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

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

• 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 $R= W(R_0)$. Aug 31, 2016 at 2:34

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.

• 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? Sep 27, 2016 at 12:14