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I would like to know what does it mean to say " a ring $R$ is complete with respect to some ideal $I \subset R$.

Is it like, we define a metric by saying two elements in $R$ is "close" if the difference lies in a higher power of $I^n$ and $R$ is a complete metric space?

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  • $\begingroup$ It is quite instructive to look at the examples $R=k[x]$ and $R=k[[x]]$ with $I=(x)$. $\endgroup$ – Martin Brandenburg Sep 15 '13 at 22:47
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A ring $R$ is complete with respect to an ideal $I$ if the natural map $R\rightarrow\varprojlim_{n\geq 1}R/I^n$ is an isomorphism. This is the same as asking that $R$ is complete (in the sense that every Cauchy sequence converges) with respect to the $I$-adic topology, which is the topology for which a base of opens around $a\in R$ is given by the cosets $a+I^n$, $n\geq 1$. A sequence $(a_n)_{n\geq 1}$ in $R$ is Cauchy if for each $m\geq 1$, there is $N\geq 1$ such that $a_n-a_m\in I^m$ for $n,m\geq N$. The ring $\hat{R}=\varprojlim_{n\geq 1}R/I^n$ is called the $I$-adic completion of $R$.

In other words, your guess is basically correct, although in general the $I$-adic topology isn't metrizable, for example it isn't Hausdorff if $\bigcap_{n\geq 1}I^n\neq\{0\}$ (if this intersection is $\{0\}$ then $R$ is said to be $I$-adically separated). Completeness in the sense above includes a Hausdorff condition as well, because injectivity of the map $R\rightarrow\hat{R}$ is equivalent to the $I$-adic topology being Hausdorff.

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