Filtrations and topological ring At page $105$ of Introduction to Commutative Algebra of M. Atiyah, there is the following claim:
C. Given a ring $A$ and an ideal $I\subset A$, consider the filtration $(I^n)_{n\in\mathbb{N}}$, then we have a the $I$-adic topology on $A$. Then $A$ with this topology is a topological ring. 
I'm able to check that $A$ is a topological group, but I have some problems to check that the multiplication map $A\times A\longrightarrow A$ is continuos.
Can anyone help me?
 A: You can show this directly. Open subsets of $A$ are cosets of the form $I^n+a$, so take a pair $(b,c)$ that lies in the inverse image of $I^n+a$, i.e. $bc\in I^n+a$, and hence $I^n+bc=I^n+a$.
Now $(I^n+b)\times (I^n+c)$ is an open subset of $A\times A$, and $(I^n+b)(I^n+c)$ $=I^n+bc=I^n+a$, and hence $(I^n+b)\times (I^n+c)$ is contained in the inverse image of $I^n+a$, and it contains $(b,c)$. Therefore multiplication is continuous.
A: Over a commutative ring, any filter of ideals will define a ring topology. A filter of ideals is a set $\mathscr{F}$ of ideals such that


*

*if $I\in\mathscr{F}$ and $J$ is any ideal with $I\subseteq J$, then $J\in\mathscr{F}$;

*if $I,J\in\mathscr{F}$, then $I\cap J\in\mathscr{F}$.


If $\mathscr{S}$ is a set of ideals such that for $I,J\in\mathscr{S}$ there exists $K\in\mathscr{S}$ with $K\subseteq I\cap J$, then $\mathscr{F}$, consisting of the ideals containing some member of $\mathscr{S}$ is a filter of ideals and $\mathscr{S}$ is a basis for a filter.
A filtration, that is, a chain of ideals is certainly a basis for a filter. In particular the set of powers of a given ideal.
The topology associated to a filter $\mathscr{F}$ has as neighborhoods of $a\in A$ the sets of the form $a+I$, where $I\in\mathscr{F}$.
Continuity of addition is easy: let $a,b\in A$ and $I\in\mathscr{F}$. Then
$$
(a+I)+(b+I)=(a+b)+I
$$
Continuity of $a\mapsto-a$ is likewise obvious.
For multiplication, given $I\in\mathscr{F}$ we need to find $J,K\in\mathscr{F}$ such that
$$
(a+J)\cdot(b+K)\subseteq ab+I
$$
Notation: if $X,Y\subseteq A$, then $X\cdot Y=\{xy:x\in X, y\in Y\}$.
We can just choose $J=K=I$: indeed, if $x,y\in I$, we have
$$
(a+x)(b+y)=ab+(ay+xb+xy)
$$
and $ay+xb+xy\in I$.
