# The Stacks Project

## Tag 00P1

Lemma 10.115.3. 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 $$\dim_x(X) = \dim(S_{\mathfrak p}) + \text{trdeg}_k~\kappa(\mathfrak p).$$

Proof. By Lemma 10.115.1 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 10.113.4. 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 10.113.5 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 $$\mathfrak q_i \subset \mathfrak p'_1 \subset \ldots \subset \mathfrak p'_{s(j)} = \mathfrak p \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_r$$ 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. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 26834–26845 (see updates for more information).

\begin{lemma}
\label{lemma-dimension-at-a-point-finite-type-field}
Let $k$ be a field.
Let $S$ be a finite type $k$ algebra.
Let $X = \Spec(S)$.
Let $\mathfrak p \subset S$ be a prime ideal,
and let $x \in X$ be the corresponding point.
Then we have
$$\dim_x(X) = \dim(S_{\mathfrak p}) + \text{trdeg}_k\ \kappa(\mathfrak p).$$
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-dimension-prime-polynomial-ring} 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 \ref{lemma-dimension-spell-it-out}.
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 \ref{lemma-dimension-at-a-point-finite-type-over-field}
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
$$\mathfrak q_i \subset \mathfrak p'_1 \subset \ldots \subset \mathfrak p'_{s(j)} = \mathfrak p \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_r$$
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.
\end{proof}

## Comments (1)

Comment #715 by Keenan Kidwell on June 20, 2014 a 1:34 pm UTC

The notation $\mathfrak{p}_i$ is being used for two different sets of primes. Maybe put a prime on the second set?

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