Hi,
why the completion of a local ring $R$ can be written as an increasing union of $R$-algebras of finite type?
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1$\begingroup$ Any $R$-algebra is the increasing union of its finite type $R$-subalgebras. (After all, if $x$ is an element of the $R$-algebra, then the $R$-subalgebra generated by $x$ is of finite type, and contains $x$.) $\endgroup$– EmertonCommented Feb 23, 2011 at 15:15
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EDIT: Given a ring $R$ that is an algebra over a base ring it is always a filtering union of finite type algebras. Take a system of generators of $R$ over the base ring. The family of finite subsets of this system provides a collection of finite type subalgebras of $R$ whose filtered union is $R$.
(Some considerations on completion via Cauchy sequences deleted.)
answered Feb 23, 2011 at 11:40
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$\begingroup$ @Leo Alonso in the case $R=k[x]$ completed along $I=(x)$ what do you get as these algebras? $\endgroup$– unknownCommented Feb 23, 2011 at 11:48
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$\begingroup$ I was thinking on the description $\hat{R}=k[[x]]=k[x][[T]]/\langle T−x \rangle$. So the stages are $R_i:=k[x,T]/\langle T^i,T−x \rangle = k[x]/\langle x^i\rangle$, as usual, but this is not an increasing union... Can you provide more detail to your question? I still think that the metric is the key. $\endgroup$ Commented Feb 23, 2011 at 12:03
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$\begingroup$ @Leo Alonso I found this as I wrote as "well known fact" on a paper. The fact that this is an union is crucial in the proof. So I would like to understand this at least in easy cases. $\endgroup$– unknownCommented Feb 23, 2011 at 12:06
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$\begingroup$ Perhaps the paper you mention refers to the fact that $Spf(\hat{R})$ is the direct limit (or increasing union) of the finite type schemes $Spec(R/m^i)$, provided that $R$ is finite type over some base scheme ($m$ denotes the maximal ideal of $R$). $\endgroup$ Commented Feb 23, 2011 at 15:08