Local dimension for a reducible classical algebraic set

Choose an affine open neighborhood $U\subseteq X$ of $x$, put $A=k[U]$, let $\mathfrak{m}\subset A$ be the maximal ideal of functions vanishing at $x$, and write $R={\mathcal{O}}_{X,x}$. By [F4], $R\cong A_{\mathfrak{m}}$. The finite component decomposition of $X$ restricts to a finite decomposition of $U$ by its irreducible components $U_j=X_j\cap U$; precisely those $U_j$ containing $x$ come from the global components $X_j$ containing $x$.

For each component $U_j$, let ${\mathfrak{p}}_j=I_U(U_j)\subseteq A$. The Nullstellensatz correspondence makes ${\mathfrak{p}}_j$ prime and reverses inclusions of closed subsets. The localization correspondence identifies the primes of $R=A_{\mathfrak{m}}$ with the primes $\mathfrak{p}\subseteq \mathfrak{m}$ of $A$, preserving strict chains. In particular, if $x\in U_j$, then ${\mathfrak{q}}_j:={\mathfrak{p}}_j{A}_{\mathfrak{m}}$ is a prime of $R$.

Consider any strict prime chain ${\mathfrak{q}}_0⊊{\mathfrak{q}}_1⊊\cdots ⊊{\mathfrak{q}}_r$ in $R$, and contract it to ${\mathfrak{p}}_0⊊{\mathfrak{p}}_1⊊\cdots ⊊{\mathfrak{p}}_r$ in $A$ using [F5]. The irreducible closed subset $V_U({\mathfrak{p}}_0)$ contains $x$, since ${\mathfrak{p}}_0\subseteq \mathfrak{m}$. Because $U$ is a finite union of its irreducible components, irreducibility forces $V_U({\mathfrak{p}}_0)\subseteq U_j$ for some $j$. Thus $x\in U_j$ and ${\mathfrak{p}}_j\subseteq {\mathfrak{p}}_0$, so ${\mathfrak{q}}_j\subseteq {\mathfrak{q}}_0$. The chain therefore gives a chain of length $r$ in $R\mathrm{/}{\mathfrak{q}}_j$.

The quotient-localization isomorphism gives $R\mathrm{/}{\mathfrak{q}}_j\cong (A\mathrm{/}{\mathfrak{p}}_j{)}_{\mathfrak{m}\mathrm{/}{\mathfrak{p}}_j}$, which is the local ring ${\mathcal{O}}_{U_j,x}={\mathcal{O}}_{X_j,x}$ because $U_j=X_j\cap U$ is an open neighborhood of $x$ in $X_j$. Hence [F6] gives $\dim (R\mathrm{/}{\mathfrak{q}}_j)=\dim X_j$. Step 3.1 now bounds every chain length in $R$ by ${\max }_{x\in X_i}\dim X_i$.

Conversely, for every global component $X_i$ containing $x$, its ${\mathfrak{q}}_i$ is prime in $R$ by step 2.1 and $\dim (R\mathrm{/}{\mathfrak{q}}_i)=\dim X_i$ by step 4.1. Every prime chain in $R\mathrm{/}{\mathfrak{q}}_i$ lifts to a prime chain in $R$, so $\dim R\ge \dim X_i$. Taking the maximum gives the reverse inequality.

Steps 3.1–5.1 prove the asserted equality. The argument uses AC only through the explicitly AC-dependent component, affine-correspondence, local-ring, and irreducible local-dimension suppliers; after their finite component and prime correspondences are in hand, the chain comparison makes no further choice.