I'm learning about ring completions, and this question came to mind: If $R$ is a complete local ring with maximal ideal $\mathfrak{m}$ (e.g. $R = \mathbb{Z}_p$ or $R = k[[x]]$), is the completion of $R[t]$ with respect to $(t)$ necessarily the same as the completion of $R[t]$ with respect to $\mathfrak{m} + (t)$? How about if we replace $t$ by $t_1, \cdots, t_n$?
It seems to me that the answer is yes (at least for the $n=1$ case), because $R$ is already complete with respect to $\mathfrak{m}$ so completing with respect to $\mathfrak{m} + (t)$ should be the "same" as completing with respect to only $(t)$, but I'm unable to make this precise. The statement I want is probably $$\varprojlim R[t]/(t^n) = \varprojlim R[t]/(\mathfrak{m}, t)^n.$$
If this is false, are there any nice circumstances where this is true (e.g. $R$ Noetherian)?
There is clearly a natural map from the RHS to the LHS, the question is then whether this is an isomorphism.
THe LHS is $R[[t]]$, the ring of formal power series, and it's easy to see that the map is the limit of the maps $R[[t]]\to R[t]/(t^n)\to R[t]/(m,t)^n$.
If $f= \sum_k a_k t^k$ is sent to $0$, let's prove that $a_k = 0$ for all $k$.
$f$ is sent to $0$, so for all $n$, $\sum_{k=0}^n a_kt^k\in (m,t)^n$. Now the $t^j$'s are $R$-linearly independent, so this implies that for all $k,n, a_k \in m^{n-k}$. For $n$ large enough, this implies $a_k\in m^n$ for all $n$, so $a_k = 0$ (since $R$ is $m$-complete)
All in all, $f=0$, so the map is injective.
Now consider someone, say $x$, in the RHS, and fix $k$. Then if you look in $R[t]/(m,t)^n, n\geq k$, the coefficient of $t^k$ is some residue $\alpha_{k,n} \in R/m^{n-k}$. If you take $n_1\geq n_0$, then $\alpha_{k,n_1} = \alpha_{k,n_0}$ in $m^{n_0-k}$.
In other words, $(\alpha_{n,k})_n$ defines, up to a shift of indices, an element of $\varprojlim_n R/m^n \cong R$. Call it $\alpha_k$. Then the claim is that $x$ is the image of $\sum_k \alpha_k t^k$.
I'll leave this to you, but it should be pretty clear from the definition.
Note : I think there should be some more conceptual proof here, using some clever partial orders and cofinality arguments, but I haven't found it yet.