Lemma 10.116.3. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak p \subset S$ be a prime ideal, and let $x \in X$ be the corresponding point. Then we have

**Proof.**
By Lemma 10.116.1 we know that $r = \text{trdeg}_ k\ \kappa (\mathfrak p)$ is equal to the dimension of $V(\mathfrak p)$. Pick any maximal chain of primes $\mathfrak p \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ r$ starting with $\mathfrak p$ in $S$. This has length $r$ by Lemma 10.114.4. Let $\mathfrak q_ j$, $j \in J$ be the minimal primes of $S$ which are contained in $\mathfrak p$. These correspond $1-1$ to minimal primes in $S_{\mathfrak p}$ via the rule $\mathfrak q_ j \mapsto \mathfrak q_ jS_{\mathfrak p}$. By Lemma 10.114.5 we know that $\dim _ x(X)$ is equal to the maximum of the dimensions of the rings $S/\mathfrak q_ j$. For each $j$ pick a maximal chain of primes $\mathfrak q_ j \subset \mathfrak p'_1 \subset \ldots \subset \mathfrak p'_{s(j)} = \mathfrak p$. Then $\dim (S_{\mathfrak p}) = \max _{j \in J} s(j)$. Now, each chain

is a maximal chain in $S/\mathfrak q_ j$, and by what was said before we have $\dim _ x(X) = \max _{j \in J} r + s(j)$. The lemma follows. $\square$

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## Comments (1)

Comment #715 by Keenan Kidwell on

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