A lambda-ring $R$ is called "special" if it satisfies the $\lambda^i\left(xy\right)=...$ and $\lambda^i\left(\lambda^j\left(x\right)\right)=...$ relations, or, equivalently, if the map $\lambda\_T:R\to\Lambda\left(R\right)$ given by $\lambda\_T\left(x\right)=\sum\limits\_{i=0}^{\infty}\lambda^i\left(x\right)T^i$ (where the $\sum$ sign means addition in $R\left[\left[T\right]\right]$, not addition in $\Lambda\left(R\right)$) is a morphism of lambda-rings. If you are wondering what the hell I am talking about, most likely you belong to the school of algebraists that denote only special lambda-rings as lambda-rings at all.


Anyway, let $A$ and $B$ be two special lambda-rings, and for every $i>0$, let $\Psi\_A^i$ and $\Psi\_B^i$ be the $i$-th Adams operations on $A$ and $B$, respectively. Let $f:A\to B$ be a ring homomorphism such that $f\circ\Psi\_A^i=\Psi\_B^i\circ f$ for every $i>0$. Does this yield that $f$ is a lambda-ring homomorphism, i. e. that $f\circ\lambda\_A^i=\lambda\_B^i\circ f$ for every $i>0$ ?


Note that this is clear if both $A$ and $B$ are torsion-free as additive groups (i. e., none of the elements $1$, $2$, $3$, ... is a zero-divisor in any of the rings $A$ and $B$), but Hazewinkel, in his text [Witt vectors, part 1](http://arxiv.org/abs/0804.3888) (Lemma 16.35), claims the same result for the general case. I am writing a list of errata for his text, and I would like to know whether this should be included - well, and I'd like to know the answer anyway, as I am writing some notes on lambda-rings as well.


For the sake of completeness, here is a definition of Adams operations: These are the maps $\Psi^i:R\to R$ for every integer $i>0$ (where $R$ is a special lambda-ring) defined by the equation


$\sum\limits\_{i=1}^{\infty} \Psi^i\left(x\right)T^i = -T\frac{d}{dT}\log\left(\lambda\_{-T}\left(x\right)\right)$ in the ring $R\left[\left[T\right]\right]$ for every $x\in R$.


Here, even if the term $\log\left(\lambda\_{-T}\left(x\right)\right)$ may not make sense (since some of the fractions $\frac{1}{1}$, $\frac{1}{2}$, $\frac{1}{3}$, ... may not exist in $R$), the logarithmic derivative $\frac{d}{dT}\log\left(\lambda\_{-T}\left(x\right)\right)$ is defined formally by


$\displaystyle \frac{d}{dT}\log\left(\lambda\_{-T}\left(x\right)\right)=\frac{\frac{d}{dT}\lambda\_{-T}\left(x\right)}{\lambda\_{-T}\left(x\right)}$.



Is every Adams ring morphism a lambda-ring morphism?
Here is a more general point of view on Charles's example, which someone might find helpful.


Let $M$ be an abelian group, and let $\mathrm{Z}[M]$ denote $\mathrm{Z}\oplus M$ with the usual split ring structure $(a,m)(a',m')=(aa',am'+a'm)$. There is a simple description of the (special) $\lambda$-ring structures on such rings. For any prime number $p$, let $\theta\_p$ denote the symmetric function $(\psi\_p-e^p)/p$, where $e=x\_1+x\_2+\cdots$ and $\psi\_p=x\_1^p+x\_2^p+\cdots$. Observe that $\theta\_p$ has integral coefficients and therefore defines a natural operation on any $\lambda$-ring, and for any element $x$ in any $\lambda$-ring, we have $\psi\_p(x)=x^p+p\theta\_p(x)$. In particular, for any $\lambda$-structure on $\mathrm{Z}[M]$, the Adams operation $\psi\_p$ satisfies $\psi\_p(m)=p\theta\_p(m)$ for all $m\in M$.


Exercise: Given a commuting family of additive maps $\theta\_p:M\to M$, there is a unique $\lambda$-ring structure on $\mathrm{Z}[M]$ whose $\theta\_p$ operators on $M$ are the given ones. Conversely, suppose $\mathrm{Z}[M]$ has a $\lambda$-ring structure. Then each $\theta\_p$ preserves the ideal $M$, and resulting map $\theta\_p:M\to M$ is additive. 


Now it is a general fact that a ring map between two $\lambda$-rings is a $\lambda$-ring map if and only if it commutes with the $\theta\_p$ operators. (This is because such symmetric functions and all their compositions under plethysm generate the whole ring of symmetric functions.) Thus, in our case, a $\lambda$-ring map $\mathrm{Z}[M]\to\mathrm{Z}[N]$ is the same as a linear map $M\to N$ that commutes with each $\theta\_p$. But that is not the same as commuting with each $p\theta\_p$, i.e. the Adams operations, and it is easy to make counterexamples. For instance Charles's counterexample is with $M=N=\mathrm{Z}/2\mathrm{Z}$, where $\theta\_p$ on $M$ is the identity for all $p$, but $\theta\_p$ on $N$ is the identity for all odd $p$ but is zero for $p=2$. The identity map $M\to N$ therefore commutes with each $p\theta\_p$ and so is an Adams map. It commutes with each $\theta\_p$ with $p$ odd, but it doesn't commute with $\theta\_2$. Therefore it is not a $\lambda$-map.


