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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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    @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:$I$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.