18 votes
Accepted

Does every map from a noetherian ring to a valuation ring factor through a DVR?

The answer is no. I give two examples, which are standard non-dvr points on the Riemann-Zariski space of the plane. (1) Let $R=k[x,y]$ and let $V\subseteq k(x,y)$ be the subring consisting of rational ...
16 votes
Accepted

Extension of 2-adic valuation to the real numbers

No. The important thing to know is that, if $K \subseteq L$ is a field extension and $v: K \to \mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by ...
10 votes
Accepted

Is every field the residue field of a discretely valued field of characteristic 0?

Yes, by Hasse-Schmidt ("Die Struktur diskret bewerteter Koerper", Crelle's Journal, 1934) for any field $k$ of characteristic $p$ there exists a strict Cohen ring $A$, which is a Noetherian, ...
  • 7,714
7 votes
Accepted

Classifying Space of "Valuation Ringed Spaces over a Topos"

Since the axioms describing what a valuation ring can be put as what's called geometric sequents [*], by the fundamental theorem on classifying toposes, there is a topos $T_{val}$ with precisely the ...
7 votes
Accepted

Valuation Rings and Ultrafilters

The similarity has nothing to do with boolean algebras, but with orders in general. Filters can be defined for every partial order: A subset $\Phi$ of a poset $\Lambda$ is a filter if $\Phi\neq\...
6 votes

Torsors over complete local fields

The isomorphism classes of torsors for $\mu_{n}$ over $R$ and $k$ are classified by the fppf cohomology groups $H^{1}(R,\mu_{n})$ and $H^{1}% (k,\mu_{n})$. From the Kummer sequence, one sees that $H^{...
  • 61
5 votes

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

EDIT I may have overlooked the assumption that $L$ is contained in $\hat{K}$. In general, if $v$ is a valuation of $K$ and $L/K$ is finite separable then $L \otimes_K \hat{K}$ is reduced and thus ...
5 votes
Accepted

Torsors over complete local fields

I am rewriting my comment as an answer. That is false in characteristic $p$ for torsors for the finite, flat group scheme $\mu_p=\text{Spec}\ \mathbb{Z}[t]/\langle t^p -1 \rangle$ with the usual ...
4 votes
Accepted

Lifting Lang-Steinberg to DVR's in Characteristic 0

Yes, as long as you are careful about the hypotheses, this is a result of M. J. Greenberg [Schemata over local rings II, Section 3, Proposition 3]. You need to assume that $\mathbb{G}$ is smooth over $...
  • 3,518
3 votes
Accepted

What is K+M structure?

A "valuation ring of the form $K+M$" is a valuation ring $V$ with maximal ideal $M$ such that $V$ contains a subring $K$ which is a field and one has $V=K+M:=\{k+m\,|\,k\in K,m\in M\}$. In this case ...
  • 2,684
3 votes

Given a non-field local domain $R$, finding a dominating Valuation ring whose residue field is algebraic/finite extension of the residue field of $R$

Yes, an algebraic extension is always possible. If one applies Zorn's lemma not to $$\{R\subseteq V\subseteq Q(R) \mid V\,\text{local}\}$$ but to $$\{R\subseteq V\subseteq Q(R) \mid V\,\text{local} \...
3 votes

What does "trait" mean?

This means a little segment of line: not a point, but the smallest thing after a point. This also means "feature", which could be another reason to use this for a spectrum. See https://en.wiktionary....
  • 3,437
2 votes

A question about Dedekind schemes and proper morphisms

Injectivity is because $X$ is separated. The locus where two morphisms $S \to X$ agree is a closed subscheme and if it contains the generic point, it's everything. For surjectivity, we can "...
  • 7,142
2 votes
Accepted

Flatness criterion for $I$-adic ring: $I$-torsion free

In Section 7.3 it is assumed that $I$ is the ideal of definition or $R$. It follows that the $I$-adic topology is separated, so (because $R$ is a valuation ring) every nonzero ideal of $R$ contains ...
  • 122k
1 vote

An example of a special $1$-dimensional non-Noetherian valuation domain

This isn't possible in any $1$-dimensional quasi-local domain $D$. If $a,b\in D$ and $b$ is not a unit then consider the multiplicative set $S$ generated by $b$. Clearly $S$ is not disjoint from any ...
1 vote
Accepted

non-archimedean valuations on graded rings

Let $K$ be a field, and set $R=K[x]$ with the usual grading by degree. For each irreducible polynomial $p(x)\in R$, we get a valuation $v_p$ on $R$ given by $$ v_p(f)=2^{-\operatorname{ord}_p(f)}, $$ ...
  • 8,861

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