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If we have a commutative ring with 1, what can we say about $\operatorname{dim} S^{-1}A$ for some multiplicative subset $S$, or more specifically, what happens if $S = A \setminus \mathfrak{p}$ for a prime ideal $\mathfrak{p}$?

Do the dimensions of the localization generally coincide for distinct prime or maximal ideals? Thanks!

user26857
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argon
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1 Answers1

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What is true is that $\dim S^{-1}A\le \dim A$ because $\mathrm{Spec}(S^{-1}A)$ is homeomorphic to a subset of $\mathrm{Spec}(A)$.

For $A_{\mathfrak p}$, the dimension varies: if $\mathfrak p$ is a minimal prime, then the localization has dimension $0$. If $\mathfrak p$ is not maximal and $\dim A$ is finite, then $\dim A_{\mathfrak p}<\dim A$. On the other hand, $\dim A$ is the supremum of the $\dim A_{\mathfrak m}$ when $\mathfrak m$ runs the maximal ideals of $A$. If $A$ is an integral finitely generated algebra over a field, then $\dim A=\dim A_{\mathfrak m}$ for any maximal ideal.

  • Isn't $\dim A = \dim A_\mathfrak{m}$ always the case? What would be a counterexample? – Anakhand Nov 23 '23 at 12:12
  • $A = A_1 \times A_2$ is a counterexample, if the $A_i$ have different dimensions, say $k_1 < k_2$ because if $\mathfrak{p}1$ has height equal to the dimension of $A_1$ then $\mathfrak{m} = \mathfrak{p}_1 \times A_2$ is maximal in $A$, but $\dim A\mathfrak{m} = k_1 < k_2 = \dim A$. – David Goldberg May 16 '25 at 00:06