Let $(R, \mathfrak{m})$ be a local (Noetherian) ring containing the rationals. Then the formal power series ring $\mathbb{Q}[[x_1, \ldots, x_n]]$ naturally forms a subring of $R[[x_1, \ldots, x_n]]$. Let $\mathfrak{p}, \mathfrak{p}' \in \mathrm{Spec}\,R[[x_1, \ldots, x_n]]$ be such that neither $\mathfrak{p}$ nor $\mathfrak{p}'$ are contained in each other and \begin{equation*} \mathbb{Q}[[x_1, \ldots, x_n]] \cap \mathfrak{p} = \mathbb{Q}[[x_1, \ldots, x_n]] \cap \mathfrak{p}' \neq (0). \end{equation*} If $(x_1, \ldots, x_n) \not\subseteq \mathfrak{p}, \mathfrak{p}'$, we can choose a monomial in $x_1, \ldots, x_n$, call if $f$, such that $f \not\in \mathfrak{p}, \mathfrak{p}'$. Let $\mathfrak{q}$ be a minimal prime ideal over $(\mathfrak{p}, f)$ and $\mathfrak{q}'$ a minimal prime ideal over $(\mathfrak{p}', f)$.
I'm interested in whether $\mathbb{Q}[[x_1, \ldots, x_n]] \cap \mathfrak{q} = \mathbb{Q}[[x_1, \ldots, x_n]] \cap \mathfrak{q}'$. I suspect the answer is "not always," but can we choose $f$, $\mathfrak{q}$, and $\mathfrak{q}'$ such that equality holds?
