Exercise 46, § 1

(a) Let $A$ be a pre-$\lambda$-ring and $a \in A$. We call $\lambda$-rank of $a$ the upper bound in $\overline{\mathbf{R}}$ of the set of integers $n \in \mathbf{N}$ such that $\lambda_n(a) \neq 0$. Let $P$ be a multiplicative subset of $A$ formed of elements of $\lambda$-rank $\leq 1$. Let $a = \sum_{i \in I} a_i$ be the sum of a finite family of elements of $P$. We have

(b) Let $A$ be a ring and $a, b$ be elements of $A$. Then in $\Lambda(A)$ the elements $1 + aT$ and $(1 + aT) * (1 + bT) = 1 + abT$ are of $\lambda$-rank $\leq 1$. (c) Let $A$ be a pre-$\lambda$-ring. Suppose that there exists a multiplicative submonoid of $A$, formed of elements of $\lambda$-rank $\leq 1$, which generates the additive group of $A$. Now prove that $A$ is then a $\lambda$-ring. More generally, $A$ is a $\lambda$-ring if there exists an injective $\lambda$-morphism of $A$ into a pre-$\lambda$-ring satisfying the preceding property (“Splitting Principle”). (It amounts to showing that the $Z$-bilinear mapping $(a, b) \mapsto \lambda(a) * \lambda(b) \lambda(ab)^{-1}$ is zero and that the $Z$-linear mappings $a \mapsto \Lambda_S(\lambda_T)[\lambda_S(a)]$ and $a \mapsto \lambda^A_S(\lambda_T(a))$ from $A$ into $\Lambda_S(\Lambda_T(A))$ coincide. It suffices to verify these properties for $a, b \in P$.) (d) Let $A = \mathbf{Z}[(X_i)_{i \in I}, (X'_i)_{i' \in I'}]$ be the ring of polynomials in two finite families of indeterminates. We have where $\binom{I}{n}$ denoting the set of subsets with $n$ elements of $I$, and where $X_H = \prod_{h \in H} X_h$. By assigning the $X_i$ (and $X'_{i'}$) weight 1, $s_n$ is homogeneous of weight $n$. Every symmetric polynomial in $(X_i)_{i \in I}$ is expressed uniquely as a polynomial in the $s_n$ ($n \geq 1$) (A, IV, p. 58). In particular, for every $m \geq 0$, we have where $Q_{n,m}$ is homogeneous of weight $nm$ in the $s_r = s_r((X_i)_{i \in I})$ ($r = 1, ..., nm$). It is the coefficient of $T^m$ in $\prod_{H \in \binom{I}{n}} (1 + X_H T)$, and it is well defined and independent of $I$, provided that $\mathrm{Card}(I) \geq nm$. The coefficient of $T^n$ in $\prod_{(i, i') \in I \times I'} (1 + X_i X'_{i'} T)$ is where $s'_r = s_r((X'_{i'})_{i' \in I'})$, and where $P_n$ is homogeneous of weight $n$ in each of the families of variables $(s_r)$ and $(s'_r)$. It is well defined and independent of $I$ and $I'$ provided that $I$ and $I'$ have cardinalities $\geq n$. Now prove that in the $\lambda$-ring $\Lambda(A)$ we have and According to a) and b), we have

et

and

e) Let $A$ be a pre-$\lambda$-ring. For $A$ to be a $\lambda$-ring, it is necessary and sufficient that the following conditions be satisfied, for all $a, b$ in $A$, and $n, m$ in $\mathbf{N}$. (i) $\lambda(1) = 1 + T$, (ii) $\lambda_n(ab) = P_n(\lambda_1(a), ..., \lambda_n(a), \lambda_1(b), ..., \lambda_n(b))$, (iii) $\lambda_m(\lambda_n(a)) = Q_{n,m}(\lambda_1(a), ..., \lambda_{nm}(a))$.

f) Let $\rho : A \to B$ be a $\lambda$-morphism of pre-$\lambda$-rings. If $A$ is a $\lambda$-ring and $\rho$ is surjective, then $B$ is a $\lambda$-ring. If $B$ is a $\lambda$-ring and if $\rho$ is injective, then $A$ is a $\lambda$-ring.