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We have defined the completion of a noetherian local ring $A$ to be $$\hat{A}=\left\{(a_1,a_2,\ldots)\in\prod_{i=1}^\infty A/\mathfrak{m}^i:a_j\equiv a_i\bmod{\mathfrak{m}^i} \,\,\forall j>i\right\}.$$ I have a slight problem trying to understand the proof that then $\hat{A}$ is a complete local ring with maximal ideal $\hat{\mathfrak{m}}=\{(a_1,a_2,\ldots)\in\hat{A}:a_1=0\}$.

Proof. If $(a_1,a_2,\ldots)\in\hat{\mathfrak{m}}$, then $a_i\equiv 0\bmod{\mathfrak{m}}$ for all $i$, i.e., $a_i\in\mathfrak{m}$. Hence $$\hat{\mathfrak{m}}^i=\left\{(a_1,a_2,\ldots)\in\hat{A}:a_j=0\,\,\forall j\leq i\right\}.$$ So the canonical map $\hat{A}\to A/\mathfrak{m}^i$, $(a_1,a_2,\ldots)\mapsto a_i$, is surjective with kernel $\hat{\mathfrak{m}}^i$. Thus $\hat{A}/\hat{\mathfrak{m}}^i\cong A/\mathfrak{m}^i$. In particular, $\hat{\mathfrak{m}}$ is a maximal ideal. But why is it the only one in $\hat{A}$?

If $(a_1,a_2,\ldots)\not\in\hat{\mathfrak{m}}$, we have $a_1\neq 0$, hence $a_1\not\in\mathfrak{m}$, hence it is a unit. By the defining property of the completion, $a_j$ is a unit for all $j$. So I could choose a candidate for the inverse of $(a_1,a_2,\ldots)$ by choosing inverse elements of the $a_j$. Why would this candidate be in $\hat{A}$ then, i.e. how can I choose it properly such that the congruences on the right hand side of the definition of the completion would be fulfilled?

Regards!

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    $\begingroup$ By the way, there are local rings whose "completion" is not complete (of course they are not noetherian, and in fact quite horrible). $\endgroup$ – Martin Brandenburg Jul 10 '12 at 18:06
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Inverses are unique. [If $R$ is a ring in which $y, z$ are inverses for $x$ then $y = y(xz) = (yx)z = z$.] So if $b_2 \in A/\mathfrak m^2$ is the inverse for $a_2$ then its homomorphic image in $A/\mathfrak m$ must be the inverse of $a_1$, and so on.

You could think of this in a different way: any element not in $\hat{\mathfrak m}$ can be written as $a + x$ for some $a \in A \setminus \mathfrak m$ and $x \in \hat{\mathfrak m}$, and hence as $a(1 - y)$ for some $y \in \hat{\mathfrak m}$. Then $a^{-1}(1 + y + y^2 + \cdots )$ is an inverse.

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