Adams operations

Remarks

So $\psi^1(x) = x$ and $\psi^2(x) = x^2 - 2 \lambda^2(x)$. Newton's formula expresses the power sums through the elementary symmetric polynomials; if $x = \xi_1 + \cdots + \xi_r$ is a sum of line elements, then $\lambda^n(x) = e_n(\xi)$ and $\psi^k(x) = \xi_1^k + \cdots + \xi_r^k$.

In a lambda-ring the $\psi^k$ satisfy:

1. Each $\psi^k$ is a ring homomorphism.
2. $\psi^k \circ \psi^l = \psi^{kl}$.
3. $\psi^k(x) = x^k$ if $x$ is a line element, that is, $\lambda_t(x) = 1 + xt$.
4. $\psi^p(x) \equiv x^p \pmod{pR}$ for every prime $p$.

Example: in $\Z$ with $\lambda_t(m) = (1 + t)^m$, one gets $-t \cdot m (1 + t)^{-1} = \sum_{k \ge 1} m (-t)^k$, so $\psi^k(m) = m$ for all $k$.

Non-example: the map $x \mapsto x^2$ is not $\psi^2$ on $\Z$, since $2^2 = 4$ while $\psi^2(2) = 2$; squaring is not additive, while $\psi^2$ is.