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Questions tagged [dedekind-domain]

In abstract algebra, a Dedekind domain or Dedekind ring, named after Richard Dedekind, is an integral domain in which every nonzero proper ideal factors into a product of prime ideals.

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Lang's elementary divisors theorem is wrong

The problem comes from Serge Lang's Algebraic Number Theory: Proposition 27 Elementary divisors theorem. Let $M$ be a non-zero finitely generated projective module over a Dedekind ring $A$. Then ...
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Steinitz Isomorphism Theorem for Non-Dedekind domains

Fix a Dedekind domain $R$ and fractional ideals $I, J$. It's a classical result by Steinitz that $I\oplus J \cong R \oplus IJ$ as $R$-modules. Question Does this still hold for non-Dedekind 1-...
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Ideal Class Group of $\mathbb{Q}(\sqrt{2}+\sqrt{3})$

I'm trying to compute the ideal class group of $\mathbb Q(\sqrt{2},\sqrt{3})$, and I would like to know if my calculations are right and if I could improve my arguments. Let $K=\mathbb Q(\sqrt{2},\...
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Inverse of squares of fractional ideals

I am thinking about the following problem: Given a fractional ideal $\mathfrak{a}$ in the Dedekind domain $R$ with field of fractions $K$, its inverse is defined to be $\mathfrak{a}^{-1} = \lbrace x \...
SirToby25's user avatar
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In Dedekind domain, fractional ideals are invertible. How can we precisely compute the inverse of a given ideal in Dedekind domains [closed]

One approach I am looking at is in Dedekind domains, for any given ideal I, there exists an ideal J such that IJ is principal. We can compute this J and find the inverse of the product which is a ...
Eshan Garg's user avatar
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Is an integral domain with every ideal a product of prime ideals necessarily a Dedekind domain?

Question: Is an integral domain in which every proper ideal is a product of prime ideals necessarily a Dedekind domain? Please provide a reference. The first two sentences of the Wikipedia page for ...
jeff honky's user avatar
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Chinese Remainder Theorem and ideals generated in localizations

In Milne's notes on algebraic number theory (https://www.jmilne.org/math/CourseNotes/ANT.pdf), on page 51, Corollary 3.14 and 3.15 both used the argument "use Chinese Remainder Theorem and look ...
spiderchips's user avatar
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Counterexample of Divisibility of Ideals with Product of Ideals

Given a commutative ring $R$ with unity, we define for $I,J\subseteq R$ ideals $I\ \vert\ J\iff I\supseteq J$ $IJ=\{\sum_i a_ib_i:a_i\in I, b_i\in J\}$ For every commutative unitary ring $R$ it ...
tripaloski's user avatar
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Unique unramified ideal implies that the ramification index is equal to the degree of field extension in a galois extension

Given a Galois extension $K \supseteq \mathbb{Q} $, prove that if there is only one unramified prime number $p$ over $K$ then there is only one prime ideal $\mathfrak{p} \subseteq O_K$ containing $p$ ...
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Can polynomial rings be classified as a specific type of ring like algebraic integers are classified as Dedekind domains??

In algebraic number theory, we would like to study rings of algebraic integers but sometimes they are not PIDs and thus they don't possess good properties. Because of this, we have introduced the ...
Ubik's user avatar
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Is the integral closure of an integral domain a dedekind domain?

I am reading through Milne's number theory notes, and he shows that the ring of integers of a field is a dedekind domain. He also shows that the integral closure of a dedekind domain A in a finite ...
Ray James's user avatar
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Ad hoc proof of the fact that the localization of a ring of integers at a nonzero prime ideal is a PID

(Lot and lot of) Context. I am preparing lecture notes for a course on modules I will teach next year, and I was hoping for finding a "not-too-complicated" proof of the fact that an ideal of ...
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Dedekind domain, non PID, localization over maximal ideals.

I was going through the notes, where these two examples on Dedekind domain I couldn't prove. I don't know much about Dedekind domains. Can anyone answer or at least provide some materials and hints so ...
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Tor-dimension of $A/(a)\otimes_k B$, where $A$ and $B$ are Dedekind domains

Let $A$ and $B$ be two Dedekind domains which contain a field $k$ which is algebraically closed in both $A$ and $B$. Let $a$ be a non zero element in $A$. What is the Tor-dimension of $A\otimes_k B$? ...
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Prime of order is regular iff its decomposition in the normalization is trivial.

It's from a statement in Algebraic Number Theory by Neukirch, page 92. Example 5 "One can show..." Let ${o}$ be a one-dimensional noetherian integral domain and $\tilde{o}$ be its ...
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$M$ $\mathscr{R}$-submodule complete lattice $\iff$ $aL \subset M \subset a^{-1}L$, $a \in \mathscr{R}$

My question may be nestled in the sense that there might be confusion about several notions, please let me know if it is the case. In "The Arithmetic of Hyperbolic 3-manifolds" by Maclachlan ...
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Show that $(p, \alpha)$ is a non-invertible ideal in the quadratic order $\mathcal{O}=\mathbb{Z}[\alpha]$

Let $K=\mathbb{Q}(\alpha)$ be an imaginary quadratic field. Suppose $\mathcal{O}=\mathbb{Z}[\alpha]$ is a non-maximal order such that a prime $p$ divides its conductor. We might take $\alpha=\frac{d+\...
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Equivalence of notions of localization for Dedekind domains

Let $\mathcal{O}$ be a Dedekind domain with ring of fractions $K$. Given a cofinite set of primes $X\subseteq \operatorname{Spec}(\mathcal{O})$, we can form the localization with respect to the ...
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Why is there a fractional ideal $\mathfrak b$ such that $\mathfrak a\mathfrak b=\mathfrak o$?

For personal interest, I've been working through the exercises in Algebra(by Serge Lang) to create my own solution. Then I came across an exercise on Dedekind rings. The exercise is as follows: Prove ...
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Question about Dedekind Domains and Valuations

I've been trying to prove this claim that I think is true, but I'm getting more and more sure it's not true. Here's the set up: Let $\mathcal{O}$ be a Dedekind domain, and $K$ its field of fractions. ...
littleman's user avatar
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Inverse of integral ideal in a Dedekind domain is a specific fractional ideal

Let $R$ be a Dedekind domain with fraction field $k$. Incan show that if $p$ is a prime ideal, then $p^{-1}=\{x\in k \mid xp\subset R \}$ satisfies $pp^{-1}=R$. Moreover, if $I$ is any integral ideal ...
Victor Gustavo May's user avatar
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Question regarding divisibility with prime ideals

While I was reading this paper Lenstra page 15, for d=4729494, he says $2162+\sqrt d$ is divisible by the cube of the prime ideal $(5,2+\sqrt d)$. Can someone please help me how this works? I see that ...
John Bull's user avatar
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A definition for "non-congruency" of two ideals module another, in a ring of integers of a number field.

Let $F$ be a number field and let $O_F$ be its ring of integer. I am looking for a definition for the "non-congruency" of two ideals of $O_F$ module another, in an attempt to generalize ...
Captain Haddock's user avatar
1 vote
1 answer
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Noetherian $1$-dimensional domain with ideal generated by more than $2$ elements

It is a classical theorem that in a Dedekind domain, every ideal can be generated by two or fewer elements. I was wondering what goes wrong if we omit the integrally closed property. I thought about, ...
Anon's user avatar
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Doubts about this lemma (Fractional Ideals and Projective Modules)

This is from Conrad's notes. I agree that we can promote $\frak{a}$ and $\frak{b}$ from fractional ideals to integral ideals by multiplying by some element in $A$. i.e. For each prime ideal $\frak{p}$...
daruma's user avatar
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Neukirch's Proof of $\dim_k (\mathfrak{O}/\mathfrak{pO}) = [L:K]$

The setup is as follows. Let $\mathfrak{o}$ be a Dedekind domain, $K$ its field of fractions, and $L/K$ a finite separable field extension with $n = [L:K]$. Furthermore, let $\mathfrak{O}$ denote the ...
stoic-santiago's user avatar
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Cox's exercise 5.1 on (eventually) proving ring of integers is a dedekind domain

I am reading Cox's "Primes of the Form $x^2 + ny^2$", and am on Chapter 5 (it's a speedrun from number fields to Hilbert's class field). I am attempting exercise 5.1, and I have done some ...
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Coordinate Rings and Dedekind Domains

I am relatively new to ring theory, though I have an idea of what things kind of are. I have been learning about Dedekind domains, which are integral domains which are Noetherian, integrally closed ...
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Theorem $17$, Chapter $3$ (Marcus' Number Fields): Every ideal in a Dedekind domain is generated by at most two elements

I understand that similar questions have been asked before, but I am looking for an explanation of certain steps in Marcus' proof of the same, in Theorem $17$, Chapter $3$ of Number Fields. I shall ...
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335 views

Why is the class group torsion?

While wandering through math stack exchange I found an interesting question, namely this one: Why are the algebraic integers a Bezout Domain? (Found here: Is there an elementary way to prove that the ...
Jabberwocky's user avatar
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$I^{-1}M \cong Hom_R(I,M)$, $R$:Dedekind domain, $I$:fractional ideal of $R$, $M$:torsion free $R$module [duplicate]

Let $R$ be a Dedekind domain, let $I$ be fractional ideal of $R$. Let $M$ be torsion free $R$module. Then I want to prove $I^{-1}M \cong Hom_R(I,M)$. Let $ \phi : I^{-1}M→Hom_R(I,M)$ be given by $x→\...
Poitou-Tate's user avatar
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1 answer
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Degree of a separable extension of a Dedekind ring's quotient field

I'm studying "Algebraic number theory" written by S. Lang. In proposition 21 of chapter 1.7 it says if $A$ is a Dedekind ring and $K$ its quotient field, for a separable extension $L/K$, $[L:...
nafise modaresi's user avatar
1 vote
1 answer
67 views

the sum of the products of ramification degrees and relative degrees

I am reading Algebraic Number Fields by Gerald Janusz and I get confused about the part in the picture below. Consider two Dedekind domains $R\subset R'$ with quotient fields $K\subset L$. Let $p$ be ...
Ja_1941's user avatar
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Is the ring of formal power series $\mathbb K[[x_1, \dots, X_n]]$ in $n$ indeterminates over a field a PID?

Is the local ring of formal power series $\mathbb K[[x_1, \dots, X_n]]$ in $n$ indeterminates $x_1, \dots, x_n$, $n>1$, over a field $\mathbb K$ a principal ideal domain (PID)? This is true when $n=...
Flavius Aetius's user avatar
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1 answer
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Splitting field of $x^4+2$

While learning Galois theory, I tried to construct a splitting field for the polynomial $x^4+2$ over $\mathbb{Q}$, but I am terribly stuck. Since $x^4+2$ is irreducible by Eisenstein's criterion, I ...
Leonard's user avatar
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1 answer
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Can the decomposition of a principal ideal in a Dedekind domain contain a non-principal ideal as a factor?

If $I= \prod_{P_i\in Spec(R)}P_i$ is the decomposition into prime ideals of a principal ideal $I$ in a Dedekind domain $R$, can one of the $P_i$ be a non-principal ideal? I guess it can’t, but I don’t ...
tjdominic's user avatar
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1 answer
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Solving the equation $xy=z^n$ in a Dedekind domain

Let $xy=z^n$ where $x$, $y$ and $z$ belong to a Dedekind domain $R$, with $n>1$, and $(x,y)=1$. We can also assume that the ideal class group of $R$ is torsion-free. Then I’d like to show that $x=...
oc89's user avatar
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2 votes
1 answer
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Ramification Groups - explicitly embedding the first factor group (Marcus Number Field; Chapter 4 Exercise 21)

I've been working through some of the problems of Marcus' Number Fields and am stuck on a problem relating to ramification groups. Let $K$ be a number field, $L$ is a normal extension of $K$, $G$ is ...
tildedave's user avatar
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$(\alpha, \beta)^n=(\alpha^n, \beta^n)$ in Dedekind domains

It is a sometimes useful lemma that if $\mathfrak{a}=(\alpha, \beta)$ is an ideal of a Dedekind domain $A$, then $\mathfrak{a}^n=(\alpha^n, \beta^n)$. Of course, this is easy to prove, but I'd like to ...
mxian's user avatar
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5 votes
1 answer
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Proof of local definition of Dedekind Domain without using unique factorization of ideals

We say that an integral domain $A$ is a Dedekind domain if: $A$ is Noetherian, $A$ is integrally closed, $\dim A = 1$ (in other words, every prime ideal is maximal). I would like to show that ...
stillconfused's user avatar
1 vote
1 answer
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If $I$ is an invertible ideal, then there is $\alpha$ with $N(\alpha I^{-1})$ coprime with $N(I)$

Let $\mathcal{O}$ be an order inside a quadratic number field (not necessarily maximal). I want to show that if $I$ is an invertible $\mathcal{O}$-ideal, then there is $\alpha \in I$ such that $N(\...
MathUser123's user avatar
1 vote
0 answers
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Defining invertible ideal as projective module of rank one contained in field of fraction

For a Dedekind domain $A$ such that $\operatorname{frac}(A)=K$, we define invertible ideals to be $I\hookrightarrow K$ and rank one projective modules over $A$. (I think this definition is motivated ...
mathemather's user avatar
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2 answers
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Divisibility of Ideals [closed]

I'm an undergraduate student, who's in my final semester in university. I have a research project, but the advisor isn't the best. He said why won't we develop the notion of divisibility of ideals in ...
Boud's user avatar
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$M_p \cong N_p$ for all prime ideals $p$, but $M \not \cong N$

I am asked to find an example of a ring $R$ and $R$-module $M$ and $N$ such that $M_p \cong N_p$ for all prime ideal $P$ in $R$ but $M$ is not isomorphic to $N$. My idea is as follows: Recall that if $...
slowspider's user avatar
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1 answer
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Prove Set is a Dedekind Cut

Consider: Let $y \in \mathbb{R}. Q_y = \{q \in \mathbb{Q} \mid q < y\}$. I would like to prove that $Q_y$ is a Dedekind cut, following the definitions below. $d \neq \mathbb{Q}$ and $d \neq \...
ENV's user avatar
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1 answer
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Prove Set is Dedekind Cut

Let $G$ be a Dedekind cut. Show that $H = \{x \in \mathbb{Q} : $There exists $a \in Q_{>0}$ such that $-x-a \not \in G \}$ Prove that $H$ is a Dedekind cut. I'm not entirely sure where to begin ...
ENV's user avatar
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general theory of ideal class groups

I'm looking for good references for ideal class groups. I have studied ideal class groups in two different context; in Algebraic number theory and in Quaternion Algebra (as they have connections with ...
Andy's user avatar
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Number of ideal classes of $\mathbb{Z}\left[\frac{1+\sqrt{65}}{2}\right]$ [duplicate]

Let $K=\mathbb{Q}(\sqrt{65})$, whose number ring is $\mathcal{O}=\mathbb{Z}\left[\frac{1+\sqrt{65}}{2}\right]$. I know that the number of ideal classes in $\mathbb{Z}\left[\frac{1+\sqrt{65}}{2}\right]$...
kubo's user avatar
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2 votes
1 answer
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Show that $aI=bJ$ where $I,J$ are ideals in $\mathcal{O}_K=\mathbb{Z}[\sqrt{-6}]$

Let $K=\mathbb{Q}(\sqrt{-6})$, and therefore $\mathcal{O}_K=\mathbb{Z}[\sqrt{-6}]$. I have already proved that $[I]=[J]$, where $I=(2,\sqrt{-6})$ and $J=(3,\sqrt{-6})$. Now I want to find $a,b \in \...
kubo's user avatar
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1 vote
0 answers
162 views

Inert prime ideal in a given field extension

Let $A$ be a Dedekind domain, $K$ its field of fractions. Let $L/K$ be a finite separable extension and $B$ its ring of integers. Further, let $\theta \in B$ be an integral primitive element of $L$ ...
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