Tangent dimension bounds local dimension

Choose an affine open neighborhood $U=\mathrm{Spec}A$ of $x$ with $A$ Noetherian by [F2, F3]. Since $U$ is an open subscheme with the restricted structure sheaf [F4], neighborhoods contained in $U$ are cofinal among neighborhoods of $x$, so ${\mathcal{O}}_{X,x}={\mathcal{O}}_{U,x}$. The point $x$ corresponds to a prime $\mathfrak{p}\subset A$ by [F5], and [F6, F7] identify $R:={\mathcal{O}}_{X,x}$ with $A_{\mathfrak{p}}$. By [F8], $R$ is a nonzero local ring with maximal ideal $\mathfrak{m}=\mathfrak{p}{A}_{\mathfrak{p}}$; [F9] identifies its residue field with $\kappa (x)$.

The local ring $R=A_{\mathfrak{p}}$ is Noetherian. Let $J$ be any ideal of $A_{\mathfrak{p}}$ and contract it to $I:=\{a\in A:a\mathrm{/}1\in J\}$. Since $A$ is Noetherian, [F10] gives generators $a_1,\dots ,a_n$ of $I$. If $a\mathrm{/}s\in J$ with $s\mathrm{\notin }\mathfrak{p}$, then $a\mathrm{/}1=(s\mathrm{/}1)(a\mathrm{/}s)\in J$, so $a\in I$ and $a={\sum }_i{c}_i{a}_i$. Therefore $a\mathrm{/}s={\sum }_i(c_i\mathrm{/}s)(a_i\mathrm{/}1)$, while each $a_i\mathrm{/}1$ lies in $J$. Thus the images $a_i\mathrm{/}1$ generate $J$. As this holds for every $J$, [F10] implies that $R$ is Noetherian.

The tangent dimension equals the embedding dimension of $R$. The maximal ideal $\mathfrak{m}$ is finitely generated because $R$ is Noetherian, so $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ is a finite-dimensional vector space over $\kappa (x)$. By [F11] this quotient is $C_{x}X$, and by [F12] $T_{x}X$ is its $\kappa (x)$-linear dual; a finite-dimensional vector space and its dual have equal dimension. By [F13], this common dimension is $\mathrm{edim}R$.

Now [F8] and step 2.1 make $R={\mathcal{O}}_{X,x}$ a nonzero Noetherian local ring, so the AC-dependent bound [F14] applies. Together with step 3.1 it gives $\dim {\mathcal{O}}_{X,x}\le \mathrm{edim}R={\dim }_{\kappa (x)}{T}_{x}X$. AC is used here through [F14], whose height-theorem input requires it; it is an explicit assumption, not a consequence of finite choice.

In the stated classical closed-point specialization, [F15] gives $\dim {\mathcal{O}}_{X,x}={\max }_{x\in X_i}\dim X_i={\dim }_{x}X$. Substituting this equality into step 4.1 proves $\dim T_{x}X\ge {\dim }_{x}X$. This local-dimension supplier also assumes AC, already declared in the statement.

At $X=\mathrm{Spec}k$, the only prime is $(0)$, so the unique local ring is $k$, its maximal ideal is zero, and [F19] gives local dimension zero; [F11, F12] give tangent dimension zero. The nonreduced dual-numbers scheme $D_k=\mathrm{Spec}(k[ϵ]\mathrm{/}({ϵ}^2))$ shows that the inequality may be strict. Its ring is a two-dimensional $k$-vector space, so every ideal, as a subspace, has a finite basis that generates it as an ideal; hence $D_k$ is Noetherian. Every prime contains $ϵ$ because ${ϵ}^2=0$ by [F17]. The ideal $(ϵ)$ is proper because its elements are multiples of $ϵ$ and cannot equal $1$. Any proper ideal strictly containing $(ϵ)$ would contain $a+bϵ$ with $a\mathrm{\ne }0$, a unit with inverse $a^{-1}-a^{-2}bϵ$. Thus $(ϵ)$ is maximal by [F18] and, since every prime contains it, it is the unique prime. By [F5], $D_k$ has one point. Its local ring $D_{k,(ϵ)}$ is $D_k$ since every denominator outside $(ϵ)$ is a unit; by [F6, F7, F8, F19] its local dimension is zero. The residue field is $D_k\mathrm{/}(ϵ)\cong k$ by [F9], while the maximal ideal squares to zero, so $(ϵ)\mathrm{/}({ϵ}^2)$ is one-dimensional over the residue field. By [F11, F12], the tangent dimension is one.