Is an ordinary scheme in Borger's Absolute Geometry the same as a "scheme over 𝔽₁" with a map to Spec(ℤ)?

Question

$\newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathbb{F}_1} \newcommand{\spec}{\operatorname{Spec}}$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.

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.