What is the universal property of the thickening $Y[\varepsilon]$?

Given an $$S$$-scheme $$Y$$, let $$Y[\varepsilon]$$ denote thickening $$Y[\varepsilon]=Y \times_S D_S$$ of $$Y$$. Here $$D_S$$ is the $$S$$-scheme $$D\times_\mathbb Z S\to S$$ where $$D = \operatorname{Spec} \mathbb{Z}[\varepsilon]/(\varepsilon^2)$$. The $$S$$-scheme $$Y[\varepsilon]$$ should have a mapping out universal property in $$\mathrm{Sch}_{/S}$$, which is probably related to derivations. Unfortunately I cannot find a good reference. A morphism $$f \in \mathrm{Hom}_S(Y[\varepsilon],X)$$ is the same thing as a map $$f \in \mathrm{Hom}_S(Y,X)$$ plus what data? Help is appreciated.

Context: I am trying to understand the proof of Lemma 5.12.1 in Martin Brandenburg’s PhD thesis.

• Perhaps you might directly contact Brandenburg here? ;) Jan 12, 2023 at 20:08
• Tag me next time haha... only found it by coincidence. Please call me Martin! Jan 15, 2023 at 18:30

I only know how to say things about affine schemes so please permit me to restrict my attention to that case. Let $$k$$ be a commutative ring and $$f : R \to S$$ be a morphism of commutative $$k$$-algebras. An $$f$$-derivation (I'm making this up, I don't know what the standard term is here) is a $$k$$-linear map $$D : R \to S$$ satisfying the Leibniz rule

$$D(ab) = f(a) D(b) + D(a) f(b).$$

You can check that the data of a pair $$(f, D)$$ consisting of a morphism $$f$$ and an $$f$$-derivation $$D$$ is the same thing as the data of a morphism $$R \to S[\varepsilon]/\varepsilon^2$$; the morphism associated to such a pair is

$$R \ni a \mapsto f(a) + D(a) \varepsilon \in S$$

and the Leibniz rule is exactly equivalent to the condition that this map respects multiplication. Also equivalently, this is the data of a $$k[\varepsilon]/\varepsilon^2$$-linear morphism $$R[\varepsilon]/\varepsilon^2 \to S[\varepsilon]/\varepsilon^2$$.

Intuitively you can think of an $$f$$-derivation as an element of the tangent space of $$f^{\ast} : \text{Spec } S \to \text{Spec } R$$ in the "Hom scheme" of morphisms $$\text{Spec } S \to \text{Spec } R$$, and thickening is just a globalization of this construction.

As a special case, you can check that a morphism $$f : R \to k[\varepsilon]/\varepsilon^2$$ is the same thing as a pair of a $$k$$-point of $$\text{Spec } R$$ and a Zariski tangent vector at this $$k$$-point (at least if $$k$$ is algebraically closed). This means $$\text{Spec } k[\varepsilon]/\varepsilon^2$$ is the "walking tangent vector," so if you want to compute a tangent vector to a morphism $$f : X \to Y$$ of $$k$$-schemes, that means you want to compute

$$\text{Hom}(\text{Spec } k[\varepsilon]/\varepsilon^2, \text{Hom}(X, Y)) \cong \text{Hom}(\text{Spec } k[\varepsilon]/\varepsilon^2 \times X, Y)$$

where the "Hom scheme" $$\text{Hom}(X, Y)$$ may not always exist as a scheme but always exists at least as a presheaf $$\text{Hom}(- \times X, Y)$$.

Edit: This also means that the "Hom scheme" $$\text{Hom}(\text{Spec } k[\varepsilon]/\varepsilon^2, X)$$ itself can be interpreted as the tangent bundle of $$X$$ which is roughly the conceptual content of Lemma 5.12.1.

Thank you Qiaochu Yuan for your answer! I will, for completeness, glue your answer to the non-affine case. The key insight has been that a morphism $$(\phi^\#, D) \colon \mathcal{O}_X \to \phi_\ast \mathcal{O}_Y[\varepsilon]/(\varepsilon^2) = \phi_\ast \mathcal{O}_Y \oplus \phi_\ast \mathcal{O}_Y \varepsilon$$ is a morphism of $$p^{-1} \mathcal{O}_S$$-algebras if and only if $$D$$ is a $$p^{-1} \mathcal{O}_S$$-derivation. Knowing this, the computation looks like this: \begin{align} \mathrm{Hom}_S(Y, T(X/S)) &= \mathrm{Hom}_S(Y[\varepsilon], X) \\[0.5em] &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Hom}_{\mathrm{Alg}(p^{-1}\mathcal{O}_S)}(\mathcal{O}_X, (\phi_\ast \mathcal{O}_Y)[\varepsilon]/(\varepsilon^2)) \\[0.5em] &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Der}_{p^{-1} \mathcal{O}_S}(\mathcal{O}_X, \phi_\ast \mathcal{O}_Y) \\[0.5em] &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Hom}_{\mathcal{O}_X}(\Omega_{X/S}, \phi_\ast \mathcal{O}_Y) \\[0.5em] &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Hom}_{\mathrm{Alg}(\mathcal{O}_X)}(\operatorname{Sym} \Omega_{X/S}, \phi_\ast\mathcal{O}_Y) \\[0.5em] &= \mathrm{Hom}_S(Y, \operatorname{Spec}_X \operatorname{Sym} \Omega_{X/S}) \,. \end{align}