Example of a locally Noetherian space $X$ and a point $x$ such that $\dim_xX=\infty$

Question

There is a well-known example that a Noetherian ring $A$ with infinite Krull dimension due to Nagata, which gives a Noetherian space $\operatorname{Spec}A$ with infinite Krull dimension.

Given the definition $\dim_xX=\inf_{x\in U}\dim U$ where $U$ takes over all open neighborhood of $x$, then does there exist a locally Noetherian space $X$(not necessary a scheme) with a point $x\in X$ such that $\dim_xX=\infty$?

Edit. User leoli1 has given an example, could someone give an example of scheme?

Answer 1

With your definition of $\dim_x X$, Nagata's example satisfies $\dim_x X=\infty$ at every point. Indeed, every nonempty open subset of $X$ still contains all but finitely many of the maximal ideals (since any element of the ring is only contained in finitely many of the maximal ideals), so it contains arbitrarily long chains of prime ideals.

Let me also mention that you can get a scheme (though not a locally Noetherian scheme) whose underlying topological space is the space $\mathbb{N}\cup\{\infty\}$ described in leoli1's answer. For instance, you can just take a valuation ring with appropriate value group. Consider the ordered ring $\mathbb{Z}[t]$ where $t$ is infinitely large. Let $A$ be a valuation ring with $\mathbb{Z}[t]$ as its value group. The prime ideals of $A$ then correspond to the initial segments of the positive part of $\mathbb{Z}[t]$ which are closed under addition. For each $n$, there is such a segment consisting of elements which are less than some integer multiple of $t^n$, and the only other such segments are $\{0\}$ and the entire positive part of $\mathbb{Z}[t]$. So the set of prime ideals in $A$ under reverse inclusion is order-isomorphic to $\mathbb{N}\cup\{\infty\}$. It is the easy to see that the Zariski topology coincides with the topology described in leoli1's answer.

Answer 2

Consider the space $X=\{1,2,3\dots\}\cup\{\infty\}=\Bbb N\cup\{\infty\}$. Define the topology on $X$ so that the open sets are $\emptyset$ and sets of the form $[n,\infty]$ with $n<\infty$. Then $X$ is noetherian and $\dim X=\infty$. Now any neighborhood of $x=\infty$ is of the form $[n,\infty]$ and hence homeomorphic to $X$, so $\dim_xX=\infty$.