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Tag 00P1

Chapter 10: Commutative Algebra > Section 10.115: Dimension of finite type algebras over fields, reprise

Lemma 10.115.3. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $X = \mathop{\rm Spec}(S)$. 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 26700–26711 (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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