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Let $R$ be a ring, $F$ a free ultrafilter on a set $X$ which is not countably complete, and $R_F$ the ultrapower of $R$ with $R \not\cong R_F$. The following two results are from a masters thesis written in 1980 by James A. Nickerson, called "Ultraproducts in Commutative Algebra."

Lemma: If $R$ is a one-dimensional local integral domain, then $R_F$ will be a two or more-dimensional local integral domain.

Corollary: If $1 \leq Dim(R) < \infty$, then $Dim(R_F) > Dim(R)$.

Proof of corollary: Let $n \geq 1$ be the dimension of $R$. Then we have $n$ strict inclusions of prime ideals, as $P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_n$. Since the prime ideals of $R/P_0$ are in a one to one, order preserving correspondence with the prime ideals of $R$ which contain $P_0$, we have in the integral domain $\mathcal R = R/P_0$ $n$ strict inclusions of prime ideals, as $$ \{0\} = P_0/P_0 \subsetneq P_1/P_0 \subsetneq \cdots \subsetneq P_n/P_0$$. There are no prime ideals between $P_0/P_0$ and $P_1/P_0$; such an ideal would have to take the form $I/P_0$ with $P_0 \subsetneq I \subsetneq P_1$, which forces $Dim(R) > n$.

Relabeling the ideals, as $\mathfrak P_0 = P_0/P_0, \mathfrak P_1 = P_1/P_0$ etc., it then follows that $\mathcal R_{\mathfrak P_1}$ is a one-dimensional local integral domain: $\mathfrak P_1 \mathcal R_{\mathfrak{P_1}}$ is the unique maximal ideal of $\mathcal R_{\mathfrak P_1}$, and the prime ideals thereof are in order preserving correspondence with the prime ideals of $\mathcal R$ contained in $\mathfrak P_1$. But there are no such ideals except $\mathfrak P_0 = \{0\}$ and $\mathfrak P_1$.

Since $\mathcal R_{\mathfrak P_1}$ is a one-dimensional local integral domain, the lemma above tells us that $(\mathcal R_{\mathfrak P_1})_F$, and hence $(\mathcal R_{F})_{(\mathfrak{P_1})_F}$ (recall they are isomorphic), has dimension $> 1$, i.e. there is a prime ideal $\mathcal Q$ of $(\mathcal R_{F})_{(\mathfrak P_1)_F}$ with $\{0\} \subsetneq \mathcal Q \subsetneq (\mathfrak P_{1})_F (\mathcal R_{F})_{\mathfrak{P_{1}}_F}$.

The order-preserving correspondence of prime ideals between a ring and its localization gives us a prime ideal $\mathfrak Q$ of $\mathcal R$ with $$(\mathfrak P_0)_F \subsetneq \mathfrak Q \subsetneq (\mathfrak P_1)_F \subsetneq (\mathfrak P_2)_F \cdots \subsetneq (\mathfrak P_n)_F$$


$$ (P_0)_F/(P_0)_F \subsetneq \phi(\mathfrak Q) \subsetneq (P_1)_F/(P_0)_F \subsetneq (P_2)_F/(P_0)_F \subsetneq \cdots \subsetneq (P_n)_F/(P_0)_F$$

where $\phi: (R/P_0)_F \rightarrow R_F/(P_0)_F$ is an isomorphism. Finally, $\phi(\mathfrak Q)$ takes the form $Q/(P_0)_F$ for some prime ideal $Q$ of $R_F$, so we have a chain of $n+1$ inclusions

$$(P_0)_F \subsetneq Q \subsetneq (P_1)_F \subsetneq (P_2)_F \subsetneq \cdots \subsetneq (P_n)_F.$$

Thus $Dim(R_F) > n = Dim(R)$.

So, are there any more very general results about the prime spectrum of an ultrapower which have been discovered in the last 30 years? I'm aware of a couple of papers by B. Olberding and J. Shapiro which investigate the prime spectrum of ultraproducts (which doesn't really give new results when you restrict to ultrapowers) and another which looks at ring completions and ultrapowers. I was wondering rather whether there were some very general results, like the one above, for example whether it is possible to ever conclude $R_F$ has infinite Krull dimension when $R$ is finite dimensional.

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