For questions related to valuation functions on a field.

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4answers
666 views

Examples of Non-Noetherian Valuation Rings

For valuation rings I know examples which are Noetherian. I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind? I am very ...
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1answer
61 views

Simple property of a valuation on a field

Let $F$ be a field and $v:F\rightarrow G\cup\{\infty\}$ be a valuation on $F$ so $G$ is a totally ordered abelian group with $\infty$ having the properties $\infty+\infty=g+\infty=\infty+g=\infty$ and ...
0
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1answer
247 views

Discrete Valuation Rings problem 2

An order function on a field $K$ is a function $\phi:K\to \mathbb{Z} \cup {\{\infty}\}$ satisfying: i) $\phi(a) = \infty$ if and only if $a=0$. ii) $\phi(ab) = \phi(a) + \phi(b)$. iii) ...
0
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1answer
47 views

Extension of scalars and completions

Suppose that $A$ is a Noetherian regular (added later) local domain. Moreover $\widehat A$ is $\mathfrak m$-adic completion $\widehat A$ w.r.t the maximal ideal and $K$ is the fraction field of $A$. ...
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4answers
515 views

Why does the equation $x^2-82y^2=\pm2$ have solutions in every $\mathbb{Z}_p$ but not in $\mathbb{Z}$?

I have been working on an exercise in H. P. F. Swinnerton-Dyer's book, A Brief Guide to Algebraic Number Theory. The question is like this: Show that $x^2-82y^2=\pm2$ has solutions in every ...
6
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0answers
106 views

A slightly stranger Hensel's Lemma

I'm trying to understand the solution to this problem. It came up when doing some revision. It is essentially to show that the conclusion of Hensel's Lemma holds if we have take a valuation ring $R$ ...
5
votes
2answers
728 views

Why is the restricted direct product topology on the idele group stronger than the topology induced by the adele group?

I am confused by the saying that the restricted direct product topology on the idele group is stronger than the topology induced by the adele group. And perhaps this is elaborated by the following ...
8
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1answer
325 views

Two discrete valuation rings, one contained into another

Let $A$ and $B$ be discrete valuation rings of the same field of fractions. Suppose $A \subset B$. Then $A = B$? I came up with this problem when I was reading van der Waerden's Algebra. The ...
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1answer
627 views

How do we find prime ideals of a ring of integers of a number fileld?

For example for $F=Q(\sqrt{-5})$. the ring of integers of $F =Z[\sqrt{-5}]$.(since $-5\equiv3 \pmod 4$) but how can we determine prime ideals of this? and another problem is the corresponding ...
7
votes
1answer
140 views

A valuation-like function $w: \mathbb{N}^{+} \rightarrow \mathbb{N}$ is a $p$-adic valuation?

This question is a variant of problem 4, pg. 21, from Birkhoff and Maclane, A Survey of Modern Algebra. Given a function $w: \mathbb{N}^+ \rightarrow \mathbb{N}$ that behaves like a valuation ...
7
votes
2answers
222 views

How many absolute values are there?

My question is the following: Are there algebraic norms on the fields $\mathbb{R}, \mathbf{Q_p}$ ''other'' than the absolute value, respectively $|\cdot|_p$? Now phrasing more precisely: If generally ...
6
votes
0answers
120 views

Ramification index of infinite primes

I am reading Neukirch's Algebraic Number Theory. On page 184, Chapter 3, Neukirch defines the ramification index of infinite primes as follows: For a finite extension $L/K$ of number fields, and an ...
4
votes
2answers
201 views

Is every local ring a valuation ring?

Is every local ring a valuation ring? I know the answer is no and the first example comes to my mind was as following (I started with smallest fields, as $\mathbb{Z}_2$ and $\mathbb{Z}_3$ are ...
4
votes
1answer
189 views

A question on valuation overrings of a PID

Let $A$ be a PID and let $K$ be its quotient field. Let $V$ be a valuation ring of $K$ containing $A$ and assume $V\neq K$. Show that $V$ is a local ring $A_{(p)}$ for some prime element $p$. I ...
3
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1answer
33 views

Is the result true when the valuation is trivial and $\dim(X)=n$?

Here I proved the following result: Proposition: Let $(K,|\cdot|)$ be a valued field with non-trivial valuation and let $X$ be a vector space over $(K,|\cdot|)$. Two norms $p_1,p_2$ on $X$ are ...
2
votes
2answers
79 views

Valuations, Isomorphism, Local ring

Let $x \in \mathbb{Q}, \, x \neq 0$, such that $x=p^r \frac{a}{b}, \quad a,b, r \in \mathbb{Z}, \quad p \nmid a, \quad p\nmid b$. Let $v_p(x):=r$ and $v_p(0):= \infty$. Also, $$\mathcal O_p= \left\{ ...
2
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1answer
212 views

Formal power series ring over a valuation ring of dimension $\geq 2$ is not integrally closed.

I recently tried exercise 10.4 in Matsumura's Commutative Ring Theory, but got stuck. The question is: If $R$ is a valuation ring of Krull dimension $\geq 2$, then the formal power series ring ...
2
votes
1answer
181 views

Existence of an element of given orders at finitely many prime ideals of a Dedekind domain

Let $A$ be a Dedekind domain. Let $P$ be a non-zero prime ideal of $A$. Let $\alpha \in A$. Let $k$ be a non-negative integer. If $\alpha \in P^k$ and $\alpha\notin P^{k+1}$, we write $v_P(\alpha) = ...
1
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1answer
52 views

Localization of a valuation ring at a prime is abstractly isomorphic to the original ring

Let $A$ be an integral domain with quotient field $K$. We say that $A$ is a valuation ring if for any $0 \neq x \in K$, $x$ or $\frac{1}{x}$ lies in $A$. Then $A$ is necessarily a local ring. If ...
4
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0answers
171 views

Intuition behind “Non-Archimedean” — two senses of “non-archimedean”.

There appear to be two senses of the qualifier "Archimedean" for fields. One is for ordered fields, and one is for "valued fields" (fields with an absolute value function defined). In the first case, ...
3
votes
0answers
135 views

Archimedean places of a number field

Let $K$ be a number field with an Archimedean absolute value $|\cdot |$ and let $\bar{K}$ be the completion of $K$ wrt this valuation. Then $\bar{K}\cong \mathbb R $ or $\mathbb C$. My question is: ...
3
votes
1answer
61 views

Valuation rings and total order

Let $K$ be a field and $\mathcal{O}$ be a valuation ring of $K$ (So $\mathcal{O}\subset K$ is an integral domain with the property that either $x$ or $x^{-1}$ is in $\mathcal{O}$ for all $x\in K$). ...
3
votes
1answer
49 views

How do we extend the valuation on $K[x]$ to a valuation on $K(x)$?

Given that $v$ is a non-archimedean valuation on $K$, we can extend it to $|\cdot|:K[x]\to\mathbb{R}$ by $|a_0 + a_1x + \cdots +a_nx^n|=\max\left\{|a_1|,\ldots,|a_n|\right\}$. My question is how ...
0
votes
2answers
62 views

Discrete valuation ring and finitely generated submodules

Let $R$ be a Discrete Valuation Ring with fraction field $K$. Will this imply any proper $R$-submodule of $K$ is finitely generated (hence a fractional ideal)? I know $K$ is not finitely ...
0
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1answer
34 views

Can a valuation ring properly contains another valuation ring with the same field of fractions?

Definition of valuation ring: Let $R$ be an integral domain with $frac(R)=K$. Then $R$ is said to be a valuation ring if (1) $R \neq K$ (2) $\forall x \in K, x \in R$ or $x^{-1} \in R$. Now my ...
0
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1answer
110 views

Discrete valuations of a functional field have discrete valuation rings.

Theorem: If $\nu:F\to\mathbb R\cup\{\infty\}$ is a valuation of a functional field, then the set $$\mathfrak O_{\nu}=\{x\in F: \nu(x)\geq 0\}$$ is a local ring with maximal ideal $$\mathfrak ...