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Let $R$ be a commutative ring with identity. Let $I$ be an ideal of $R$. Suppose, we give a topology on $R$ where a set is open if and only if it is a union of cosets of powers of $I$. Then, is $R$ a topological ring?

EDIT: The question has been edited in the light of comments below.

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    A set as you describe, that is a translation of a set containing a power of $I$ (so a set of the form $r+A$ where $A\subseteq R$ contains $I^n$ for some $n\geq 0$) is not obviously open in the $I$-adic topology. It clearly contains the open neighborhood $r+I^n$ of $r$, but that just means it's a neighborhood of $r$. To be open it would need to be a union of such sets.2011-02-20
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    @Keenan: Thanks for the comment. I thought by neighbourhood of a point x, the authors meant an open set containing x. This was the convention used in the other books I have read. But perhaps this is not the case.2011-02-20
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    Yes, in A and M they use the term "neighborhood of $0$" to mean a set $U$ containing an _open_ set containing $0$. Such a set need not actually be open; however, if $U$ is a subgroup of $R$, then being a neighborhood in this sense is the same as being open, because for each $r\in U$, $r+U\subseteq U$.2011-02-20
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    @Keenan: This is all too annoying. None of these things are clearly defined anywhere in the book. I still don't understand how you construct open sets in this topology given a fundamental system of neighbourhoods.2011-02-20
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    It's important that in A and M they are considering topologies defined by a fundamental system of neighborhoods consisting of subgroups. The open sets in the topology are precisely the unions of translates of these subgroups.2011-02-20
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    @Keenan: OK, I have modified my question now. Hopefully it makes sense now.2011-02-20

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