Lemma 10.61.2. A Noetherian ring with finitely many primes has dimension $\leq 1$.
Proof. Let $R$ be a Noetherian ring with finitely many primes. If $R$ is a local domain, then the lemma follows from Lemma 10.61.1. If $R$ is a domain, then $R_\mathfrak m$ has dimension $\leq 1$ for all maximal ideals $\mathfrak m$ by the local case. Hence $\dim (R) \leq 1$ by Lemma 10.60.4. If $R$ is general, then $\dim (R/\mathfrak q) \leq 1$ for every minimal prime $\mathfrak q$ of $R$. Since every prime contains a minimal prime (Lemma 10.17.2), this implies $\dim (R) \leq 1$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)