A tangent direction is realized by a local smooth curve

Setup and dimensions. By [F2] and [F3] the closed subvariety $X\subseteq {\mathbf{A}}_k^n$ is the affine model with its reduced finite-type structure, $k[X]=P\mathrm{/}I(X)$ is reduced and generated by the finitely many classes ${\bar{x}}_1,\dots ,{\bar{x}}_n$, and the closed points of $X$ have residue field $k$; thus $x$ is a $k$-rational point with coordinates $a=(a_1,\dots ,a_n)$. Smoothness at $x$ means that the structure morphism is locally standard smooth at $x$ [F6], so some principal shrinking around $x$ has a standard smooth presentation [F7]; applying [F8] with $R=k$, $\mathfrak{p}=(0)$ and $K=k$ shows that ${\mathcal{O}}_{X,x}$ is a regular local ring, so $x$ is a regular point and ${\dim }_k{T}_{x}X=\dim {\mathcal{O}}_{X,x}$ [F9], while $\dim {\mathcal{O}}_{X,x}={\max }_{x\in X_i}\dim X_i={\dim }_{x}X=d$ by [F5] with [F4]. A based dual-number point of ${\mathbf{A}}_k^n$ at $a$ is, by the affine anti-equivalence [F15] and the polynomial universal property [F12], a map $P\to k[ϵ]\mathrm{/}({ϵ}^2)$ whose reduction modulo $ϵ$ sends $x_i$ to $a_i$. Its variable images are therefore uniquely $x_i\mapsto a_i+ϵw_i$ for a vector $w\in k^n$, and conversely every such vector determines a based point. Under the $k$-linear dual-number bijection [F11], the tangent space $T_a{\mathbf{A}}_k^n$ is thus $k^n$ in these coordinates, and the composite with $X\hookrightarrow {\mathbf{A}}_k^n$ sends a tangent vector of $X$ to the unique $w\in k^n$ whose map factors through $X$; write $T=\{w\in k^n:\text{the based dual-number point }{X}_i\mapsto a_i+ϵw_i\text{ lies in }X\}\subseteq k^n\mathrm{.}$ Since the correspondence is $k$-linear and bijective, $T$ is a linear subspace [F21] with ${\dim }_{k}T=d$, so $T\mathrm{\ne }0$ because $d\ge 1$; let $v\in T$ denote the image of the given tangent vector $v\in T_{x}X$, and note that all tangent-space identifications below are made with these coordinates, so that $T=T_{x}X$ as subspaces of $k^n$.

Reduction does not change the tangent subspace. Let $Y\subseteq {\mathbf{A}}_k^n$ be a closed subscheme of finite type over $k$ with $x\in Y(k)$, say $Y=\mathrm{Spec}(P\mathrm{/}J)$ for its ideal $J$ [F15], and suppose that the local ring ${\mathcal{O}}_{Y,x}$ is reduced. Then the closed immersion $Y_{\mathrm{r}\mathrm{e}\mathrm{d}}\hookrightarrow Y$ has ${\mathcal{O}}_{Y_{\mathrm{r}\mathrm{e}\mathrm{d}},x}={\mathcal{O}}_{Y,x}\mathrm{/}\mathrm{nil}({\mathcal{O}}_{Y,x})={\mathcal{O}}_{Y,x}$ [F14], so it induces an isomorphism of local rings at $x$, hence an isomorphism of cotangent spaces ${\mathfrak{m}}_x\mathrm{/}{\mathfrak{m}}_x^2$ [F10] and, dualizing, an isomorphism $T_x{Y}_{\mathrm{r}\mathrm{e}\mathrm{d}}\to T_{x}Y$ [F13]; since the composite $Y_{\mathrm{r}\mathrm{e}\mathrm{d}}\to Y\to {\mathbf{A}}_k^n$ is the closed immersion $Y_{\mathrm{r}\mathrm{e}\mathrm{d}}\hookrightarrow {\mathbf{A}}_k^n$, the chain rule of [F13] shows that the identification of $T_x{Y}_{\mathrm{r}\mathrm{e}\mathrm{d}}$ with a subspace of $T_x{\mathbf{A}}_k^n=k^n$ [F11, F12] agrees with the composite of the identifications for $Y_{\mathrm{r}\mathrm{e}\mathrm{d}}\to Y$ and $Y\to {\mathbf{A}}_k^n$, and since the first of these is an isomorphism the two subspaces of $k^n$ coincide: $T_x{Y}_{\mathrm{r}\mathrm{e}\mathrm{d}}=T_{x}Y$. This equality is the form in which the invariance under reduction is used below, and it also shows that a reduced closed subscheme with regular local ring at $x$ is smooth at $x$: if ${\mathcal{O}}_{Y,x}$ is regular, then $x$ is a regular point of $Y$, and $Y=\mathrm{Spec}(P\mathrm{/}J)$ is locally standard smooth at $x$ by [F19] because $k$ is perfect [F18]; by [F6] that is smoothness of $Y$ at $x$.

The linear forms. Assume first that $v\mathrm{\ne }0$; since $v\mathrm{\ne }0$, there is a least index $s\in \{1,\dots ,n\}$ with $v_s\mathrm{\ne }0$ in the relabelled coordinates. For every index $j\mathrm{\ne }s$ define ${\mathrm{\ell }}_j(u)=u_j-\frac{v_j}{v_s}{u}_s(u\in k^n),$ a $k$-linear functional on $k^n$ [F20, F21] satisfying ${\mathrm{\ell }}_j(v)=v_j-\frac{v_j}{v_s}{v}_s=0$; consequently $kv\subseteq \ker {\mathrm{\ell }}_j$ for every $j\mathrm{\ne }s$, and conversely if $u\in T$ satisfies ${\mathrm{\ell }}_j(u)=0$ for all $j\mathrm{\ne }s$, then $u_j=\frac{v_j}{v_s}{u}_s$ for all $j\mathrm{\ne }s$ and therefore $u=\frac{u_s}{v_s}v$, so $T\cap {\bigcap }_{j\mathrm{\ne }s}\ker {\mathrm{\ell }}_j=kv\mathrm{.}$ For each $j\mathrm{\ne }s$ put $h_j=x_j-\frac{v_j}{v_s}{x}_s-(a_j-\frac{v_j}{v_s}{a}_s)\in P$ [F24]; then $h_j(a)=0$ and $h_j$ is affine-linear with linear part ${\mathrm{\ell }}_j$, because $h_j(a+ϵw)=h_j(a)+ϵ{\mathrm{\ell }}_j(w)$ for every $w\in k^n$, and ${\mathrm{\ell }}_j\mathrm{\ne }0$ because ${\mathrm{\ell }}_j(v)=0$ and ${\mathrm{\ell }}_j(e_j)=1$ for $j\mathrm{\ne }s$.

The active indices. Recursively for $j=1,\dots ,n$, put $T^{(0)}:=T$ and $T^{(j)}:=T^{(j-1)}\cap \ker {\mathrm{\ell }}_j\text{ if }j\mathrm{\ne }s\text{ and }{\mathrm{\ell }}_j\mathrm{\ne }0\text{ on }{T}^{(j-1)},T^{(j)}:=T^{(j-1)}\text{ otherwise},$ the first case being called active at $j$; let $J$ be the finite set of active indices, listed in increasing order as $J=\{j_1<\cdots <j_m\}$. At an active index the restriction ${\mathrm{\ell }}_j{\mathrm{\mid }}_{T^{(j-1)}}$ is a nonzero linear map to $k=k^1$, so its image has dimension $1$ [F20] and rank-nullity [F22] gives ${\dim }_k{T}^{(j)}={\dim }_k{T}^{(j-1)}-1$, while at an inactive index $T^{(j)}=T^{(j-1)}$; hence ${\dim }_k{T}^{(j)}=d-\mathrm{\#}\{i:j_i\le j\}$ for every $j$, and ${\dim }_k{T}^{(n)}=d-m$. Moreover $T^{(n)}=kv$: on the one hand every $T^{(j)}$ contains $v$ because $v\in T$ and ${\mathrm{\ell }}_{j^{\mathrm{\prime }}}(v)=0$ for all $j^{\mathrm{\prime }}\mathrm{\ne }s$ [step 1.3], so $kv\subseteq T^{(n)}$; on the other hand if $u\in T^{(n)}$ and $j\mathrm{\ne }s$, then either $j$ is active, in which case $T^{(n)}\subseteq T^{(j)}\subseteq \ker {\mathrm{\ell }}_j$, or $j$ is inactive, in which case ${\mathrm{\ell }}_j$ vanishes on $T^{(j-1)}\supseteq T^{(n)}$; so $u\in T\cap {\bigcap }_{j\mathrm{\ne }s}\ker {\mathrm{\ell }}_j=kv$ by step 1.3, giving $T^{(n)}\subseteq kv$. Therefore ${\dim }_k{T}^{(n)}=1$ and $m=d-1$.

The induction on the active slices. Put $Y_0:=X$, and for $i=1,\dots ,m$ define $Z_i:=Y_{i-1}{\times }_{{\mathbf{A}}_k^n}{H}_i$, where $H_i=V(h_{j_i})$ is the hyperplane cut out by the affine-linear polynomial of step 1.3 for the index $j_i$, and put $Y_i:=(Z_i{)}_{\mathrm{r}\mathrm{e}\mathrm{d}}$ [F14, F15, F16]. Each $Z_i$ is a closed subscheme of $Y_{i-1}$ (base change of the closed immersion $H_i\hookrightarrow {\mathbf{A}}_k^n$) and each $Y_i$ is a reduced closed subvariety of $X$ containing $x$, because $x\in X=Y_0$ and each $h_{j_i}$ vanishes at $x$ [step 1.3] so the $k$-point $x$ lifts to $Z_i$ and to $Y_i$. The induction claim is: $T_x{Y}_i=T^{(j_i)}$ as subspaces of $k^n$, ${\dim }_k{T}_x{Y}_i=d-i$, ${\mathcal{O}}_{Y_i,x}$ is a regular local ring of dimension $d-i$, ${\dim }_x{Y}_i=d-i$, and $Y_i$ is smooth at $x$. For $i=0$ this is step 1.1 together with the given smoothness. Assume the claim for $i-1$ with $1\le i\le m=d-1$; then $d-i+1\ge 2\ge 1$, so the slice lemma [F17] applies to the variety $Y_{i-1}$, smooth at $x$ by the induction claim, with the affine-linear form $h_{j_i}$ of linear part ${\mathrm{\ell }}_{j_i}$: the index $j_i$ is active, which means ${\mathrm{\ell }}_{j_i}$ is nonzero on $T^{(j_i-1)}$, and $T^{(j_i-1)}=T^{(j_{i-1})}=T_x{Y}_{i-1}$ (for $i=1$, $T^{(j_1-1)}=T^{(0)}=T_{x}X$ because all indices below $j_1$ are inactive), so the transversality hypothesis holds. The slice lemma gives that $Z_i$ is smooth at $x$ over $k$ with ${\mathcal{O}}_{Z_i,x}$ a regular local ring of dimension $(d-i+1)-1=d-i$ and $T_x{Z}_i=\ker (d_x(h_{j_i}{\mathrm{\mid }}_{Y_{i-1}}))=\{u\in T_x{Y}_{i-1}:{\mathrm{\ell }}_{j_i}(u)=0\}=T^{(j_i-1)}\cap \ker {\mathrm{\ell }}_{j_i}=T^{(j_i)}\mathrm{.}$ Since ${\mathcal{O}}_{Z_i,x}$ is regular, it is reduced by [F25], so step 1.2 applied to $Y=Z_i$ gives $T_x{Y}_i=T_x(Z_i{)}_{\mathrm{r}\mathrm{e}\mathrm{d}}=T_x{Z}_i=T^{(j_i)}$ and ${\mathcal{O}}_{Y_i,x}={\mathcal{O}}_{Z_i,x}$, a regular local ring of dimension $d-i$; moreover $Y_i$ is reduced, so [F5] with [F4] gives ${\dim }_x{Y}_i=\dim {\mathcal{O}}_{Y_i,x}=d-i$, and $Y_i$ is smooth at $x$ by the second part of step 1.2. This proves the claim for $i$ and completes the induction.

The case $v=0$. If $v=0$, then $T\mathrm{\ne }0$ by step 1.1, so choose any nonzero $u\in T$ and apply the construction of steps 1.3, 2.1 and 3.1 to $u$ in place of $v$; it yields a curve $C$ with $T_{x}C$ mapped isomorphically onto the line $ku$, and $v=0\in ku=d_x{\iota }_C(T_{x}C)$.

Conclusion for nonzero $v$. Let $v\mathrm{\ne }0$ and put $C:=Y_m=Y_{d-1}$ with the notation of step 3.1; then $C\subseteq X$ is a reduced closed subvariety with $x\in C$, smooth at $x$, and step 3.1 at $i=m=d-1$ gives $T_{x}C=T^{(n)}=kv$ (if $m>0$, the last active slice has $T_{x}C=T^{(j_m)}=T^{(n)}$; if $m=0$, no slice occurs and $T_{x}C=T=T^{(n)}$) [step 2.1] and ${\dim }_{x}C=1$. Under the closed immersion $C\hookrightarrow X$ the differential $d_x{\iota }_C$ is injective and the diagram with the two ambient identifications commutes [F13, step 1.1, step 1.2], so $d_x{\iota }_C$ carries $T_{x}C$ isomorphically onto the subspace of $T_{x}X$ whose ambient image is $kv$, namely onto $kv$ itself; in particular $v\in d_x{\iota }_C(T_{x}C)$.

Boundaries. If $d=1$ then $m=0$ [step 2.1] and $C:=Y_0=X$ works: $X$ is reduced with $x\in X$, smooth at $x$ by hypothesis with ${\dim }_{x}X=1$, and $T_{x}X=T$ is one-dimensional, so $T=kv$ for the nonzero $v$ [step 1.1]. If $d=0$ no such curve exists: a reduced closed subvariety $C\subseteq X$ with $x\in C$ and ${\dim }_{x}C=1$ has, by [F5] with [F4], an irreducible component of $C$ containing $x$ of dimension $1$, which is an irreducible closed subset of $X$ passing through $x$ and hence is contained in an irreducible component of $X$ containing $x$ [F23], so that ${\dim }_{x}X\ge 1$ by [F4], a contradiction; this is why the hypothesis $d\ge 1$ is stated. The construction divides only by $v_s\mathrm{\ne }0$ [step 1.3], so no characteristic hypothesis is needed and the affine-linear forms $h_j$ are available in every characteristic; the ambient dimension $n\ge 1$ may equal $1$, in which case $d\le 1$ and the case $d=1$ above applies; $X$ may be reducible, and the curve $C$ produced is closed in $X$, hence locally closed. The Axiom of Choice enters the statement through [F1] and is used only through the suppliers that assume it, namely [F4], [F5], [F6], [F8], [F17], [F18 as used through F19] and [F23], [F25], each cited at the step that uses it; the explicit linear forms ${\mathrm{\ell }}_j$, the finite recursion defining the active set, the polynomials $h_j$, the enumerations and all tangent identifications involve no selection, and [F20], [F22] and the reductions of [F14] are choice-free.