Let $R$ Be Noetherian local ring with unique maximal ideal $\mathfrak{m}$. We define the completion $\hat{R}$ of $R$ to be the inverse limit of $R_m := R/\mathfrak{m}^{m+1}$. More explicitly, the ring $\hat{R}$ is given by $$\hat{R}:= \{ (a_1, a_2 ,\cdots )\in \prod_i R_m : a_j \equiv a_i \text{ mod } \mathfrak{m}^i, \forall j > i\} $$
Moreover, we have a natural injection $i: R\rightarrow \hat{R}$. One can show that $\hat{R}$ is again, a local Noetherian ring with maximal ideal $\mathfrak{m}\hat{R}$. A lemma states that the natural map $i_m: R_m \rightarrow \hat{R}/ \mathfrak{m}^{m+1}\hat{R}$ is in fact an isomorphism for all $m$.
I was able to understand why this induced map is a surjection. However, the proof of $i_m$ being an injection confuses me. It involves an induction on $m$. The case for $m=0$ is clear. Now assume $i_{m-1}$ is an isomorphism. From here, we conclude that $\ker i_m$ must be contained in $\mathfrak{m}/\mathfrak{m}^{m+1}$.
My question is: Why is the claim “$\ker i_m$ must be contained in $\mathfrak{m}/\mathfrak{m}^{m+1}$“ true? I tried to write out the workings but was unable to see why. Would greatly appreciate any help given!
Best Answer
The key claim is the following one: $\mathfrak{m}^k \hat{R}$ is the set $S_k$ of $x \in \hat{R}$ such that $x_1,\ldots,x_k=0$. Indeed, as $i^{-1}(S_k)=\mathfrak{m}^k$, we get the injectivity of $i_k$.
Clearly, $\subset$ holds. Let’s see $\supset$.
Let $y_p$ be a sequence in $R$ such that each $y_i$ reduces to $x_i$ mod $\mathfrak{m}^i$, and $y_1=\ldots,y_k=0$. Let $b_1,\ldots,b_s$ be a system of generators for $\mathfrak{m}^k$.
We then choose, for all $t \geq 1$, $1 \leq i \leq s$, the $c^t_i$ as follows: $\sum_{i=1}^s{(c^{t}_i-c^{t-1}_i)b_i}=y_{t+k}-y_{t+k-1} \in \mathfrak{m}^{t+k-1}$, with the $c^t_i-c^{t-1}_i$ chosen in $\mathfrak{m}^{t-1}$, and $c^0_i=0$. Thus, for each $i$, the sequence $(c^t_i)_{t \geq 1}$ can be reduced to an element of $\hat{R}$, and it follows $x=\sum_i{b_i(c^t_i)_t} \in \mathfrak{m}^k\hat{R}$.