
Lemma 10.113.5. 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. The following numbers are equal

1. $\dim _ x(X)$,

2. $\max \dim (Z)$ where the maximum is over those irreducible components $Z$ of $X$ passing through $x$, and

3. $\min \dim (S_{\mathfrak m})$ where the minimum is over maximal ideals $\mathfrak m$ with $\mathfrak p \subset \mathfrak m$.

Proof. Let $X = \bigcup _{i \in I} Z_ i$ be the decomposition of $X$ into its irreducible components. There are finitely many of them (see Lemmas 10.30.3 and 10.30.5). Let $I' = \{ i \mid x \in Z_ i\}$, and let $T = \bigcup _{i \not\in I'} Z_ i$. Then $U = X \setminus T$ is an open subset of $X$ containing the point $x$. The number (2) is $\max _{i \in I'} \dim (Z_ i)$. For any open $W \subset U$ with $x \in W$ the irreducible components of $W$ are the irreducible sets $W_ i = Z_ i \cap W$ for $i \in I'$ and $x$ is contained in each of these. Note that each $W_ i$, $i \in I'$ contains a closed point because $X$ is Jacobson, see Section 10.34. Since $W_ i \subset Z_ i$ we have $\dim (W_ i) \leq \dim (Z_ i)$. The existence of a closed point implies, via Lemma 10.113.4, that there is a chain of irreducible closed subsets of length equal to $\dim (Z_ i)$ in the open $W_ i$. Thus $\dim (W_ i) = \dim (Z_ i)$ for any $i \in I'$. Hence $\dim (W)$ is equal to the number (2). This proves that (1) $=$ (2).

Let $\mathfrak m \supset \mathfrak p$ be any maximal ideal containing $\mathfrak p$. Let $x_0 \in X$ be the corresponding point. First of all, $x_0$ is contained in all the irreducible components $Z_ i$, $i \in I'$. Let $\mathfrak q_ i$ denote the minimal primes of $S$ corresponding to the irreducible components $Z_ i$. For each $i$ such that $x_0 \in Z_ i$ (which is equivalent to $\mathfrak m \supset \mathfrak q_ i$) we have a surjection

$S_{\mathfrak m} \longrightarrow S_\mathfrak m/\mathfrak q_ i S_\mathfrak m =(S/\mathfrak q_ i)_{\mathfrak m}$

Moreover, the primes $\mathfrak q_ i S_\mathfrak m$ so obtained exhaust the minimal primes of the Noetherian local ring $S_{\mathfrak m}$, see Lemma 10.25.3. We conclude, using Lemma 10.113.4, that the dimension of $S_{\mathfrak m}$ is the maximum of the dimensions of the $Z_ i$ passing through $x_0$. To finish the proof of the lemma it suffices to show that we can choose $x_0$ such that $x_0 \in Z_ i \Rightarrow i \in I'$. Because $S$ is Jacobson (as we saw above) it is enough to show that $V(\mathfrak p) \setminus T$ (with $T$ as above) is nonempty. And this is clear since it contains the point $x$ (i.e. $\mathfrak p$). $\square$

Comment #712 by Keenan Kidwell on

I have two comments. First, in the third line of the proof, the union should be over $i\notin I^\prime$ instead of $x\notin I^\prime$. Second, I don't understand how the 00OS implies that $\dim(W_i)=\dim(Z_i)$, though perhaps I'm just missing something basic about the notion of dimension. By the Jacobson-ness of $S$, there is a closed point $\mathfrak{m}$ of $X$ in $W\cap Z_i=W_i$, and for such a point we have $\dim(Z_i)=\dim(S_\mathfrak{m}/\mathfrak{q}_iS_\mathfrak{m})$ in the notation from later in the proof. But why is this $\dim(W_i)$?

Comment #713 by on

OK, I fixed the index. Because $W_i \subset Z_i$ we see that $\dim(W_i) \leq \dim(Z_i)$. The existence of a closed point implies, via Lemma 10.113.4, that there is a chain of irreducible closed of length equal to $\dim(Z_i)$ in $W_i$. So $\dim(W_i) = \dim(Z_i)$. I've also added a couple of lines explaining this in this commit. Thanks!

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