Vtyjkyuk

Let $R$ be a commutative ring, $mathcalM_R$ the set of monic polynomials in $R[q]$, $M subset mathcalM_R$ 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 $M subset mathcalM_R$, let
$$mathcalS_R = M'$$
If $M' subset M$ we have a natural map
$$mu_M',M colon R[q]^M to R[q]^M'$$
Hence, the sets in $mathcalS_R$ 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
$$R[q]^S = varprojlim_M' in mathcalS_R R[q]^M'$$

Why is $R[q]^M cong R[q]^S$?

Intuitively it seems so, since
$$R[q]^s = varprojlim_M' in mathcalS_RR[q]^M' = varprojlim_M' in mathcalS_R left( varprojlim_f in (M')^* R[q]/(f) right)$$
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
$$R[q]^S = varprojlim_f in AR[q]/(f)$$
where
$$A = M'$$
so I tried to show that $A$ is cofinal in $M^*$. I need to prove that for any $f in M^*$ there is some $overlineƒ$ such that $overlinefin (M')^*$ 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 $overlinef$ 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 $f_1mid f_2$




• $mathcal Q$ be the poset category with objects are pairs $(f,M')$ such that $fin M'inmathcal S_R$ and with morphisms $(f_1,M')to (f_2,M'')$ iff $f_1mid f_2$ and $M'subseteq M''$

(I.e. $mathcal Q$ agrees with $B$.) Consider the diagrams $$X:mathcal P^textopto textbfRingtext mapping leftf_1to f_2rightoversetXlongmapsto leftR/(f_1)leftarrow R/(f_2)right$$ $$Y:mathcal Q^textopto textbfRingtext mapping left(f_1,M')to (f_2,M'')rightoversetYlongmapsto leftR/(f_1)leftarrow R/(f_2)right$$ Where $R/(f_1)leftarrow R/(f_2)$ is in both cases the obvious quotient morphism. By definition of $X$ and $Y$, $$lim X=varprojlim_f in M^*R[q]/(f)=R[q]^M$$ $$lim Y=varprojlim_fin (M')^*, M' in mathcalS_R R[q]/(f)=R[q]^S$$ Now consider the functor $F:mathcal Qto mathcal P$ mapping $left(f_1,M')to (f_2,M'')rightoversetFlongmapsto leftf_1to f_2right$, and observe the natural isomorphism $$Xcirc F^textopsimeq Y$$ (They're the exact same map.) Furthermore $F$'s fullness and essential surjectivity imply the equivalence of the images of $Xcirc F^textop$ and $X$. (The two actually turn out to be the exact same.) If follows that $$lim Xcong lim Xcirc F^textopcong lim Y$$ As desired $blacksquare$

Note that the "fullness and essential surjectivity" of $F$ follows from the fact that for any two $f_1,f_2in M^*$ for which $f_1mid f_2$, one may choose an $M'in mathcal S_R$ such that $f_1,f_2in M'$. Some care is also required in justifying $$varprojlim_M' in mathcalS_R left( varprojlim_f in (M')^* R[q]/(f) right)=varprojlim_fin (M')^*, M' in mathcalS_R R[q]/(f)$$ The main idea is to describe a natural bijection between cones over the former diagram and those over the latter. (Drawing a picture helps.)

I think you have the correct intuition, here is a hint following your update. Instead of using $A$, try using the poset $B$:
$$
B = M'
$$
where the poset structure is given by $(f, M') leq (g, M'')$ whenever $f mid g$ and $M' subset M''$.