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Why is the dimension of a ring, with finite spectrum, $\leq 1$?

It is clear that it works for the case if the ring is noetherian, due to the fact that every ideal of height $2$ is an infinite union of prime ideals of height $1$. It seems like one could use the prime avoidance lemma, but I am really unsure how to work it out.

Jendrik Stelzner
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Aaron Wild
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1 Answers1

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The statement is not true if $R$ is not Noetherian. Indeed, for every $n$, there exists a ring with finite spectrum and Krull dimension $n$.

To see this, we can use valuation rings: For $n \in \Bbb N$, let $\Gamma$ be a linearly ordered group of rank $n$ (one can take for example $\Bbb Z^n$ with the lexicographical order), then there exists a valuation ring $R$ with value group $\Gamma$ (compare for example this question) and the Krull dimension of $R$ will be equal to the rank of $\Gamma$. But the prime ideals in any valuation ring are totally ordered, so for valuation rings, having finite Krull dimension is in fact equivalent to having finite spectrum.

Lukas Heger
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    Nice! I have forgotten about the Krull-Kaplansky-Jaffard-Ohm construction of such rings since my dissertation. It is extraordinarily useful. I've make this ring 99 and attributed it to the account that bears your name. If you wish me to change that, or if it's not you, just let me know and I'll disassociate the name from it. (Also, it looks a little empty now, but a lot of properties will be filled out when I hit the processing button later today.) – rschwieb Jul 06 '18 at 13:56
  • @rschwieb thanks! I'm glad to have contributed something to your very useful and interesting database. I learned the construction from exercises in Atiyah-Macdonald. – Lukas Heger Jul 06 '18 at 14:02
  • Glad to hear it. It is a nice counterpart to $k[x_1,\ldots x_n]$. Always looking for more examples of course... the site is just now burgeoning. – rschwieb Jul 06 '18 at 14:11