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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!

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  • $\begingroup$ Are you sure you copied that correctly? It is true but kind of vacuously true (if it is false then the kernel contains an element not in that ideal, but anything not in that ideal is a unit and so the map is the zero map, which we can see is not the case by looking at $1$) $\endgroup$
    – user208649
    Sep 2, 2020 at 21:41

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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 $x=\sum_i{b_i(c^t_i)_t} \in \mathfrak{m}^k\hat{R}$.

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  • $\begingroup$ Thank you very much for your answer, @mindlack. Your answer helped is very clear! So it appears that there is no need to consider any induction at all? $\endgroup$
    – Soby
    Sep 3, 2020 at 3:38
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    $\begingroup$ Indeed, it looks like it (apart maybe from the definition of the sequences $c^t_i$). $\endgroup$
    – Aphelli
    Sep 3, 2020 at 7:10
  • $\begingroup$ Just one more silly question from me. I would like to clarify some notations. For simplicity we assume $\mathfrak{m}^k$ is generated by an element, say $b$. So when we write $x = b(c^t)_t$ we are actually referring to the multiplication of the element $(b + \mathfrak{m}^i)_i = (0, \cdots , 0 , b+ \mathfrak{m}^{k+1}, \cdots )$ with $(c^1 + \mathfrak{m}, c^2 + \mathfrak{m}^2 ,\cdots )$ in $\hat{R}$, right? (I am still new to this whole completion concept so pardon me if this sounds really elementary.) $\endgroup$
    – Soby
    Sep 3, 2020 at 14:49
  • $\begingroup$ @thedilated: please don’t apologize about that, asking questions, not understanding exactly everything at once is absolutely natural. Here, I am taking the point of view of seeing $\hat{R}$ as a $R$-module (by $x \cdot y=i(x)y$), and $\mathfrak{m}\hat{R}$ is (formally) the $R$-submodule generated by the $i(c)y$, for $y \in \hat{R}$, $c\in \mathfrak{m}$. I hope this answers your question? $\endgroup$
    – Aphelli
    Sep 3, 2020 at 15:10
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    $\begingroup$ Your point isn't incorrect, but you are missing that as $b \in \mathfrak{m}^k$, $bc^{k+1}+\mathfrak{m}^{k+1}=bc^1+\mathfrak{m}^{k+1}$ by construction of the $c$ sequences. The product in $\hat{R}$ is the componentwise product. $\endgroup$
    – Aphelli
    Sep 3, 2020 at 16:26

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