$\newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathbb{F}_1} \newcommand{\spec}{\operatorname{Spec}}$If I understand correctly, in Borger's paper $\Lambda$-rings and the field with one element about the field with one element, the category of "affine schemes over $\spec\F$" is just the opposite of the category of $\Lambda$-rings. The base change from $\spec\F$ to $\spec\Z$ is forgetting the $\Lambda$-structure. I.e. we can view a $\Lambda$-structure on a commutative ring as a descent data from $\spec \Z$ to $\spec\F$ (there is a generalization to the non-affine case, but let's keep it simple for now).

The base change has adjoints from both sides. The one that comes from the composition $$\spec R \to \spec \Z \to \spec \F$$

takes a ring $R$ to the $\Lambda$-ring of (big) Witt vectors over $R$ denoted $W(R)$.

Now, it seems to me that a desirable property of this setup would be that an (affine) scheme over $\spec\Z$ would be "the same as" an (affine) scheme over $\spec\F$ with a morphism to $\spec\Z$ over $\spec\F$. So for a $\Lambda$-ring $R$, a map of $\Lambda$-rings $W(\Z)\to R$ should be induced by a ring $R_0$ by applying $W$ to the structure morphism $\Z \to R_0$. Moreover, there should be a natural bijection between the two types of data (up to isomorphisms in the obvious way). It seems that the elements of $R_0$ should be something like elements of $R$ of rank $\le 1$ (i.e. those for which $\lambda^n$ vanishes for n $> 1$) and the sum should be something like the "rank one approximation of the sum in $R$".

Question: Is this indeed true?

Perhaps I am wrong in assuming that this is desirable, so an explanation of that would be great too. By the way, I am not an expert on $\Lambda$-rings and I ask this purely out of curiosity.

  • $\begingroup$ Presumably we want to let $I$ be the kernel of some natural map $W(\mathbb Z) \to \mathbb Z$ (I believe the one where we take the first term of the power series) and $R_0 = R/IR$. But I don't see why we should then have $R= W(R_0)$. $\endgroup$
    – Will Sawin
    Commented Aug 31, 2016 at 2:34

1 Answer 1


I guess I should be the one to answer this!

Unfortunately, the answer is no.

There is a natural adjunction between the two types of data: given a $\Lambda$-ring $R$, set $R_0$ to be $R\otimes_{W(\mathbb{Z})}\mathbb{Z}$, where the map $W(\mathbb{Z})\to\mathbb{Z}$ is the projection on the first component, i.e. the co-unit of the adjunction between $W$ and the forgetful functor from $\Lambda$-rings to rings. (This is as in Will Sawin's comment.) For instance, if $R$ is already of the form $W(S)$, then $R_0$ would be $W(S)\otimes_{\mathbb{Z}}\mathbb{Z}$, which does indeed map to $S$ but not isomorphically, in general. It is if $S$ is etale over $\mathbb{Z}$, but it fails for $S=\mathbb{Z}[x]$, if I remember. (Also, note that $R_0$ is naturally a quotient ring of $R$, not a subring, as you proposed in your question. This is as it should be, since $S$ is a quotient ring of $W(S)$, not a subring.)

But it's also true that there are some $R$'s that are not of the form $W(S)$ for any $S$. To see this it's nice to warm up with a toy example. Instead of considering $\Lambda$-structures on rings, let's consider $G$-actions, where $G$ is a monoid. Then the analogue of the Witt vector functor is given by $S\mapsto S^G$, where $G$ acts on $S^G$ by translation in the exponent. (In fact, this is more than an analogy. If $G$ is the monoid $\mathbb{Z}_{\geq 1}$ of integers $\geq 1$ under multiplication, then a $G$-action is also called a $\Psi$-ring structure, and the analogue of the Witt vector functor $S\mapsto S^G$ is the ghost-component functor.) Then a $G$-equivariant $S^G$-algebra $R$ is canonically of the form $R=\prod_{g\in G} R_g$ and an element $h\in G$ sends $R_g$ to $R_{hg}$.

If $G$ is a group, then these maps are isomorphisms and hence all the $R_g$ can be recovered from $R_1$. More precisely, $R_1$ is exactly $R\otimes_{\mathbb{Z}^G}\mathbb{Z}$, and the map $R\to (R\otimes_{\mathbb{Z}^G}\mathbb{Z})^G$ is an isomorphism. But if $G$ is not a group, then the maps $R_g\to R_{hg}$ are not in general isomorphisms. For instance, you can take $G=\mathbb{N}$ (under addition) and $R_0$ a nonzero ring but $R_n$ the zero ring for all $n\geq 0$.

In fact, as alluded to above, a $\Lambda$-structure on $R$ is just an action of the monoid $\mathbb{Z}_{\geq 1}$ such that certain congruence conditions are satisfied. (More precisely, this is true if $R$ is torsion-free.) So if $R$ is a $\mathbb{Q}$-algebra, then all congruence conditions are vacuously true, and so a $\Lambda$-structure is equivalent to an action of the monoid $\mathbb{Z}_{\geq 1}$. So then we're really in monoid land, and we can make a counterexample as above, by setting $R_1=\mathbb{Q}$ and $R_n=\{0\}$ for all $n>1$, and $R=\prod_{n\geq 1} R_n$. Then the map $R\to (R\otimes_{\mathbb{Z}^G}\mathbb{Z})^G$ is identified with the diagonal map $\mathbb{Q}\to \mathbb{Q}\times\mathbb{Q}\times\cdots$.

Another way of putting this is that the answer to your question is already no if we work with the ghost-component functor, instead of the Witt vector functor, because the Adams operators / Frobenius maps are not necessarily isomorphisms. But you might then ask your question again in the world of 'perfect $\Lambda$-rings' and 'perfect Witt vectors'. These are where you require the Adams / Frobenius maps to be isomorphisms. I don't see the answer immediately, but it shouldn't be too hard to work out.

  • $\begingroup$ Thank you for the answer, it really clarifies things (I hoped this question will draw your attention :) ) Perhaps you can say in a few words what the failure of this "means" in the conceptual picture of absolute algebraic geometry? $\endgroup$
    – KotelKanim
    Commented Sep 27, 2016 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.