2
votes
1answer
68 views

Infinite primes of a number field

Let $K$ be a number field. I know that to each real and to each complex conjugate pair of embeddings of $K$ there corresponds exactly one prime (equivalence class of absolute values) of $K$. How do I ...
0
votes
2answers
34 views

Discrete valuations of the rational numbers

I'm trying to find every discrete valuation on the field of rational numbers. If $a\in \mathbb Q$, we can write $a=p^j\frac{x}{y}$, where $p$ is a prime number and $p\nmid x$ and $p\nmid y$. We can ...
2
votes
1answer
41 views

Places of this extension

I'm reading this book. I'm trying to find the degree of the places of the extension $\mathbb C(X)\mid\mathbb R$. I know the places of the extension $\mathbb R(X)\mid\mathbb R$ and I've already ...
3
votes
1answer
60 views

real closure of an archimedean field

my question is: Is an archimedean field dense in its real closure? I know that in the non-archimedean case, this does not have to be true (e.g., rational fucntions). Thanks!
1
vote
1answer
51 views

When Is there no Local Power Integral Basis?

Let $A$ be a Dedekind domain, $K, L, B$ the usual designations of $A$'s quotient field, a finite separable extension of $K$, and integral closure of $A$ in $L$ respectively. If $\alpha \in B$ ...
1
vote
1answer
22 views

Extensions of nonarchimedean valuations

This is a question from Janusz 'Algebraic Number Theory'. Let $R$ be a DVR with maximal ideal $\mathfrak p=\pi R$. Let $K$ be the quotient field of $R$ and $\mid\cdot\mid_{\mathfrak p}$ the ...
3
votes
1answer
41 views

Examples where there is no power integral basis

Let $A$ be a completion discrete valuation ring with quotient field $K$, $L/K$ finite and separable, and $B$ the integral closure of $A$ in $L$. Let $P, \mathfrak P$ be the unique maximal ideals of ...
4
votes
1answer
90 views

Question on extensions of discrete valuation fields

Let $F$ be a discrete valuation field. Let $L$ be a finite extension of $F$. Let $L=F(\alpha)$ where $\alpha$ belongs to ring of integers of $L$, denoted by $O_L$. Is it always true that ...
2
votes
0answers
30 views

automorphism of $\Bbb{Q_p}$closed [duplicate]

How to show that there does not exist an automorphism of $\Bbb{Q_p}$ except identity. Please help. Is any automorphism of $\Bbb{Q_p}$ already continuous.?
1
vote
1answer
85 views

Question on complete discrete valuation field.

Let $F$ be a complete discrete valuation field and $f(X) = X^n + a_{n-1}X^{n-1} +\cdots+ a_0$ is an irreducible polynomial over $F$. How to show that a) $ v(a_0) > 0$ implies $v(a_i) > 0$ for ...
2
votes
0answers
38 views

The number of abolute value on $\mathbb{Q}(\sqrt{2})$

Let $|.|$ be the usual absolute value on $\mathbb{Q}$. The number of absolute value on $Q(\sqrt{2})$ extending |.| is 2 since $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ in $\mathbb{R}[x]$. Let ...
4
votes
1answer
42 views

Why is the order of a prime element well-defined?

This is in relation to the $p$-adic valuation on the field of fractions $F$ of an integral domain $D$. The idea is that for each $x \in F$ there is a unique maximal $k$ such that $x = p^k u v^{-1}$ ...
0
votes
0answers
36 views

Showing that a set is Discrete Valuation Ring

An order function on a field $K$ is a function $ϕ:K→Z∪{∞}$ satisfying: i) $ϕ(a)=∞$ if and only if $a=0$. ii) $ϕ(ab)=ϕ(a)+ϕ(b)$. iii) $ϕ(a+b)≥min(ϕ(a),ϕ(b))$. Show that R=$\{z∈K∣ϕ(z)≥0\}$ is a DVR ...
0
votes
0answers
18 views

A valued field is complete iff its ring of intergers is complete

Let $K$ be a field, and let $v$ be a valuation defined on K, and let $O$ be the ring of integers, and $M$ be the maximal ideal. How to show that $K$ is complete (i.e $K$ is equal to its completion) ...
2
votes
1answer
26 views

Valuation rings (confusion with arithmetics)

I am reading Goldschmidt's Algebraic functions and projective curves. From the book: Let $K$ be a field. An integral domain $\mathcal{O}\subset K$ is a valuation ring if for all $x\in K$ either $x$ ...
0
votes
1answer
37 views

Annihilators in discrete valuation rings

Let $A$ be a discrete valuation ring and $M$ be an $A$-module. Let $a \in A$ and $m \in M$ such that $am \neq 0$. Is it true that $\operatorname{Ann}(m) = a \operatorname{Ann}(am)$?
3
votes
1answer
37 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
56 views

Integral Closure in an Unramified Extension is Generated by a Single Element

Let $R$ be a discrete valuation ring with quotient field $K$, and $L/K$ a finite separable extension which is unramified over $K$. Also suppose that $K$ is complete with respect to the valuation of ...
1
vote
1answer
42 views

Subring of a field with a discrete valuation is a euclidean domain

I am confused by the following problem in Aluffi's Algebra Chapter Zero... A discrete valuation on a field $k$ is a surjective group homomorphism $v : k^* \to (\Bbb Z,+)$ such that $v(a + b) \geq ...
0
votes
1answer
43 views

Help in this proof on DVRs

I'm trying to understand this proof: Anyone could clarify the converse please? I really need help. Thanks a lot
4
votes
1answer
147 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 ...
0
votes
0answers
59 views

Application of Hensel's lemma

Show that the polynomial $\Phi(x)=x^2 -2 \in O(\widehat{\Bbb Q_2})[x] $ has no root in $\widehat{\Bbb Q_2}$, even though $\bar\Phi(x)\in E(\widehat{\Bbb Q_2})[x]$ has a root in $E(\widehat{\Bbb Q_2}) ...
0
votes
1answer
100 views

Non-archimedean valuation on a field

It was a exercise that one of our professors gave it to us and I don't know the solution of it: Suppose that $v$ is a non-archimedean valuation on the field $F$ and $o(v)$ is the valuation ring of ...
3
votes
2answers
115 views

Finding a Uniformizer of a Discrete Valuation Ring

Suppose I have a discrete valuation ring. Then what are some techniques for explicitly finding a uniformizer? I'm especially interested in situations where the ring is given similarly to the following ...
4
votes
1answer
43 views

Discrete valuation on a field - equivalent statements

I have a question and I am stuck, although it should not be too difficult. We consider $K$ a field, $v$ a discrete valuation on $K$ and $O=\{x \in K:v(x)\geq 0\}$ the valuation ring of $v$. Let ...
7
votes
1answer
122 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 ...
2
votes
2answers
115 views

Proving that a discrete valuation-like function $w: \mathbb{Z}\backslash\{0\} \rightarrow \mathbb{N}$ is a $p$-adic valuation

This problem is from Birkhoff and Maclane, A Survey of Modern Algebra, pg 21, problem 4*. Given a function $w: \mathbb{Z}\backslash\{0\} \rightarrow \mathbb{N}$ that behaves like a discrete ...
1
vote
2answers
41 views

Show $m_v$ is maximal in $R_v$, valuation ring ($v:K\rightarrow G\cup\{\infty\}$ a valuation on $K$)

Let $K$ be a field, $G$ a totally ordered group and $v:K \rightarrow G\cup\{\infty\}$ be a valuation on $K$. I am trying to show that $m_v=\{k \in K: v(k)>0\}$ is a (the?) maximal ideal in ...
1
vote
1answer
43 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 ...
3
votes
1answer
62 views

Regarding valuation on function field

Notations: Let $k$ be a field of characteristic $\neq 2$, $k(X)$ be a function field. Suppose $p(X)\in k[X]$ is an irreducible polynomial and $\alpha$ be a root of $p(X)$. Let ...
0
votes
1answer
70 views

discrete valuation ring

I am struggling to understand the proof of the following proposition Let $A=\{x\in K|v(x)\ge 0\}$ for a field $K$ be a discrete valuation ring. Let $t\in A$ s.t. $v(t)=1$. Then any element $x\in A$ ...
3
votes
2answers
137 views

Learn about valuations, valuation rings, value group

I am reading a paper for a summer research project (Example of an interpolation domain ). I am unfamiliar with some of the terms used here and I have tried searching on google for definitions but I am ...
6
votes
1answer
110 views

Valuations on a field and ramification

For $K\subseteq L$, where $L$ is a finite number field extension of $K$, we consider $p\subset R_K$ and $p'\subset R_L$ where $p'$ lies over $p$, where $R_K$ is the ring of integers of $K$ and $R_L$ ...
1
vote
1answer
197 views

An application of the Weak Approximation theorem - Artin-Whaples Approximation Theorem

Let us recall the weak approximation theorem from Valuation theory in Algebraic Number Theory. Let K be a field, and let $|\cdot|_{1},\cdots, |\cdot|_n$ be pairwise non-equivalent nontrivial ...
3
votes
2answers
230 views

What is a “normalized valuation” corresponding to a valuation ring?

I encountered the phrase "normalized valuation" similar to the following: Let $A_i$ be the valuation ring $k[x_1,...,x_n]_{\langle x_i\rangle}$ and $v_i$ be the normalized valuation defined by ...
5
votes
3answers
595 views

Algebraic Closure of Puiseux Series

Using the construction $R_N = K[t^\frac1N]$ $L_N = Quot(R_N)$ and $P = \bigcup_{N\in \mathbb{N}} L_N$ one automatically gets that the puiseux series are a field. Nevertheless they are also an ...
13
votes
1answer
337 views

Why is $\mathbb{C}_p$ isomorphic to $\mathbb{C}$?

I know that two closed fields of caracteristic $0$ and uncountable are isomorphic iff they have the same cardinality. But I don't know why $\mathbb{C}_p$ has the same cardinality as $\mathbb{C}$. Can ...
5
votes
1answer
145 views

Lang's “General Integrality Criterion”

Theorem 3.7 in the chapter on ring extension on page 352 of the latest edition of Lang's "Algebra" appears redundant in its phrasing to me. Specifically, if $g_s$ is a polynomial of total degree ...
9
votes
2answers
320 views

Concrete examples of valuation rings of rank two.

Let $A$ be a valuation ring of rank two. Then $A$ gives an example of a commutative ring such that $\mathrm{Spec}(A)$ is a noetherian topological space, but $A$ is non-noetherian. (Indeed, otherwise ...