Linked Questions

0 votes
2 answers
340 views

Questions of Proposition 9.2 from Atiyah's Introduction to Commutative Algebra

Proposition 9.2. Let $A$ be a Noetherian local domain of dimension one, $\mathfrak{m}$ its maximal ideal, $k = A / \mathfrak{m}$ its residue field. Then the following are equivalent: i) $A$ is a ...
Judy Judy's user avatar
5 votes
1 answer
868 views

Intuition behind discrete valuation rings

I'm trying to understand what DVRs are. I have seen two formulations, one in terms group homomorphisms and a discrete valuation function satisfying an axiom of the sort: $$ v(x+y) \geq \text{min} \...
Cathartic Encephalopathy's user avatar
5 votes
1 answer
232 views

Noetherian domain with unique principal prime ideal that is not a DVR

The question is whether such a thing exists. Namely, a discrete valuation ring (DVR), in whatever way you define it, is quite obviously a domain, Noetherian, and has a unique prime element up to ...
Torsten Schoeneberg's user avatar
1 vote
0 answers
73 views

Explanation on discrete valuation rings

Let $A$ be a noetherian domain. We want to prove that if $A$ is normal then, for every prime ideal $p$ associated to a principal ideal of $A$, the localization $A_p$ is a discrete valuation ring. It ...
Dr. Scotti's user avatar
  • 2,523
1 vote
0 answers
599 views

What makes a Principal Ideal Domain a Euclidean Domain?

The definitions of Principal Ideal Domain (PID) and Euclidean Domain (ED) are both from integral domain: A non-trivial ring $R$ with no zero divisors is said to be entire; a commutative entire ring ...
athos's user avatar
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0 votes
0 answers
66 views

DVR is PID proof [duplicate]

I already know that if $R$ is a Noetherian local ring with Krull dimension $1$, then $R$ is DVR if and only if its maximal ideal $\mathfrak{m}$ is principal ideal if and only if every nonzero ideal is ...
one potato two potato's user avatar
3 votes
0 answers
219 views

Every Principal Ideal Domain is a Unique Factorization Domain

I am interested in verifying the existence aspect of the theorem asserting that every Principal Ideal Domain is a Unique Factorization Domain. In the first paragraph, I (think that I) have provided an ...
user74973's user avatar
  • 706
2 votes
1 answer
3k views

Proof of every PID is Noetherian

I saw the proof of this proposition in here, but I have a question about this. Definition of Noetherian ring is that ring is commutative, and every ideal of R is finitely generated, right? Principal ...
종민이's user avatar
  • 105
4 votes
1 answer
3k views

Krull Intersection Theorem

In this proof I know since R is noetherian it can be written as descending sequence of ideals which stabilizes after finite steps. Also I know since R is noetherian implies every ideal is finitely ...
maths student's user avatar
3 votes
1 answer
1k views

Why is a P.I.D. Dedekind domain?

Definition A Noetherian integrally closed domain of Krull dimension $1$ is said to be a Dedekind domain. Since fields are of Krull dimension $0$, fields are not Dedekind domain. However, it is ...
Rubertos's user avatar
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3 votes
1 answer
1k views

The proof of Krull's Principal Ideal Theorem

Theorem: Let $R$ be Noetherian and $P$ be a minimal prime ideal over $(a)$ for some nonunit $a$ of $R$. Then $\operatorname{ht}(P)\leq 1$. My lecture notes prove this as follows. WLOG $R$ is local ...
user3131035's user avatar
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2 votes
1 answer
518 views

Characterization of Discrete Valuation Rings

Let $R$ be a Noetherian local domain with unique maximal ideal $M$. Then I want to show that if every $M$-primary ideal is a power of $M$, then $R$ is a Discrete Valuation Ring. I know I'll be done ...
Nishant's user avatar
  • 9,205
3 votes
2 answers
1k views

How to show that a valuation ring has a unique maximal ideal?

A subring $R$ of a fi eld $K$ is said to be a valuation ring of $K$ if for each $x$ $\in$ $K^{*}$ we have either $x$ $\in$ R or $x^{-1}$ $\in$ $R$. How can I show that the valuation ring has a unique ...
amdandy's user avatar
  • 105
6 votes
1 answer
2k views

Deduce that a Noetherian valuation ring is either a field or a Discrete Valuation Ring.

I'm trying to solve this question from a book and I have already proved 1. Let $R$ be a local domain which is not a field. Suppose that the maximal ideal $M$ of $R$ is principal and satisfies $\cap_{n=...
Mathproof P.'s user avatar
37 votes
4 answers
7k views

Intuitive explanation of Nakayama's Lemma

Nakayama's lemma states that given a finitely generated $A$-module $M$, and $J(A)$ the Jacobson radical of $A$, with $I\subseteq J(A)$ some ideal, then if $IM=M$, we have $M=0$. I've read the proof, ...
Pandora's user avatar
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