Let $R$ be a commutative ring, $\mathcal{M}_R$ the set of monic polynomials in $R[q]$, $M \subset \mathcal{M}_R$ and $M^*$ the multiplicative closed set associated to $M$. Order the polynomials in $M^*$ by divisibility. Then, $$\left(R[q]/(f_i), f_{ij} \right)$$ is an inverse system of rings and we can take the inverse limit $$R[q]^M = \varprojlim_{f \in M^*}R[q]/(f)$$ On the other hand, for a fixed $M \subset \mathcal{M}_R$, let $$\mathcal{S}_R = \{ M' \subset M \mid |M'|<\infty\}$$ If $M' \subset M$ we have a natural map $$\mu_{M',M} \colon R[q]^M \to R[q]^{M'}$$ Hence, the sets in $\mathcal{S}_R$ along with the maps $\mu_M'',M' \colon R[q]^{M'} \to R[q]^{M''}$ form an inverse system of rings, so we can take the inverse limit $$R[q]^S = \varprojlim_{M' \in \mathcal{S}_R} R[q]^{M'}$$
Why is $R[q]^M \cong R[q]^S$?
Intuitively it seems so, since $$R[q]^s = \varprojlim_{M' \in \mathcal{S}_R}R[q]^{M'} = \varprojlim_{M' \in \mathcal{S}_R} \left( \varprojlim_{f \in (M')^*} R[q]/(f) \right)$$ but I have not been able to prove it. Any help would be appreciated. Thank you.
UPDATE: I believe I came up with an answer but I'm not sure if it works. I assume that $R[q]^S$ can be written as $$R[q]^S = \varprojlim_{f \in A}R[q]/(f)$$ where $$A = \{ f \mid f \in (M')^*, M' \subset M, |M'|<\infty \}$$ so I tried to show that $A$ is cofinal in $M^*$. I need to prove that for any $f \in M^*$ there is some $\overline{ƒ}$ such that $\overline{f}\in (M')^*$ for some $M' \subset M$ with $|M'|<\infty$. Let $f \in M^*$, then $$f = \prod_{i \in I} f_i^{n_i}$$ where each $f_i \in M$. Consider any $i_0 \in I$ and $M' = \{ f_{i_0} \}\subset M$ which is finite. Then the same polynomial $f_{i_0}$ works as the $\overline{f}$ in the definition, and creamy $f_{i_0} \vert f$. Therefore $A \subset M^*$ is cofinal so both inverse limits are isomorphic. The problem is then the fact that I'm not entirely sure that $R[q]^S = \varprojlim_{f \in A} R[q]/(f)$.
Someone might want to translate this into ring-theoretic terms, but to expand on Damien L's pos(e)t, let
$\mathcal P$ be the poset category with objects polynomials $f$ in $M^*$ and with morphisms $f_1 \to f_2$ iff $f_1\mid f_2$
$\mathcal Q$ be the poset category with objects are pairs $(f,M')$ such that $f\in M'\in\mathcal S_R$ and with morphisms $(f_1,M')\to (f_2,M'')$ iff $f_1\mid f_2$ and $M'\subseteq M''$
(I.e. $\mathcal Q$ agrees with $B$.) Consider the diagrams $$X:\mathcal P^{\text{op}}\to \textbf{Ring}\text{ mapping }\left\{f_1\to f_2\right\}\overset{X}{\longmapsto} \left\{R/(f_1)\leftarrow R/(f_2)\right\}$$ $$Y:\mathcal Q^{\text{op}}\to \textbf{Ring}\text{ mapping }\left\{(f_1,M')\to (f_2,M'')\right\}\overset{Y}{\longmapsto} \left\{R/(f_1)\leftarrow R/(f_2)\right\}$$ Where $R/(f_1)\leftarrow R/(f_2)$ is in both cases the obvious quotient morphism. By definition of $X$ and $Y$, $$\lim X=\varprojlim_{f \in M^*}R[q]/(f)=R[q]^M$$ $$\lim Y=\varprojlim_{f\in (M')^*,\ M' \in \mathcal{S}_R} R[q]/(f)=R[q]^S$$ Now consider the functor $F:\mathcal Q\to \mathcal P$ mapping $\left\{(f_1,M')\to (f_2,M'')\right\}\overset{F}{\longmapsto} \left\{f_1\to f_2\right\}$, and observe the natural isomorphism $$X\circ F^{\text{op}}\simeq Y$$ (They're the exact same map.) Furthermore $F$'s fullness and essential surjectivity imply the equivalence of the images of $X\circ F^\text{op}$ and $X$. (The two actually turn out to be the exact same.) If follows that $$\lim X\cong \lim X\circ F^\text{op}\cong \lim Y$$ As desired $\blacksquare$
Note that the "fullness and essential surjectivity" of $F$ follows from the fact that for any two $f_1,f_2\in M^*$ for which $f_1\mid f_2$, one may choose an $M'\in \mathcal S_R$ such that $f_1,f_2\in M'$. Some care is also required in justifying $$\varprojlim_{M' \in \mathcal{S}_R} \left( \varprojlim_{f \in (M')^*} R[q]/(f) \right)=\varprojlim_{f\in (M')^*,\ M' \in \mathcal{S}_R} R[q]/(f)$$ The main idea is to describe a natural bijection between cones over the former diagram and those over the latter. (Drawing a picture helps.)