Exercise 45, § 1

By a pre-$\lambda$-ring is meant a ring A endowed with mappings $\lambda_n : A \to A$ ($n \in \mathbf{N}$) such that (i) $\lambda_0(a) = 1$, (ii) $\lambda_1(a) = a$, (iii) $\lambda_n(a + b) = \sum_{p=0}^n \lambda_p(a) \lambda_{n-p}(b)$, for all $a, b$ in A. Conditions (i) and (iii) can also be expressed by saying that

defines a homomorphism of the additive group of A into the multiplicative group $\Lambda(A)$.

By $\lambda$-morphisms from a pre-$\lambda$-ring A into another B is meant a ring homomorphism $\rho : A \to B$ such that $\rho(\lambda_n(a)) = \lambda_n(\rho(a))$ for all $a \in A, n \in \mathbf{N}$, or in other words such that the diagram

is commutative (the mapping $\Lambda(\rho)$ transforms the series $1 + \sum_{n \geq 1} a_n T^n$ into the series $1 + \sum_{n \geq 1} \rho(a_n) T^n$).

Let A be a ring. We propose to endow $\Lambda(A)$ with a structure of pre-$\lambda$-ring. Let $E^A : U(A) \to \Lambda_T(A)$ denote the ring isomorphism defined in Exercise 43. Let S be another indeterminate. a) Show that the two composite isomorphisms

and coincide; they will be denoted by $\tilde{E}^A$. The mapping $\lambda^A$ is defined by the commutativity of the diagram

where $\mu^A$ denotes the homomorphism defined in Exercise 41, d), p. 55.

EXERCISES

AC IX. 59

b) The ring $\Lambda(A)$, endowed with $\lambda^A$, is a pre-$\lambda$-ring. For every ring homomorphism $\rho : A \to B, \Lambda(\rho) : \Lambda(A) \to \Lambda(B)$ is a $\lambda$-morphism. c) A pre-$\lambda$-ring $A$ is called a $\lambda$-ring if $\lambda : A \to \Lambda(A)$ is a $\lambda$-morphism (for the pre-$\lambda$-ring structure on $\Lambda(A)$ just defined). Let $A$ be a ring. Show that $(\Lambda(A), \lambda^A)$ is a $\lambda$-ring. (Use Exercise 41, f), p. 56.)