Ultraproduct with no long descending sequence I have a countably infinite well-ordered structure $M$ (over a countable language if it helps), and an uncountable regular cardinal $κ$, and I wanted to construct an elementarily equivalent structure with certain properties. I could not find a way to control the model that the compactness theorem gives, so I thought of following the proof of the compactness theorem based on ultraproducts.
Let $C$ be the collection of finite subsets of $κ$ and let $F = \{ \{ S : S \in C \land S \supseteq I \} : I \in C \}$. Then $F$ is closed under binary intersection, and hence can be extended to an ultrafilter $U$. Let $(c_i)_{i \in κ}$ be distinct constant symbols that are not in the language of $M$. For each $I \in C$ let $M_I$ be $M$ extended by interpreting $(c_i)_{i \in I}$ as the sequence of the first $\#(I)$ elements in $M$. Let $N$ be the ultraproduct $\prod_{I \in C} M_I / U$. Then for any $i,j \in κ$ such that $i < j$, we have $M_S \vDash c_i < c_j$ for every $S \in C$ such that $S \supseteq \{i,j\}$, and hence $N \vDash c_i < c_j$.
Now I would also like $N$ to have no decreasing sequence of length $κ$, but I do not know whether that can be ensured by some suitable choice of $U$. So my questions are:


*

*Is there some choice of $U$ such that $N$ has no decreasing sequence of length $κ$?
a. If so, is there some choice of $U$ such that $N$ has no uncountable decreasing sequence?
b. If not, what is the smallest $κ$ such that there is a structure $M$ where every choice of $U$ makes $N$ have a decreasing sequence of length $κ$?
My random guess is that (1) is true and (1b) is false but I can't even guess about (1a).
I tried searching for various types of ultrafilters. I found out that if there is a countably-complete ultrafilter on $κ$, then $κ$ must be a measurable cardinal, and for such $κ$ this proves the strong result that $N$ is well-ordered. But I would like to know if the weaker (1a) for arbitrary $κ$ can be answered in ZFC alone. Similarly I hope that ZFC is enough to settle my other questions.
Finally, I am curious to know if the answers change if $κ$ is not regular.
 A: In the comments (but not in the question), you clarify that you want the ultrapower to have an increasing sequence of length $\kappa$, but no decreasing sequence of length $\kappa$. I don't know if this is possible (for a general ultrapower of a well-order), but I'll explain why it seems hard. The following theorem is from Keisler's paper Ultraproducts which are not saturated.
Theorem: Let $M$ be a structure, and let $U$ be a regular ultrafilter on $\kappa$. Let $\Sigma$ be a set of formulas such that $|\Sigma| = \kappa$ and every finite subset of $\Sigma$ is satisfiable in $M$. Then $\Sigma$ is satisfiable in $M^\kappa/U$. 
Proof: Since $U$ is regular, there is some set $X\subseteq U$ such that $|X| = \kappa$, and for all $\alpha\in \kappa$, $X_\alpha = \{S\in X\mid \alpha\in S\}$ is a finite subset of $X$. Let $h\colon \Sigma\to X$ be a bijection. For all $\alpha$, let $\Sigma_\alpha = h^{-1}(X_\alpha)$. Since $\Sigma_\alpha$ is finite, it is satisfiable in $M$.
To show that $\Sigma$ is satisfiable in $M^\kappa/U$, we have to assign the variables used in $\Sigma$ to element of $M^\kappa/U$, i.e. for each $\alpha\in \kappa$, we have to assign them to elements of $M_\alpha$, the copy of $M$ indexed by $\alpha$ in the product. In each $M_\alpha$, do this so that the variables used in $\Sigma_\alpha$ are assigned to elements satisfying $\Sigma_\alpha$, and assign the other variables arbitrarily. Now, with this variable assignment, for each $\varphi\in \Sigma$, $\{\alpha\in \kappa\mid M_\alpha \models \varphi\}\supseteq \{\alpha\in \kappa\mid \varphi\in \Sigma_\alpha\} = \{\alpha\in \kappa\mid h(\varphi)\in X_\alpha\} = h(\varphi) \in X\subseteq U$, so $M^\kappa/U$ satisfies $\Sigma$. $\square$
As a consequence, if $M$ is any infinite linear order and $U$ is a regular ultrafilter on $\kappa$, then the ultrapower $M^\kappa/U$ has increasing and descreasing sequences of length $\kappa$. To see this, introduce variables $(x_\alpha)_{\alpha<\kappa}$, and let $\Sigma = \{x_\alpha<x_\beta\mid \alpha<\beta<\kappa\}$ or $\{x_\alpha>x_\beta\mid \alpha<\beta<\kappa\}$.
This symmetry between increasing and decreasing sequences, and the difficulty of coming up with non-regular ultrafilters makes me think that your desired construction will be difficult to accomplish, if not impossible.
Edit: As noted in the comments below, the theorem really just generalizes the construction given in the question to arbitrary sets of formulas of size $\kappa$ and arbitrary regular ultrafilters. 
If you're only interested in ultrafilters $U$ on $C$ the set of all finite subsets of $\kappa$ extending the set $F$ given in the question, then your question is answered. Let $X_\alpha = \{S\in C\mid \{\alpha\}\subseteq S\}$, and let $X = \{X_\alpha\mid \alpha < \kappa\}\subseteq F$. This $X$ witnesses regularity of any ultrafilter $U$ extending $F$. So there is no ultrafilter $U$ on $C$ extending $F$, such that the ultrapower has no decreasing sequence of length $\kappa$. The choice of $\kappa$ is irrelevant. 
Of course, there may still be ultrapowers of $M$ by non-regular ultrafilters or ultrafilters on sets of size $<\kappa$ such that $M$ has an increasing sequence of length $\kappa$ but no decreasing sequence of length $\kappa$.
