Tangent vectors at rational points are dual-number points

The dual-numbers scheme has one point. If $\mathfrak{p}$ is a prime ideal of $D$, then ${ϵ}^2=0\in \mathfrak{p}$ implies $ϵ\in \mathfrak{p}$ by [F11]. The quotient $D\mathrm{/}(ϵ)$ is $k$, so $(ϵ)$ is maximal by [F12, F13]. Every prime containing this maximal ideal equals it. Thus $\mathrm{Spec}D$ has the single point defined by $(ϵ)$, and the map $\mathrm{Spec}k\to \mathrm{Spec}D$ induced by $\rho$ selects that point.

Based morphisms can be computed in an affine neighborhood. Choose an affine open $U=\mathrm{Spec}A$ containing $x$ by [F5, F6]. Its inclusion into $X$ is an open immersion by [F7]. Since $\mathrm{Spec}D$ has only one point, every morphism in the fibre over $x$ factors uniquely through $U$. The affine anti-equivalence [F3], together with the $k$-morphism condition [F10], identifies such maps with $k$-algebra homomorphisms $\phi :A\to D$ whose reduction $\rho \circ \phi :A\to k$ is the point $x$. Conversely every such homomorphism gives a based morphism. If $\mathfrak{m}=\ker (A\to k)$, then $\mathfrak{m}$ is the point of $U$ by [F8].

These homomorphisms are exactly derivations. Each $a\in A$ has a unique image $\phi (a)=\bar{a}+ϵ\delta (a)$, where $\bar{a}=x^{\mathrm{\#}}(a)$. Since $\phi$ is a $k$-algebra homomorphism, $\delta$ is $k$-linear and vanishes on $k$. Comparing the $ϵ$-coefficients of $\phi (ab)=\phi (a)\phi (b)$ gives $\delta (ab)=\bar{a}\delta (b)+\bar{b}\delta (a)$, so $\delta$ is a derivation for the $A$-module structure on $k$ given by evaluation at $x$. Conversely each such derivation defines a homomorphism by this formula, since ${ϵ}^2=0$. These constructions are inverse.

Derivations on $A$ are the dual of its cotangent space at $x$. The structure map $k\to A$ splits evaluation $A\to k$, so $A=k\oplus \mathfrak{m}$ as $k$-vector spaces. The derivation identity makes $\delta$ vanish on ${\mathfrak{m}}^2$, and restriction gives a linear form on $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$. Conversely, for $\mathrm{\ell }\in {\mathrm{Hom}}_k(\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2,k)$, define $\delta (c+u)=\mathrm{\ell }(u+{\mathfrak{m}}^2)$ for $c\in k$, $u\in \mathfrak{m}$. For $c+u,c^{\mathrm{\prime }}+u^{\mathrm{\prime }}\in A$, the product has $\mathfrak{m}$-part $c{u}^{\mathrm{\prime }}+c^{\mathrm{\prime }}u+u{u}^{\mathrm{\prime }}$, and $u{u}^{\mathrm{\prime }}\in {\mathfrak{m}}^2$; hence this formula satisfies the Leibniz rule. It is inverse to restriction.

Passing to the stalk gives the claimed intrinsic tangent and local derivation formulation. By [F9], ${\mathcal{O}}_{X,x}\cong A_{\mathfrak{m}}$ with maximal ideal ${\mathfrak{m}}_x=\mathfrak{m}{A}_{\mathfrak{m}}$. The rational-point localization isomorphism [F4] identifies $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ with ${\mathfrak{m}}_x\mathrm{/}{\mathfrak{m}}_x^2$, so their $k$-linear duals agree; [F1] identifies the latter dual with $T_{x}X$. Also every $s\in A\setminus \mathfrak{m}$ has $\phi (s)=\bar{s}+ϵ\delta (s)$ with $\bar{s}\mathrm{\ne }0$, which is a unit in $D$ with inverse ${\bar{s}}^{-1}-{\bar{s}}^{-2}\delta (s)ϵ$. Thus $\phi$ extends uniquely to a local $k$-algebra map ${\mathcal{O}}_{X,x}=A_{\mathfrak{m}}\to D$. Conversely, every local $k$-algebra map $\alpha :{\mathcal{O}}_{X,x}\to D$ has a unique form $\alpha (u)=u(x)+ϵd(u)$, where multiplicativity makes $d$ a $k$-derivation through the residue action. Every such derivation defines a local map by this formula, since units have nonzero residue. Applying the decomposition ${\mathcal{O}}_{X,x}=k\oplus {\mathfrak{m}}_x$ as in step 4.1 gives ${\mathrm{Der}}_k({\mathcal{O}}_{X,x},k)\cong {\mathrm{Hom}}_k({\mathfrak{m}}_x\mathrm{/}{\mathfrak{m}}_x^2,k)$. The localization, extension, and restriction maps commute on smaller affine neighborhoods, so the identifications are independent of $U$ and natural. Scaling $ϵ$ by $c\in k$ scales the derivation and tangent vector by $c$.