid:96): (cid:86)k(V) → W such that m = (cid:96)◦(cid:86)k, that is, such that the following diagram
commutes:
∧k
Vk (cid:86)k(V)
(cid:96)
m
W
LAMBDA-RINGS 5
Lemma 3.4. Let V be a vector space, and v ,...,v ∈ V. Then,
1 k
v ∧···∧v = −v ∧···∧v ∧v ∧v ∧···∧v ∧v ∧v ∧···∧v
1 k 1 i−1 j i+1 j−1 i j+1 k
for all 1 ≤ i < j ≤ k.
Proof. The element
x = v ∧···∧v ∧(v −v )∧v ∧···∧v ∧(v −v )∧v ∧···∧v
1 i−1 i j i+1 j−1 i j j+1 k
= −v ∧···∧v ∧(v −v )∧v ∧···∧v ∧(v −v )∧v ∧···∧v
1 i−1 i j i+1 j−1 j i j+1 k
has v −v at positions i and j where i (cid:54)= j. So, by the definition of Ak, x = 0 in (cid:86)k(V).
i j
Then,
0 = (v ∧···∧v )−(−v ∧···∧v ∧v ∧v ∧···∧v ∧v ∧v ∧···∧v )
1 k 1 i−1 i i+1 j−1 i j+1 k
−(v ∧···∧v ∧v ∧v ∧···∧v ∧v ∧v ∧···∧v )
1 i−1 j i+1 j−1 j j+1 k
+(v ∧···∧v ∧v ∧v ∧···∧v ∧v ∧v ∧···∧v )
1 i−1 j i+1 j−1 i j+1 k
= (v ∧···∧v )−0−0+(v ∧···∧v ∧v ∧v ∧···∧v ∧v ∧v ∧···∧v )
1 k 1 i−1 j i+1 j−1 i j+1 k
= (v ∧···∧v )+(v ∧···∧v ∧v ∧v ∧···∧v ∧v ∧v ∧···∧v ).
1 k 1 i−1 j i+1 j−1 i j+1 k
Therefore, v ∧···∧v = −v ∧···∧v ∧v ∧v ∧···∧v ∧v ∧v ∧···∧v . (cid:3)
1 k 1 i−1 j i+1 j−1 i j+1 k
Theorem 3.5. Let V be an n-dimensional vector space with a basis {e ,e ,...,e }. Then
1 2 n
the set S = {e ∧ e ∧ ··· ∧ e | 1 ≤ i < i < ... < i ≤ n} is a basis of (cid:86)k(V), and
dim(cid:86)k(V) = (cid:0)ni1(cid:1). i2 ik 1 2 k
k
Proof. For any element that can be written as v ∧ ··· ∧ v where v ,...,v are vectors in
1 k 1 k
V, let v = α e + α e + ... + α e for 1 ≤ i ≤ n and α ∈ F. Then, v is in span of
i i,1 1 i,2 2 i,n n i
the set S(cid:48) = {e ∧ ··· ∧ e | 1 ≤ i ≤ n for all 1 ≤ j ≤ k} and we have S ⊆ S(cid:48). For any
i1 ik j
element in v = e ∧···∧e ∈ S(cid:48) \S, v has either i = i or i > i for some a < b. The
i1 ik a b a b
former case leads v to be 0, and the later case can be reduced to ±v(cid:48) for some v(cid:48) ∈ S by
applying Lemma 3.4. So, any element u ∧u ∧···∧u +v ∧v ∧···∧v +... in (cid:86)k(V)
1 2 k 1 2 k
is in Span(S(cid:48)) = Span(S). Obviously, S ⊆ (cid:86)k(V), so Span(S) ⊆ (cid:86)k(V). Thus, we have
(cid:86)k(V) = Span(S).
Let ψ ,...,ψ be the dual basis to {e ,e ,...,e }. We can define the n × k matrix
1 n 1 2 n
A(v ,...,v )tobeA = [a ]wherea = ψ (v ). Then, wedefinethemapsm : Vk → F
1 k i,j i,j i j i1,...,ik
for 1 ≤ i < ... < i ≤ n to be m (v ,...,v ) = detB(v ,...,v ) where B(v ,...,v ) is
1 k i1,...,ik 1 k 1 k 1 k
the k×k submatrix of A(v ,...,v ) consisting of the i ,...,i rows of A(v ,...,v ). Then,
1 k 1 k 1 k
the maps m are multilinear by properties of the determinant. If v = v for some 1 ≤
i1,...,ik i i+1
i ≤ n−1, then the matrix A(v ,...,v ) will have the same i-th and (i+1)-th column. So, the
1 k
submatrix B(v ,...,v ) have two identical columns. Since B(v ,...,v ) is a singular matrix,
1 k 1 k
it has determinant of 0. Then, we have m (v ,...,v ) = 0. Hence, the maps m are
i1,...,ik 1 k i1,...,ik
alternating. By the universal property, these induce maps on the exterior power, i.e. for all
1 ≤ i < ... < i ≤ n, there exist (cid:96) such that (cid:96) (v ∧···∧v ) = m (v ,...,v ).
1 k i1,...,ik i1,...,ik 1 k i1,...,ik 1 k
Now, fix 1 ≤ j < ... < j ≤ n, and suppose
1 k
(cid:88)
c e ∧···∧e = 0 for some c ∈ F.
i1,...,ik i1 ik i1,...,ik
1≤i1<...<ik≤n
6 ZEHUI CHEN
Then, by the universal property, we have
(cid:88)
c m (e ,...,e ) = 0.
i1,...,ik j1,...,jk i1 ik
1≤i1<...<ik≤n
However, m (e ,...,e ) = 1 if i = j ,...,i = j , and m (e ,...,e ) = 0
j1,...,jk i1 ik 1 1 k k j1,...,jk i1 ik
otherwise. Hence,
(cid:88)
c m (e ,...,e ) = c = 0.
i1,...,ik j1,...,jk i1 ik j1,...,jk
1≤i1<...<ik≤n
Thus, c = 0 for all 1 ≤ j < ... < j ≤ n. So, the set S is linearly independent.
j1,...,jk 1 k
Since Span(S) = (cid:86)k(V) and S is linearly independent, we have that S is a basis of (cid:86)k(V).
Since the basis S has cardinality (cid:0)n(cid:1), it follows that dim(cid:86)k(V) = (cid:0)n(cid:1). (cid:3)
k k
4. Graded Vector Spaces
Definition 4.1. (Graded vector space) A graded vector space, is a vector space V together
with a decomposition into a direct sum of the form
(cid:77)
V = V .
i
i∈Z
We say that elements of V are homogeneous of degree i.
i
Definition 4.2. (Graded dimension) Let V be a graded vector space, then the graded di-
mension of V is defined as
(cid:88)
grdimV = (dimV )qi
i
i∈Z
where q is an indeterminate.
Example 4.3. Let V be an n-dimensional vector space, then W = (cid:86)∗(V) = (cid:76) (cid:86)k(V)
k∈Z
is a graded vector space, and
n (cid:18) (cid:19)
(cid:88) n
grdimW = qi = (1+q)n.
i
i=0
Example 4.4. Let V be an n-dimensional vector space, then W = T∗(V) = (cid:76) Tk(V) is
k∈Z
a graded vector space, and
(cid:88) (cid:88)
grdimW = dimTk(V)qk = nkqk.
k≥0 k≥0
Definition 4.5. (Direct sum of graded vector spaces) Let V,W be two graded vector spaces.
Then, the direct sum of V and W is a graded vector space with the decomposition
(cid:77)
V ⊕W = (V ⊕W)
n
n∈Z
where
(V ⊕W) = V ⊕W .
n n n
Theorem 4.6. Let V,W be two graded vector spaces with grdimV = f (q) and grdimW =
1
f (q), then grdim(V ⊕W) = f (q)+f (q).
2 1 2
LAMBDA-RINGS 7
Proof. Let f (q) = a +a q +a q−1 +a q2 +a q−2 +... where a = dimV for all i ∈ Z.
1 0 1 −1 2 −2 i i
Let f (q) = b +b q +b q−1 +b q2 +b q−2 +... where b = dimW for all j ∈ Z. Then,
2 0 1 −1 2 −2 j j
dim(V ⊕W) = dim(V ⊕W ) = a +b . Thus, grdim(V ⊕W) = f (q)+f (q). (cid:3)
n n n n n 1 2
Definition 4.7. (Tensor product of graded vector spaces) Let V,W be two graded vector
spaces. Then, thetensor product ofV andW isagradedvectorspacewiththedecomposition
(cid:77)
V ⊗W = (V ⊗W)
n
n∈Z
where
(cid:77)
(V ⊗W) = (V ⊗W ).
n i j
i+j=n
Theorem 4.8. Let V,W be two graded vector spaces with grdimV = f (q) and grdimW =
1
f (q), then grdim(V ⊗W) = f (q)f (q).
2 1 2
Proof. Let f (q) = a +a q +a q−1 +a q2 +a q−2 +... where a = dimV for all i ∈ Z.
1 0 1 −1 2 −2 i i
Let f (q) = b +b q +b q−1 +b q2 +b q−2 +... where b = dimW for all j ∈ Z. Now,
2 0 1 −1 2 −2 j j
we regard (V ⊗W) for all n ∈ Z as direct sum and compute its dimension as
n
(cid:88) (cid:88) (cid:88)
dim(V ⊗W) = dim(V ⊗W ) = dimV dimW = a b ,
n i j i j i j
i+j=n i+j=n i+j=n
and all elements in (V ⊗W) are homogeneous of degree n. Now, the graded dimension of
n
V ⊗W is
(cid:88) (cid:88) (cid:88)
grdim(V ⊗W) = dim(V ⊗W) qn = a b qn.
n i j
n∈Z n∈Zi+j=n
Thus, we have grdim(V ⊗W) = f (q)f (q). (cid:3)
1 2
∼
Proposition 4.9. Let U,V,W be graded vector spaces, then U⊗(V⊕W) = (U⊗V)⊕(U⊗W).
Proof. The proof is exactly same as Proposition 2.9. (cid:3)
Lemma 4.10. Let V,W be vector spaces, then
k
(cid:77)
∧k(V ⊕W) ∼= ∧i(V)⊗∧k−i(W).
i=0
Proof. Let V be an m-dimensional vector space with basis {v ,...,v } and W be an n-
1 m
dimensional vector space with basis {w ,...,w }. Then, a basis of V ⊕W is
1 n
{e ,...,e ,e ,...,e }
1 m m+1 m+n
where e = v ,...,e = v ,e = w ,...,e = w . By Theorem 3.5, a basis of
1 1 m m m+1 1 m+n n
∧k(V ⊕W) is S = {e ∧e ∧···∧e | 1 ≤ i < i < ... < i ≤ m+n}.
i1 i2 ik 1 2 k
Now, we define a bijective linear transformation
k
(cid:77)
T: ∧k (V ⊕W) → ∧i(V)⊗∧k−i(W)
i=0
whereT(e ∧e ∧···∧e ) = (e ∧···∧e )⊗(e ∧···∧e )foreachs = e ∧e ∧···∧e ∈ S
i1 i2 ik i1 ij ij+1 ik i1 i2 ik
and 1 ≤ i < ... < i ≤ m,m+1 ≤ i < ... < i ≤ m+n and extend linearly. Then, T is
1 j i+1 k
an isomorphism, so the two vector spaces are isomorphic. (cid:3)
8 ZEHUI CHEN
Definition 4.11. (Exterior power of graded vector spaces) Let V be a graded vector space.
Then, the k-th exterior power of V is a graded vector space with the decomposition
(cid:77)
∧k(V) = (∧k(V))
n
n∈Z
where
(cid:77)
(∧k(V)) = ∧k1(V )⊗···⊗∧km(V ).
n i1 im
k1i1+···+kmim=n,
k1+···+km=k
Theorem 4.12. Let V be a graded vector space with grdimV = f(q)a + a q + a q−1 +
0 1 −1
a q2 +a q−2 +··· where a = dimV for all i ∈ Z, then the graded dimension of the k-th
2 −2 i i
exterior power of V is
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) a a
grdim∧k(V) = i1 ··· im qi1k1+···imkm
k k
1 m
k1+···+km=k
.
Proof. Let f(q) = a +a q +a q−1 +a q2 +a q−2 +··· where a = dimV for all i ∈ Z.
0 1 −1 2 −2 i i
Then, we have
(cid:77)
dim(∧k(V)) = dim ∧k1(V )⊗···⊗∧km(V )
n i1 im
k1i1+···+kmim=n,
k1+···+km=k
(cid:88)
= dim(∧k1(V )⊗···⊗∧km(V ))
i1 im
k1i1+···+kmim=n,
k1+···+km=k
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) a a
= i1 ··· im
k k
1 m
k1i1+···+kmim=n,
k1+···+km=k
Therefore,
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) a a
grdim∧k(V) = i1 ··· im qi1k1+···imkm
k k
1 m
k1+···+km=k
(cid:3)
5. λ-Rings
Definition5.1. (λ-ring)Aλ-ring isacommutativeringRtogetherwithoperationsλn: R →
R for every non-negative integer n such that
(a) λ0(r) = 1 for all r ∈ R
(b) λ1(r) = r for all r ∈ R
(c) λn(1) = 0 for all n > 1
(d) λn(r+s) = (cid:80)n λk(r)λn−k(s) for all r,s ∈ R
k=0
(e) λn(rs) = P (λ1(r),...,λn(r),λ1(s),...,λn(s)) for all r,s ∈ R
n
(f) λm(λn(r)) = P (λ1(r),...,λmn(r)) for all r ∈ R
m,n
LAMBDA-RINGS 9
where P and P are certain universal polynomials with integer coefficients such that if
n m,n
x = x +x +···, y = y +y +···, then
1 2 1 2
(cid:88) (cid:89)
P (λ1(x),...,λn(x),λ1(y),...,λn(y))tn = (1+tx y ),
n i j
n i,j
i.e. each P (λ1(x),...,λn(x),λ1(y),...,λn(y)) is the coefficient of tn in the expansion of the
n
right hand side. So, we have P = 1, P (λ1(x),λ1(y)) = (cid:80) x y , and so on.
0 1 i,j i j
(cid:88) (cid:89)
P (λ1(x),...,λmn(x))tm = (1+tx x ···x ),
m,n i1 i2 in
m i1<i2<···<in
i.e. each P (λ1(r),...,λmn(r)) is the coefficient of tm in the expansion of the right hand
m,n
side. So, we have P = 1, P (λ1(x),...,λn(x)) = (cid:80) x x ···x , and so on.
0,n 1,n i1<i2<···<in i1 i2 in
Lemma 5.2. Given any λ-ring R and x ∈ R, P (x) = 1 and P (x) = 0 for all m > 1.
1,0 m,0
Proof. Let x be an arbitrary element in R. By axiom (f), P (x) = λ1(λ0(x)). Then, by
1,0
applying axioms (a) and (b), λ1(λ0(x)) = λ1(1) = 1. So, P (x) = 1. Similarly, let m > 1.
1,0
Then, by axiom (f), (a), and (c), P (x) = λm(λ0(x)) = λm(1) = 0. (cid:3)
m,0
Example 5.3. The ring (Z,+,·) together with the operation of binomial coefficients λn(x)
where
(cid:0)x(cid:1) if 0 ≤ x ≤ n
n
λn(x) = (−1)n(cid:0)−x(cid:1) if −n ≤ x < 0
n
0 elsewhere
is a λ-ring.
The binomial coefficient λn(x) = (cid:0)x(cid:1) satisfies the axioms: (a), (b), (c), and (d) hold by
n
definition and properties of binomial coefficient.
For (e), let r = r +r +··· where
1 2
1 if r ≥ 0,1 ≤ i ≤ r,
r = −1 if r < 0,1 ≤ i ≤ −r,
i
0 elsewhere.
Similarly, we define s = s +s +··· where
1 2
1 if s ≥ 0,1 ≤ j ≤ s,
s = −1 if s < 0,1 ≤ j ≤ −s,
j
0 elsewhere.
Then, when r ≥ 0,s ≥ 0 or r < 0,s < 0,
(cid:18) (cid:19)
(cid:88) (cid:89) (cid:88) rs
P (λ1(r),...,λn(r),λ1(s),...,λn(s))tn = (1+tr s ) = (1+t)rs = tn.
n i j
n
n i,j n
When r < 0,s ≥ 0 or r ≥ 0,s < 0,
(cid:18) (cid:19)
(cid:88) (cid:89) (cid:88) −rs
P (λ1(r),...,λn(r),λ1(s),...,λn(s))tn = (1+tr s ) = (1−t)−rs = (−1)n tn.
n i j
n
n i,j n
10 ZEHUI CHEN
Hence, by equating the coefficients of each tn from both sides, (e) holds.
For (f), when r ≥ 0, we have
(cid:18)(cid:0)r(cid:1)(cid:19)
(cid:88)Pm,n(λ1(r),...,λmn(r))tm = (cid:89) (1+tri1ri2···rin) = (1+t)(nr) = (cid:88) mn tm.
m i1<i2<···<in m
When r < 0, we have
(cid:88)Pm,n(λ1(r),...,λmn(r))tm = (cid:89) (1+tri1ri2···rin) = (1+(−1)nt)(−nr) = (cid:88)(cid:18)(−1)mn(cid:0)−nr(cid:1)(cid:19)tm.
m i1<i2<···<in m
Hence, by equating the coefficients of each tm from both sides, (f) holds.
Thus, all axioms of a λ-ring hold for the binomial coefficient, so (Z,+,·) with the λn(x) =
(cid:0)x(cid:1) is a λ-ring.
n
Definition 5.4. (Torsion free) In a ring R, an element r is torsion free if nr = 0 implies
n = 0.
Definition 5.5. (Binomial ring) A ring R is a binomial ring if R is torsion free and the
binomial coefficients
(cid:18) (cid:19)
r r(r−1)···(r−n+1)
= ∈ R
n n!
for all r ∈ R, and n a positive integer.
Remark 5.6. According to [2, Proposition 8.4], any binomial ring with the operations of
binomial coefficients λn(x) = (cid:0)x(cid:1) is a λ-ring. Hence, the ring (Z,+,·) together with the
n
operation of binomial coefficients is a λ-ring as a special case of binomial rings.
Definition 5.7. (Power series) A power series in a ring R, denoted R t , is a series of the
form (cid:80)∞ a tn where a ∈ R for all n ≥ 0 and t is an indeterminate. (cid:74) (cid:75)
n=0 n n
Example 5.8. Let R be a ring, then the ring 1+tR t of power series with constant term
1 where (cid:74) (cid:75)
(a) addition is defined to be multiplication of power series, i.e. r⊕s = rs for all r,s,
(b) multiplication is defined to be
(cid:32) (cid:33) (cid:32) (cid:33)
∞ ∞ ∞
(cid:88) (cid:88) (cid:88)
1+ r tn (cid:12) 1+ s tn = 1+ P (r ,...,r ,s ,...,s )tn,
n n n 1 n 1 n
n=1 n=1 n=1
(c) λ-operations are defined to be
(cid:32) (cid:33)
∞ ∞
(cid:88) (cid:88)
λn 1+ r tm = 1+ P (r ,...,r )tm
m m,n 1 mn
m=1 m=1
is a λ-ring.
The ring 1+tR t is a ring under the defined addition and multiplication in (a) and (b).
The additive ident(cid:74)ity(cid:75) is 1 and the additive inverse of an arbitrary element 1 + (cid:80)∞ a xn
n=1 n
is 1+(cid:80)∞ b xn, where b = −a +(cid:80)n−1a b . Hence, since R is closed under addition
n=1 n n n i=1 i n−i
and multiplication, b = −a and by induction on n, b ∈ R, so the additive inverse 1 +
1 1 n
(cid:80)∞ b xn ∈ 1+tR t .
n=1 n
(cid:74) (cid:75)
