Exercise 51, § 1

Let $A$ be a $\lambda$-ring and let $(X_i)_{i \in I}$ be a family of indeterminates. Introduce the family of indeterminates $(\lambda_n X_i)_{(n,i) \in J \times I}$ such that $\lambda_1 X_i = X_i$ for every $i \in I$. Consider the polynomial algebra $B = A[(\lambda_n X_i)_{(n,i) \in J \times I}]$.

The ring homomorphism $\lambda : A \to \Lambda(A) \subset \Lambda(B)$ extends uniquely to a ring homomorphism $\lambda : B \to \Lambda(B)$ such that

and, for $q \geq 2$,

a) Now prove that $(B, \lambda)$ is a $\lambda$-ring. (What has to be shown is that the diagram of ring homomorphisms

is commutative. It is enough to verify this on a generating system of the ring $B$.)

b) Let $C$ be a $\lambda$-ring, $\rho : A \to C$ a $\lambda$-morphism, and $(c_i)_{i \in I}$ a family of elements of $C$. There exists one and only one $\lambda$-morphism $\rho' : B \to C$ extending $\rho$ and such that $\rho'(X_i) = c_i$ for every $i \in I$.