$\newcommand{\U}{\mathcal{U}}$ $\newcommand{\P}{\mathbb{P}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\F}{\mathcal{F}}$ Recall the following equivalent definitions of a Ramsey ultrafilter over $\omega$:

Theorem (Ramsey Ultrafilter).Let $\U$ be a non-principal ultrafilter over $\omega$. TFAE:

For any partition $F : [\omega]^n \to k$, there is a homogeneous set $H \in \U$.

For any partition $F : [\omega]^2 \to 2$, there is a homogeneous set $H \in \U$.

For every partition of $\omega$, $\{A_n : n \in \omega\}$ such that $A_n \notin \U$ for all $n$, there is a sequence $\langle{x_n : n \in \omega}\rangle$ such that $x_n \in A_n$ and $\{x_n : n \in \omega\} \in U$.

Given a decreasing sequence of sets $A_0 \supseteq A_1 \supseteq \cdots$ in $\U$, there exists a set $\{x_i\}_{i \in \omega} \in \U$ such that for all $n \in \omega$, $x_{n+1} \in A_{x_n}$.

I wish to generalise this notion to countable posets, and am in particular more interested in criterion (4). More specifically, let $(\P,\leq)$ be a countable poset, and let $\U$ be an ultrafilter over $\P$. Let's suppose we say that $\U$ is a *Ramsey ultrafilter* over $\P$ if for any partition $F : [\P]^n \to k$, there is a homogeneous set $H \in \U$.

I wish to attain an equivalence that is of something like the following:

Conjecture.Let $\U$ be a non-principal ultrafilter over a countable poset $(\P,\leq)$ (with possibly more assumptions required on $(\P,\leq)$). TFAE:

$\U$ is Ramsey. That is, for any partition $F : [\P]^n \to k$, there is a homogeneous set $H \in \U$.

(?) Let $\F = \{A_p : p \in \P\}$ be a family of sets in $\U$ such that $p < q \implies A_p \supseteq A_q$. Then there exists a set $\Q \subseteq \P$, $\Q \in \U$, such that for all $p,q \in \Q$, if $p < q$ then $q \in A_p$.

If this is false in full generality, what additional assumptions should we impose on $(\P,\leq)$? Several assumptions I'm more than willing to impose are:

Every chain in $\P$ is well-ordered.

For all $p,q \in \P$, there are only finitely many $r$ such that $p < r < q$.

But I wish to stay away from the linear order case.

**EDIT**: Allow me to explain the main obstacle I face in generalising the equivalence: The most accessible proof of (3) $\implies$ (4) appears to be Jech's book, Lemma 9.2. His proof is as follows:

[Let] $X_0 \supseteq X_1 \supseteq \cdots$ be sets in $D$ [where $D$ is a Ramsey ultrafilter]. Since $D$ is a p-point, there exists $Y \in D$ such that each $Y - X_n$ is finite. Let us define a sequence $y_0 < y_1 < \cdots$ in $Y$ as follows:

$y_0 = $ the least $y_0 \in Y$ such that $\{y \in Y : y > y_0\} \subseteq X_0$.

$y_1 = $ the least $y_1 \in Y$ such that $\{y \in Y : y > y_2\} \subseteq X_{y_0}$.

$\dots$

$y_n = $ the least $y_n \in Y$ such that $\{y \in Y : y > y_{n-1}\} \subseteq X_{y_{n-1}}$.

For each $n$, let $A_n = \{y \in Y : y_n < y \leq y_{n+1}\}$. Since $D$ is Ramsey, there exists a set $\{z_n\}_{n=0}^\infty$ such that $z_n \in A_n$ for all $n$.

We observe that for each $n$, $z_{n+2} \in X_{z_n}$: Since $z_{n+2} > y_{n+2}$, we have $z_{n+2} \in X_{y_{n+2}}$, and since $y_{n+1} \geq z_n$, we have $X_{y_{n+1}} \subseteq X_{z_n}$ and hence $z_{n+2} \in X_{z_n}$.

Thus if we let $a_n = z_{2n}$ and $b_n = z_{2n+1}$, for all $n$, then either $\{a_n\}_{n=0}^\infty \in D$ or $\{b_n\}_{n=0}^\infty \in D$; and in either case we get a sequence that satisfies [the property (4)].

For general posets, we cannot so simply split to two subsets $a_n = z_{2n}$ and $b_n = z_{2n+1}$. If there are infinitely many branches, then we cannot guarantee one such subset will be in $D$. It also makes no sense to consider $\sigma$-complete Ramsey ultrafilters.