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5 votes
0 answers
144 views

Is there a good notion of higher-rank archimedean norm?

Let $K$ be a field. I think I know what a norm (archimedean or not) $|-| : K \to \mathbb R_{\geq 0}$ is. In the case where the norm is nonarchimedean, it's equivalent to the data of a valuation of ...
Tim Campion's user avatar
19 votes
2 answers
566 views

Ostrowski's Theorem for topological rings?

Ostrowski's theorem classifies all absolute values on a number field $K$. Questions: More generally, can one classify all Hausdorff topologies on $K$ making $K$ into a topological field? In ...
Tim Campion's user avatar
4 votes
1 answer
800 views

Is the integral closure of a valuation ring in a finite separable extension of its fraction field étale?

Let $K$ be a field endowed with a rank (height) one valuation with completion $\hat{K}$, which is not discrete. Let $R$ be the valuation ring of $K$. Let $L \subset \hat{K}$ be a separable finite ...
userm's user avatar
  • 41
3 votes
1 answer
136 views

Rank 1 valuations that are not discrete on finite transcendental extensions of the rationals

Suppose $K=\mathbb{Q}(X_1,\dots,X_n)$ is a purely transcendental extension of the rationals on finitely many indeterminates. Can anyone give an example of a rank $1$ valuation on $K$ that fails to be ...
Harry Gindi's user avatar
  • 19.6k
6 votes
2 answers
563 views

If the discriminant of a binary quadratic form has high valuation, is the form "almost a square".

For a binary quadratic form $ax^2+bxy+cy^2$ over a field (characteristic not 2), the discriminant $b^2-4ac$ is 0 if and only if that form is the square of a linear form. I am curious about an ...
Daniel Erman's user avatar
  • 2,955
3 votes
1 answer
1k views

Does totally ramified extension really exist?

The answer is certainly "Yes", but this is the problem I met in Algebraic Number Theory by Neukirch. I guess that I must be doing something wrong, since otherwise I will get the statement "There are ...
Li Zhan's user avatar
  • 95
2 votes
1 answer
399 views

Quotient field extension for an incomplete DVR

Let $R$ be a DVR with maximal ideal $xR$, and assume that $R$ is not complete in the $xR$-adic topology. Let $\hat{R}$ be the completion of $R$ in the $xR$-adic topology. Set $K=Q(R)$, the fraction ...
Sean Sather-Wagstaff's user avatar