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Assume $X$ is a scheme over the base-ring $\mathbb Z$ and $t:\text{Spec}\,\mathbb Z \to X$ is an integer shaped point of $X$. Is there necessarily an affine open subscheme $U$ of $X$ through which $t$ factors?

Context: Richard Garner claims on page 5 first paragraph of An embedding theorem for tangent categories that all schemes are microlinear (in the sense of synthetic differential geometry). They claim that I can reduce the proof to the affine case by using that open embeddings of schemes are formally etale. Formally etale means that if I have some $t: \text{Spec } \mathbb Z = \{0\} \to X$, lying in some open affine subspace $U$, then any infinitesimal thickening of $\{0\}$ such as e.g. $D_2(2) = \text{Spec } \mathbb Z[x,y]/(x,y)^3$ lies also in $U$. But to use this I have to know that $t$ lies in some open affine in the first place.

I am equally happy if you can tell me why any scheme is infinitesimally linear (the weaker version of microlinearity).

Nico
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  • This thread answers your question in the positive under some mild assumptions on X: https://math.stackexchange.com/questions/1621733/is-every-integral-point-of-an-arithmetic-scheme-contained-in-an-affine-open-set – Nathan Lowry Oct 18 '22 at 19:06
  • @NathanLowry I have seen it, I should have linked to the question. Doesn't it only answer the question when $X$ is projective over $\text{Spec }\mathbb Z$? – Nico Oct 18 '22 at 19:11
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    Quasi-projective, but yes. If you would like this to be true for any scheme $X$ over $\mathbb{Z}$ I doubt it will be true. I will try to find a counterexample – Nathan Lowry Oct 18 '22 at 20:14
  • @NathanLowry Ouch. Do you know how I can show that schemes are microlinear? – Nico Oct 18 '22 at 20:30
  • Unfortunately I do not know what this means. I took a quick look at the source you provided, but I wasn't really able to make anything out. If it helps at all, your claim is true if you replace $mathbb{Z}$ by $\mathbb{Z}[\epsilon]/(\epsilon^2)$. In fact it will be true for any local ring. – Nathan Lowry Oct 18 '22 at 21:12
  • @NathanLowry Ahh, the version of microlinearity which I need is not complicated at all. Maybe I should open a new question where I describe it explicitly. I will link to it from this question. I know that it works for local rings, since a local ring shaped point lies in every open space in which its underlying point lies (local rings are a point + fuzz, and the fuzz always fits into any open neighbourhood of the point by the very concept of openness :) – Nico Oct 18 '22 at 21:45
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    @Nico : Garner, page 672: "The category $Sch$ of schemes over $Spec(\mathbb{Z})$ is a tangent category under the structure which sends a scheme $X$ to its Zariski tangent space $TX$." Given a scheme $X/\mathbb{Z}$ there is no canonical way to choose a "Zariski tangent space $TX$" of $X$. You should explain what Garner is talking about. – hm2020 Dec 19 '22 at 17:45
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    @hm2020 Write $D$ for $Spec (\mathbb Z[x]/(x^2))$. Then $TX$ is the exponential object $X^D$ in the appropriate sheaf category $Sh(Ring^{op}, topology)$. The functor $X^D$ is representable by a scheme, and it makes no difference which surrounding sheaf topos is chosen to form the exponential. – Nico Dec 19 '22 at 19:41
  • @Nico: In the language of the paper: The functor $F(X):=TX$ should be an endofunctor of the category $Sch(\mathbb{Z})$ of "schemes over $Spec(\mathbb{Z})$". How does the "scheme" $F(X):=TX:=X^D$ look like? – hm2020 Dec 20 '22 at 09:55
  • @hm2020 The category of schemes embeds fully faithful into $Sh(Ring^{op},Zariski)$ via the functor of points construction. $X^D$ is a functor, and it is a theorem that it is the functor of points of a scheme when $X$ is a scheme. As a functor, you can use the Yoneda trick to compute it. $X^D(R) = Hom(Spec R, X^D) = Hom(Spec R \times D, X) = Hom (Spec(R\otimes \mathbb Z[x]/(x^2)),X) =X(R[x]/(x^2))$. – Nico Dec 20 '22 at 10:11
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    This makes more sense. – hm2020 Dec 20 '22 at 10:14

1 Answers1

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This answer is somewhat complementary to the discussion in the comments – while they show the statement holds for a reasonable class of schemes, I find a rather ugly one where the statement fails (as Nathan Lowry expected).

Consider the following scheme $X$: it is the union of two copies of $\mathbb{A}^1_{\mathbb{Z}}$, glued together on the open subset corresponding to all the primes except $(2,T)$ and $(3,T)$. Denote as $X_1$ the (isomorphic and open) image of the first affine line in $X$ (with coordonate $T_1$), and $X_2$ the image of the second one (with coordinate $T_2$).

Then take the following morphism $t: \mathrm{Spec}\,\mathbb{Z} \rightarrow X$: on $R_1=\mathbb{Z}[1/2]$, it is the map $R_1 \rightarrow X_1$ given by $T_1\longmapsto 0$. On $R_2=\mathbb{Z}[1/3]$, it is the map $R_2 \rightarrow X_2$ given by $T_2\longmapsto 0$.

(indeed, the two morphisms agree on the spectrum of $\mathbb{Z}[1/6]$).

What happens here is that we have a lack of separatedness – geometrically, distinct morphisms that somehow still agree on the generic fiber.

But I’m not sure that this is the only obstruction, although by Chow’s lemma separatedness (along with finite type, which is rather natural in this case) implies a property which doesn’t look so much weaker than quasi-projectiveness.

Aphelli
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