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Dec 18, 2021 at 18:02 comment added Johan Krull hauptidealsatz is always a fun result to use.
Dec 18, 2021 at 14:00 comment added Jason Starr @TakagiBenseki You did not write that your directed system of rings is the system of all finite type $\mathbb{Z}$-subalgebras, so Johan is free to choose a different directed system. At any rate, the directed system of finite type $\mathbb{Z}$-subalgebras that contain $t$ is cofinal in the directed system of all finite type $\mathbb{Z}$-subalgebras.
Dec 18, 2021 at 8:43 comment added Takagi Benseki @Johan, thank you very much, Johan. I now feel that descending codim is not easy because of your example. For the ring $A$ you mentioned, I consider it as a valuation ring of rank two (it is a normal local domain). But I still don't know how you make sure that all $\mathbb{Z}$-finite type subrings of $A$ satisfy that $V(t)$ has codim=1. I appreciate it if you could explain a little or give a reference.
Dec 17, 2021 at 23:22 comment added Johan No this isn't true: take a local domain with exactly three prime ideals $(0) \subset \mathfrak p \subset \mathfrak m$. (By Hochster such a ring exists.) Take $t \in \mathfrak m$, $t \not \in \mathfrak p$. Then write $A = \colim A_i$ as a filtered colimit with $A_i \subset A$ of finite type over $\mathbf{Z}$ with $t \in A_i$ for all $i$. Then $V(t)$ has codim 1 in the spectrum of $A_i$ but $V(t)$ has codimen $2$ in the spectrum of $A$.
Dec 17, 2021 at 23:11 comment added Johan Just making sure: $\text{codim}(Z, X) = \inf \text{codim}(Z', X)$ where $Z'$ runs over the irreducible closed subschemes of $Z$.
S Dec 17, 2021 at 15:24 review First questions
Dec 17, 2021 at 16:08
S Dec 17, 2021 at 15:24 history asked Takagi Benseki CC BY-SA 4.0