Exercise 48, § 1

Let $A$ be a pre-$\lambda$-ring. Let $\Psi$ denote the composite $A \xrightarrow{\lambda} \Lambda(A) \xrightarrow{L} A^J$, so that $\Psi(a) = (\psi_n(a))_{n \in J}$, where

More explicitly,

for every $n \in J$. In particular,

The mappings $\psi_n : A \to A$ are called Adams operations.

When $(A, \lambda) = (\Lambda(B), \lambda^B)$, where $B$ is a ring, we shall also write $\Psi^B$ for $\Psi$.

a) Let $B$ be a ring. Show that for every $n \in J$, one has

(Use the commutative diagram)

!Commutative diagram

where all the horizontal mappings are bijective.)

b) Let $A$ be a pre-$\lambda$-ring. For every $n \in J$, $\psi_n$ is an endomorphism of the additive group of $A$, and $\psi_1 = \mathrm{Id}_A$. If $A$ is a $\lambda$-ring, $\psi_n$ is an endomorphism of the ring $A$, and $\psi_n \circ \psi_m = \psi_{nm}$ for every $n, m$ in $J$. Moreover, for every prime number $p$ one has $\psi_p(a) \equiv a^p \mod pA$ for every $a \in A$. (Use the injectivity of $\lambda : A \to \Lambda(A)$, and the analogous properties of $\psi_n^A = F_n$.)

c) Let $A$ be a pre-$\lambda$-ring whose additive group is $\mathbf{Z}$-torsion-free. In order that $A$ be a $\lambda$-ring, it is necessary and sufficient that the following conditions be satisfied for all integers $n \geq 2$ and $m \geq 2$:

(i) $\psi_n(1) = 1$; (ii) $\psi_n(ab) = \psi_n(a) \psi_n(b)$ for every $a, b$ in $A$; (iii) $\psi_n \circ \psi_m = \psi_{nm}$.

(Since $L : \Lambda(A) \to A^J$ is an injective homomorphism of rings, $\Psi = L \circ \lambda$ is a homomorphism of rings if and only if $\lambda$ is one. Likewise, from the diagram

one deduces that

Now $L \circ \lambda = \Psi$ and $L \circ \psi_n^A = L \circ F_n = f_n \circ L$. Hence

d) Let $A$ be a $\lambda$-ring. Let $a \in A$ be an element of $\lambda$-rank $\leq 1$ (i.e. $\lambda(a) = 1 + aT$). Then $\psi_n(a) = a^n$ for every $n \in J$. If $P$ is a multiplicative subset of $A$ consisting of elements of $\lambda$-rank $\leq 1$ and if $a_1, ..., a_r$ belong to $P$, one has $\psi_n(a_1 + \cdots + a_r) = a_1^n + \cdots + a_r^n$.

In the polynomial algebra $\mathbf{Z}[(X_i)_{i \in I}]$, let $(s_n)$ denote the elementary symmetric polynomials,

and let be the $n$-th Newton polynomial. Now prove that, for every $a \in A$, one has (First verify this relation when $a$ is of the form $a_1 + \cdots + a_r$ as above. Proceed from there to the case where $a = 1 + \sum_{h \geq 1} s_h T^h \in \Lambda(\mathbf{Z}[(X_i)_{i \in I}])$. Then deduce the general case by embedding $A$ in the $\lambda$-ring $\Lambda(A)$ and using the homomorphism $\mathbf{Z}[(s_h)_{h=1,...,n}] \to A$ which sends $s_h$ to $\lambda_h(a)$.)