Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
By the way, there are local rings whose "completion" is not complete (of course they are not noetherian, and in fact quite horrible). –  Martin Brandenburg Jul 10 '12 at 18:06

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.