One small question on the book 'Real and Functional Analysis' by S. Lang:
Example 6 on page 21: Let $R$ be a commutative ring. We define a subset $U$ of $R$ to be open if for each $x\in U$ there exists an ideal $J$ in $R$ such that $x+J\subseteq U$. This is called the ideal topology.
It seems to me that this definition is useless: All sets will be open because we can always pick $J=0$.
Unfortunately I couldn't find anything related to this 'ideal topology' via a Google search, so I don't know what is meant in the book. Perhaps requiring $J\ne0$?
Best Answer
This is clearly a mistake.
Usually one assumes a family $\mathscr{I}$ of ideals of $R$ which is directed with respect to $\supseteq$. Then we define a subset $U$ of $R$ to be open if for every $x \in U$ there is some $I \in \mathscr{I}$ with $x+I \subseteq U$. This defines a topology: the union axiom is clear, and if $U,V$ are open and $x \in U \cap V$, choose $I,J \in \mathscr{I}$ with $x+I \subseteq U$, $x+J \subseteq V$. By assumption there is some $K \in \mathscr{I}$ with $K \subseteq I$, $K \subseteq J$, thus $x+K \subseteq U \cap V$, showing that $U \cap V$ is open.
Alternatively, consider the natural map $$\tau : R \to \prod_{I \in \mathscr{I}} R/I.$$ Endow each $R/I$ with the discrete topology and the product with the product topology. Then $R$ carries the initial topology with respect to $\tau$. Actually, $\tau$ factors through the projective limit $\varprojlim_{I \in \mathscr{J}} R/I$.
In general, the topology makes $R$ into a topological ring. Topological rings of this kind are precisely the linear topological rings. A classical reference is EGA I, Chap. 0, §7 ("Compléments d'algèbre topologique"). The topology is Hausdorff (separated) if and only if $\bigcap \mathscr{J} = 0$. It is called complete when $R \to \varprojlim_{I \in \mathscr{I}} R/I$ is an isomorphism.
The most important case is $\mathscr{I} = \{I^n : n \geq 0\}$ for some fixed ideal $I \subseteq R$. In this case the topology is called the $I$-adic topology.
If $0 \in \mathscr{I}$ however, the topology is discrete and not very interesting. I do not think that the set of non-zero ideals is a natural example, since it is not always directed with respect to $\supseteq$.
If $\mathscr{I}$ consists of all ideals $I$ such that $R/I$ is a finite ring, then we call it the profinite topology. Since this is analogous to the profinite topology on a group in the previous example in Lang's book, I assume that this is what he had in mind. The profinite topology on $\mathbf{Z}$ is known as the Fürstenberg topology, which can be used to give a "topological proof" that there are infinitely many prime numbers.