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I'm wondering if there is some generic topology that can be put on any field of characteristic zero which is similar to those induced by a norm on the field. I know that for vector spaces you can take the terminal topology induced by the family of all semi-norms on the space, but I don't think that this topology is also a field topology in general. Can one take the join of all topologies induced by a norm on the field and hope for a field topology?

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  • $\begingroup$ Maybe first figure out what the "generic" topology should be for $\mathbb Q[\sqrt{2}]$. There are (at least) three different field topologies that extend the usual topology on $\mathbb Q$. $\endgroup$ – Gerald Edgar Mar 15 '14 at 13:13
  • $\begingroup$ For the fields which are finite dimensional over Q, I'd like the topology to be that induced by considering it as a subset of $\mathbb{R}^n$. $\endgroup$ – Liam Baker Mar 15 '14 at 21:09
  • $\begingroup$ So, for example, you want $|a+b\sqrt{2}| = |a|+|b|$ or equivalent. A good choice. For infinte-dimensional over $\mathbb Q$ do you still want $\mathbb Q$ itself a closed subset? $\endgroup$ – Gerald Edgar Mar 15 '14 at 21:20
  • $\begingroup$ I hadn't thought of that before, but I suppose I would like Q to be closed $\endgroup$ – Liam Baker Mar 15 '14 at 21:50
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Any locally compact field, which is not discrete, is a non-archimedean or archimedean field.

Wiki: http://en.wikipedia.org/wiki/Local_field

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Too long for a comment.

You might find useful the two books of S.Warner [Topological fields; topological rings], and the book N.Shell, Topological fields and near valuation. In Chap. 3 (The lattice of ring topologies) you find that the field (resp. ring, group) topologies form a complete lattice (with the same Sup and different meets).

On a algebraically closed field $K$ of characteristic 0, consider the collection of all involutions (equivalently, the set of all real closed subfields $R$ whose algebraic closure $R(i)$ in $K$ is $K$). Each such element gives, like in the case of the real numbers inside the complex numbers, a absolute value (with values in $R$) on $K$ whose topology is that of $R^2$. The sup of all such topologies is a "intrinsic" field topology on $K$ (quite analogue to the finest locally convex topology on a real vector space), but I know no explicit study of it (is this infinite sup of minimal field topologies discrete? It should be not reducible to a finite sup of such valuation topologies, hence not locally bounded).

In general, embed a field (of characteristic 0) in its algebraic closure (or a formally real field in its real closures, one for each ordering, and consider the sup of all order topologies).

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  • $\begingroup$ In the topology you mentioned, $K$ has the topology of $R^2$, but what topology is on $R$? $\endgroup$ – Liam Baker Mar 20 '14 at 22:26
  • $\begingroup$ The order topology (with basis the open intervals for the unique field order of a real closed field), i.e. the topology of the "metric" (with generally non-archimedean values, in $R$ itself) of its "absolute value" $|x|=\mbox{Max}(x,-x)$. Note that even the complex field $K$ has infinitely many non-isomorphic such non-archimedean $R$ with $R(i)=K$. $\endgroup$ – user46855 Mar 20 '14 at 23:03

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