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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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That cannot work: the topology you want has as a basis the set of all translations of all powers of I, while not all sets you describe in the first paragraph are unions of such things.

For example, suppose $I$ is such that $I^2=0$. Then the open sets are just the cosets of $I$ in $R$. But there are sets which contain a coset but are not a union of cosets. We can construct a concrete example as follows: Let $k$ be a field, and let $R=k[x]/(x^2)$. Let $\varepsilon$ be the image of $x\in k[x]$ in the quotient $R$. Then $I=(\varepsilon)$ is an ideal which squares to zero. And $I\cup(k\setminus\{0\})$ is a set which contains a power of $I$, but it is not a union of cosets of $I$.

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    @Mariano: Thanks for your response. I don't quite understand what you mean. I have not specified a basis. I have just described all the open sets in the topology.2011-02-20
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    @Chitti: The basis *I* specified is a basis for the $I$-adic topology. It makes it easy for me to check that the other topology you defined (it may not even be a topology, really...) does not coincide with it.2011-02-20
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    @Mariano: Perhaps I don't quite understand this topology then. On page 101 of Atiyah-Macdonald, they start with a topological group. Then show that translations are homemomorphisms. Then conclude that the topology is defined by specifying neighbourhoods of $0$. On the next page, they describe a topology with respect to a filtration of subgroups of a group and they call it a fundamental system of neighbourhoods. Then declare that a set containing zero is open if and only it contains one of the groups in the filtration. I don't see what I am missing here.2011-02-20
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    @Chitti: I don't have the book at heand, but I am pretty sure that is not what they do, for that would be wrong. They surely declare a set to be *a neighborhood of zero* if it contains one of the groups in the filtration...2011-02-20
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    @Mariano: I found a spanish copy of the book on google books (English version does not have access to the chapter on completions) http://books.google.com/books?id=cQE-nnLL-OkC&printsec=frontcover&dq=atiyah+macdonald&hl=en&ei=6UthTY7qLYjegQfY6Y2DAg&sa=X&oi=book_result&ct=result&resnum=2&ved=0CDEQ6AEwAQ#v=onepage&q&f=false2011-02-20
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    @Chitti: it does not show me the relevant pages. But trust me: the $I$-adic topology is *not* defined like that.2011-02-20
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    In other words, the set of subsets of $R$ which contain a translation of $I$ is *not* a topology on $R$ (in general)2011-02-20
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    @Mariano: I trust you. But I am trying to understand what this topology really is. If you click on chapter 10 in the table of contents, it will take you to the chapter on completions. The first few pages are available where their definitions appear. I think now that my error was assuming neighbourhood means "open neighbourhood". I am rereading everything assuming this.2011-02-20
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In another, now deleted, venue you mentioned a more specific question of just showing the operations are continuous. You mentioned you could handle negation, but had trouble with addition and multiplication. Here is one way to verify them:

Let J = In.

The inverse image of x+J under addition is { (y,z) : y+z-x in J }. This is a union of A(y) = (y+J)⊕(x-y+J) ≤ R⊕R as y varies over R. Each A(y) is open (being a direct product of open sets), and so the union is open. In other words, the preimage under + of an open set is open.

The inverse image x+J under multiplication is { (y,z) : yz-x in J }. This is a union of B(y) = (y+J)⊕(∪{ z+J : yz-x in J }). Each B(y) is open (being a direct product of an open set and a union of open sets). In other words the preimage under multiplication of an open set is open.

The only trick to it is (1) knowing some open sets of a direct product and (2) noticing that you could have just worked in a quotient ring R/J where every set (every!) is open, since it is a union of singletons, x+J. In plainer language, (y+j)*(z+j') = yz + jz+yj'+jj' = yz + (something in J) is still in yz+J.