Definition 32.16.1. For any ring $R$ the *dual numbers* over $R$ is the $R$-algebra denoted $R[\epsilon ]$. As an $R$-module it is free with basis $1$, $\epsilon $ and the $R$-algebra structure comes from setting $\epsilon ^2 = 0$.

## 32.16 Tangent spaces

In this section we define the tangent space of a morphism of schemes at a point of the source using points with values in dual numbers.

Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point with image $s = f(x)$ in $S$. Consider the solid commutative diagram

with the curved arrow being the canonical morphism of $\mathop{\mathrm{Spec}}(\kappa (x))$ into $X$.

Lemma 32.16.2. The set of dotted arrows making (32.16.1.1) commute has a canonical $\kappa (x)$-vector space structure.

**Proof.**
Set $\kappa = \kappa (x)$. Observe that we have a pushout in the category of schemes

where $\kappa [\epsilon _1, \epsilon _2]$ is the $\kappa $-algebra with basis $1, \epsilon _1, \epsilon _2$ and $\epsilon _1^2 = \epsilon _1\epsilon _2 = \epsilon _2^2 = 0$. This follows immediately from the corresponding result for rings and the description of morphisms from spectra of local rings to schemes in Schemes, Lemma 25.13.1. Given two arrows $\theta _1, \theta _2 : \mathop{\mathrm{Spec}}(\kappa [\epsilon ]) \to X$ we can consider the morphism

where the first arrow is given by $\epsilon _ i \mapsto \epsilon $. On the other hand, given $\lambda \in \kappa $ there is a self map of $\mathop{\mathrm{Spec}}(\kappa [\epsilon ])$ corresponding to the $\kappa $-algebra endomorphism of $\kappa [\epsilon ]$ which sends $\epsilon $ to $\lambda \epsilon $. Precomposing $\theta : \mathop{\mathrm{Spec}}(\kappa [\epsilon ]) \to X$ by this selfmap gives $\lambda \theta $. The reader can verify the axioms of a vector space by verifying the existence of suitable commutative diagrams of schemes. We omit the details. (An alternative proof would be to express everything in terms of local rings and then verify the vector space axioms on the level of ring maps.) $\square$

Definition 32.16.3. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. The set of dotted arrows making (32.16.1.1) commute with its canonical $\kappa (x)$-vector space structure is called the *tangent space of $X$ over $S$ at $x$* and we denote it $T_{X/S, x}$. An element of this space is called a *tangent vector* of $X/S$ at $x$.

Since tangent vectors at $x \in X$ live in the scheme theoretic fibre $X_ s$ of $f : X \to S$ over $s = f(x)$, we get a canonical identification

This pleasing definition involving the functor of points has the following algebraic description, which suggests defining the *cotangent space of $X$ over $S$ at $x$* as the $\kappa (x)$-vector space

simply because it is canonically $\kappa (x)$-dual to the tangent space of $X$ over $S$ at $x$.

Lemma 32.16.4. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. There is a canonical isomorphism

of vector spaces over $\kappa (x)$.

**Proof.**
Set $\kappa = \kappa (x)$. Given $\theta \in T_{X/S, x}$ we obtain a map

Taking sections we obtain an $\mathcal{O}_{X, x}$-linear map $\xi _\theta : \Omega _{X/S, x} \to \kappa \text{d}\epsilon $, i.e., an element of the right hand side of the formula of the lemma. To show that $\theta \mapsto \xi _\theta $ is an isomorphism we can replace $S$ by $s$ and $X$ by the scheme theoretic fibre $X_ s$. Indeed, both sides of the formula only depend on the scheme theoretic fibre; this is clear for $T_{X/S, x}$ and for the RHS see Morphisms, Lemma 28.31.10. We may also replace $X$ by the spectrum of $\mathcal{O}_{X, x}$ as this does not change $T_{X/S, x}$ (Schemes, Lemma 25.13.1) nor $\Omega _{X/S, x}$ (Modules, Lemma 17.25.7).

Let $(A, \mathfrak m, \kappa )$ be a local ring over a field $k$. To finish the proof we have to show that any $A$-linear map $\xi : \Omega _{A/k} \to \kappa $ comes from a unique $k$-algebra map $\varphi : A \to \kappa [\epsilon ]$ agreeing with the canonical map $c : A \to \kappa $ modulo $\epsilon $. Write $\varphi (a) = c(a) + D(a) \epsilon $ the reader sees that $a \mapsto D(a)$ is a $k$-derivation. Using the universal property of $\Omega _{A/k}$ we see that each $D$ corresponds to a unique $\xi $ and vice versa. This finishes the proof. $\square$

Lemma 32.16.5. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point and let $s = f(x) \in S$. Assume that $\kappa (x) = \kappa (s)$. Then there are canonical isomorphisms

and

This works more generally if $\kappa (x)/\kappa (s)$ is a separable algebraic extension.

**Proof.**
The second isomorphism follows from the first by Lemma 32.16.4. For the first, we can replace $S$ by $s$ and $X$ by $X_ s$, see Morphisms, Lemma 28.31.10. We may also replace $X$ by the spectrum of $\mathcal{O}_{X, x}$, see Modules, Lemma 17.25.7. Thus we have to show the following algebra fact: let $(A, \mathfrak m, \kappa )$ be a local ring over a field $k$ such that $\kappa /k$ is separable algebraic. Then the canonical map

is an isomorphism. Observe that $\mathfrak m/\mathfrak m^2 = H_1(\mathop{N\! L}\nolimits _{\kappa /A})$. By Algebra, Lemma 10.132.4 it suffices to show that $\Omega _{\kappa /k} = 0$ and $H_1(\mathop{N\! L}\nolimits _{\kappa /k}) = 0$. Since $\kappa $ is the union of its finite separable extensions in $k$ it suffices to prove this when $\kappa $ is a finite separable extension of $k$ (Algebra, Lemma 10.132.9). In this case the ring map $k \to \kappa $ is étale and hence $\mathop{N\! L}\nolimits _{\kappa /k} = 0$ (more or less by definition, see Algebra, Section 10.141). $\square$

Lemma 32.16.6. Let $f : X \to Y$ be a morphism of schemes over a base scheme $S$. Let $x \in X$ be a point. Set $y = f(x)$. If $\kappa (y) = \kappa (x)$, then $f$ induces a natural linear map

which is dual to the linear map $\Omega _{Y/S, y} \otimes \kappa (y) \to \Omega _{X/S, x}$ via the identifications of Lemma 32.16.4.

**Proof.**
Omitted.
$\square$

Lemma 32.16.7. Let $X$, $Y$ be schemes over a base $S$. Let $x \in X$ and $y \in Y$ with the same image point $s \in S$ such that $\kappa (s) = \kappa (x)$ and $\kappa (s) = \kappa (y)$. There is a canonical isomorphism

The map from left to right is induced by the maps on tangent spaces coming from the projections $X \times _ S Y \to X$ and $X \times _ S Y \to Y$. The map from right to left is induced by the maps $1 \times y : X_ s \to X_ s \times _ s Y_ s$ and $x \times 1 : Y_ s \to X_ s \times _ s Y_ s$ via the identification (32.16.3.1) of tangent spaces with tangent spaces of fibres.

**Proof.**
The direct sum decomposition follows from Morphisms, Lemma 28.31.11 via Lemma 32.16.5. Compatibility with the maps comes from Lemma 32.16.6.
$\square$

Lemma 32.16.8. Let $f : X \to Y$ be a morphism of schemes locally of finite type over a base scheme $S$. Let $x \in X$ be a point. Set $y = f(x)$ and assume that $\kappa (y) = \kappa (x)$. Then the following are equivalent

$\text{d}f : T_{X/S, x} \longrightarrow T_{Y/S, y}$ is injective, and

$f$ is unramified at $x$.

**Proof.**
The morphism $f$ is locally of finite type by Morphisms, Lemma 28.14.8. The map $\text{d}f$ is injective, if and only if $\Omega _{Y/S, y} \otimes \kappa (y) \to \Omega _{X/S, x} \otimes \kappa (x)$ is surjective (Lemma 32.16.6). The exact sequence $f^*\Omega _{Y/S} \to \Omega _{X/S} \to \Omega _{X/Y} \to 0$ (Morphisms, Lemma 28.31.9) then shows that this happens if and only if $\Omega _{X/Y, x} \otimes \kappa (x) = 0$. Hence the result follows from Morphisms, Lemma 28.33.14.
$\square$

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