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Tag 0ALW

Chapter 10: Commutative Algebra > Section 10.60: Applications of dimension theory

Lemma 10.60.3. Let $S$ be a nonzero finite type algebra over a field $k$. Then $\dim(S) = 0$ if and only if $S$ has finitely many primes.

Proof. Recall that $\mathop{\rm Spec}(S)$ is sober, Noetherian, and Jacobson, see Lemmas 10.25.2, 10.30.5, 10.34.2, and 10.34.4. If it has dimension $0$, then every point defines an irreducible component and there are only a finite number of irreducible components (Topology, Lemma 5.9.2). Conversely, if $\mathop{\rm Spec}(S)$ is finite, then it is discrete by Topology, Lemma 5.18.6 and hence the dimension is $0$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 14145–14150 (see updates for more information).

    \begin{lemma}
    \label{lemma-finite-type-algebra-finite-nr-primes}
    Let $S$ be a nonzero finite type algebra over a field $k$.
    Then $\dim(S) = 0$ if and only if $S$ has
    finitely many primes.
    \end{lemma}
    
    \begin{proof}
    Recall that $\Spec(S)$ is sober, Noetherian, and Jacobson, see
    Lemmas \ref{lemma-spec-spectral}, \ref{lemma-Noetherian-topology},
    \ref{lemma-finite-type-field-Jacobson}, and \ref{lemma-jacobson}.
    If it has dimension $0$, then every point defines an
    irreducible component and there are only a finite number
    of irreducible components (Topology, Lemma \ref{topology-lemma-Noetherian}).
    Conversely, if $\Spec(S)$ is finite, then it is discrete
    by Topology, Lemma \ref{topology-lemma-finite-jacobson}
    and hence the dimension is $0$.
    \end{proof}

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