For questions about principle ideal domains: rings without zero divisors where every ideal is principle.

learn more… | top users | synonyms

0
votes
0answers
9 views

What can we say about the function $f(x)$ in this case?

Alright, I'm little bit confused about what's happening here to the function $f(x)$, i thought that the formula of $f(x)$, have nothing to do with its behavior or domain. there are two or many ...
0
votes
1answer
41 views

continuous poset w.r.t. Scott topology

I am learning continuous poset by myself. I have conclusion as follows: If $P$ is a continuous poset w.r.t. Scott topology then there is $x\in P$ s.t. for any $y\in P$ and for any open sets $U_x$ and ...
0
votes
2answers
45 views

Looking for help, Aluffi Exercise 5.13, Chapter 6: characterization of PIDs

I quote: "Let $M$ be a finitely generated module over an integral domain $R$. Prove that if $R$ is a PID, then $M$ is torsion-free if and only if it is free. Prove that this property characterizes ...
3
votes
1answer
79 views

Proof for Unique Factorization Domain

Prove that the quotient ring $\mathbb{C}[x,y]/(x^2+y^2-1)$ is a unique factorization domain. I am trying to prove first it is a principal ideal domain. However I am really stuck on this problem
2
votes
1answer
28 views

Questions about ring of smooth functions

First of all, this is a homework problem. Let $C^{\infty}(\mathbb{R})$ denote the ring of smooth functions. Let $I_n$ denote the set of $f\in C^{\infty}(\mathbb{R})$ such that $$f^{(k)}(0)=0, \ 0 ...
1
vote
0answers
45 views

When is a holomorphy ring a PID?

I will use the notation and language of Stichtenoth, Algebraic Function Fields and Codes. Let $F$ be a function field over a finite field $\mathbb F_q$, $S$ a non empty set of places (possibly ...
3
votes
2answers
87 views

PID and finitely generated module

I am trying to prove the following statements: Let $R$ be a PID and $M$ a finitely generated $R$-module. Prove: (a) $M$ is torsion module iff $\operatorname{Hom}_R(M,R)=0$ (b) $M$ is an ...
2
votes
3answers
21 views

Torsion-free module morphism

I am trying to prove the statement: Let $R$ be a PID but not a field and let $M$ be an $R$-module. Then $$ M \space \text{is torsion-free $R$-module} \space \text{iff} \space ...
2
votes
1answer
49 views

PID and module problem

Let $R$ be a principal ideal domain but not a field, and let $M$ be an $R$-module. Show the following: (i) Let $p \in R$ be an irreducible element and $r \in R \setminus \{0\}$. Then $(R/ ...
-3
votes
1answer
52 views
1
vote
1answer
72 views

Challenging problem in Hungerford's $\textit{Algebra}$

Here is a problem from the text $\textit{Algebra}$ by Hungerford, which I seem to be stuck on for quite some time now: Let $A$ be a cyclic $R$-module ($R$ is assumed to be a principal ideal domain) ...
2
votes
2answers
67 views

Does the statement ''PID of dimension $0$ $\Longrightarrow$ field'' actually use Zorn's Lemma?

Everywhere I look seems to blow by the statement that PIDs which are not fields have Krull dimension $1$. This relies on the fact: A PID with Krull dimension $0$ is a field. (*) It seems that the ...
3
votes
1answer
715 views

For which $d$ is $\mathbb Z[\sqrt d]$ a principal ideal domain?

Is there any general idea about for which $d$, $\mathbb Z[\sqrt d]$ a principal ideal domain (PID)? As for example $\mathbb Z[\sqrt{-1}]$ and $\mathbb Z[\sqrt 2] $ are PIDs, but $\mathbb Z[\sqrt{-5}] ...
5
votes
1answer
57 views

Proof that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a PID

How would one prove that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a principal ideal domain (PID)? It isn't a Euclidean domain according to the Wikipedia article on PIDs.
0
votes
0answers
144 views

Homology out of Smith normal form: simultaneous or independent diagonalization?

Let $R$ be a PID and $R^m\overset{A}{\longrightarrow} R^n\overset{B}{\longrightarrow} R^o$ matrices with $BA=0$ and Smith normal forms ...
1
vote
1answer
19 views

What exactly does it mean for a maximal ideal to be unique in a principal ideal domain?

I'm currently reading about PIDs and have come across a question involving maximal ideals which at one point reads "Suppose that a Euclidean domain $R$ had a unique maxima ideal $P$". Does this mean ...
0
votes
1answer
47 views

Suppose that $K$ is a field and that $f$ and $g$ are relatively prime in $K[x]$. Show that $f - Yg$ is irreducible in $K(y)[x]$.

I'm a bit confused of the notation $K(y)[x]$, is that simply $K[y][x]$ so... $K[y,x]?$ Anyways, here's my attempt at trying this before I get stuck. Since $f$ and $g$ are relatively prime, that ...
0
votes
0answers
27 views

Domain of reciprocal of log in complex plane

Ok so what i already know is that the function f(z) = log(z) is undefined when z = negative or zero which is quite a basic concept. Can someone explain the mystery ...
2
votes
1answer
52 views

Principal Ideal Domains with Many Primes of Index 2

By the index of a prime $p$ in a principal ideal domain $R$, I mean the number of cosets of the ideal $(p)$ in $R$ (i.e. $\vert R/(p)\vert$). If I am given a positive integer $m$, I am wondering ...
0
votes
1answer
54 views

$\Bbb Z[x]$ is not a principal domain

I know this is already answered here but I am wondering that if the following way to prove is also correct - let $$f(x) = 4x^2+4x+1$$ $$g(x)=4x^2-1$$ Since this is a UFD so a unique gcd will ...
3
votes
1answer
112 views

Problem with Smith normal form over a PID that is not an Euclidean domain

This is an homework exercise of the Algebra lecture. I need to evaluate the Smith normal form of the following matrix $$A:=\begin{pmatrix}1 & -\xi & \xi-1\\2 ...
0
votes
1answer
23 views

algebraic poset

I learn domain theory and stack in definition of algebraic poset. Recall $P$ is algebraic if for every $x\in P$,the set of compact element $y$ below $x$ is directed and has $x$ as least upper bound. ...
0
votes
1answer
29 views

Principal Ideal Domain Modulo a Power of a Prime Ideal

If $R$ is a PID, $p$ is a prime element of $R$, $R/(p)$ is finite, and $\alpha$ is a positive integer, is it true that $\vert R/(p^{\alpha})\vert=\vert R/(p)\vert^{\alpha}$? I seem to recall seeing ...
1
vote
1answer
38 views

The number 2 in a PID

Let $R$ be a PID. Then $R$ is a commutative ring with multiplicative identity $1$. We can then define $2=1+1$. From here, what is known about $2$ and its prime factors? I suppose this breaks into two ...
0
votes
1answer
40 views

$R$ is a commutative integral ring, $R[X]$ is a principal ideal domain imply $R$ is a field

I've just read a proof of the statement: Let $R$ be a commutative integral ring. If $R[x]$ is a principal ideal domain, then $R$ is a field. In one part of the proof there is a step which I don't ...
1
vote
1answer
50 views

Difference between PID and principal ideal ring

All rings are commutative, associative and with 1. Wikipedia states that the difference between PID and Principal Ideal Ring is that the former has to be integral domain while the latter does not. ...
12
votes
2answers
2k views

Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal

Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal. The forward direction is standard, and the reverse direction is giving me trouble. In particular, I can prove that ...
6
votes
2answers
471 views

Prime elements in $\mathbb{Z}[\sqrt{2}]$

What are the prime elements in the ring $\mathbb{Z}[\sqrt{2}]$? Note that since the ring is a PID (and thus a UFD) then prime = irreducible. Even more, it is Euclidean with respect to the absolute ...
3
votes
2answers
1k views

Every prime ideal is either zero or maximal in a PID.

$(1)$ Let $R$ be a commutative ring with $1\neq 0.$ If $R$ is a PID, show that every prime ideal is either zero or maximal. In many books I have found the proof of the above statement where they ...
5
votes
2answers
222 views

Characterization of primary ideals in a principal ideal domain

On the commutative algebra wiki, a table of properties lists that "for a PID, the primary ideals coincide with the powers of prime ideals." I played around with it, couldn't produce a proof, ...
1
vote
1answer
26 views

Help understanding statement relating to structure of modules over PIDs

Lemma IV.6.11 of Hungerford's Algebra is Lemma 6.11. Let $R$ be a principal ideal domain. If $r \in R$ factors as $r = p_1^{n_1} \cdots p_k^{n_k}$ with $p_1,\ldots,p_k \in R$ distinct primes and ...
5
votes
3answers
1k views

If F is a field, then $F[x,y]$ is a Principal Ideal Domain?

Let $F$ be a field, and $F[x,y]$ be a ring of polynomials in two variables. Is $F[x,y]$ a Principal Ideal Domain? Also show that $F[x,y]/(y^2-x)$ and $F[x,y]/(y^2-x^2)$ are not isomorphic for any ...
3
votes
1answer
93 views

Finitely generated graded modules over $K[x]$

I need some help on this exercise from A Course in Ring Theory by Donald S. Passman The result is supposely similar to the well-known structure theorem in the non-graded case. So let $M$ be a ...
2
votes
2answers
43 views

Is the ideal $(2,x^4+x^2+1)<\mathbb{Z}[x]$ maximal?, principal?

I'm trying to solve the following problem: Let I=$(2,x^4+x^2+1)<\mathbb{Z}[x]$ be an ideal. Is $I$ maximal? Is $I$ principal? Any help would be appreciated.
8
votes
1answer
592 views

Proofs of the structure theorem for finitely generated modules over a PID

I'm looking for different proofs (references or sketch of main ideas) of the structure theorem for finitely generated modules over a PID. If possible, a comparison in terms of clarity, elegance or ...
1
vote
2answers
158 views

Is $\mathbb{C}[x,y] / (y^2-x^3)$ a PID?

First, I'd like to show $\mathbb{C}[x,y] / (y^2-x^3)$ is an integral domain. Then I need to find out whether or not it is a PID. For the first part, I want to show $y^2-x^3 \: | \: fg \implies ...
3
votes
2answers
84 views

module over a quotient of a principal ideal domain

The Statement I suspect the following proposition is well known, but I found no reference. Proposition If $A$ is a principal ideal domain, if $I$ is a nonzero ideal of $A$, and if $M$ is an ...
-2
votes
2answers
32 views

R is a PID, and a is a nonzero nonunit in R. How can we show R/Ra is an injective module over R?

If we use Baer's criterion then it suffices to show that if there exist a map from an ideal $I$ to $R/Ra$ we must find a map $g$ such that $g\circ i=f$.
18
votes
2answers
382 views

Quotient of polynomials, PID but not Euclidean domain?

While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the Euclidean domain page of Wikipedia that $$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$ is ...
2
votes
3answers
104 views

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal. What could be wrong in this approach?

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which ...
3
votes
2answers
57 views

If $N\cap rM=rN$ for all $r\in R$, then is $M=N\oplus K$ for some $K$?

Suppose $M$ is a finitely generated free module over a principal ideal domain $R$, and $N$ a submodule. Why does the condition $N\cap rM=rN$ for all $r\in R$ implies that $M=N\oplus K$ for some ...
11
votes
2answers
374 views

Ring of integers is a PID but not a Euclidean domain

I have noticed that to prove fields like $\mathbb{Q}(i)$ and $\mathbb{Q}(e^{\frac{2\pi i}{3}})$ have class number one, we show they are Euclidean domains by tessalating the complex plane with the ...
2
votes
2answers
346 views

Specific way of showing $\Bbb Z[\sqrt{-d}]$ is not a Euclidean Domain when $d>2$

Is it true that if a ring is not a UFD then it's not a Euclidean Domain? I have a ring $R=\mathbb{Z}[\sqrt{-d}]=\{ a+b\sqrt{-d} \mid a,b \in \mathbb{Z} \}$ where $d$ is a square free integer. I want ...
0
votes
0answers
28 views

Smith normal forms and a math program

I am interested to know the Smith normal form of $4 \times 2$ matrices $M$: The two cases of my interests are: (1). $$M_1= \begin{pmatrix} 3 & 0\\ -5 & 4\\ 4 & -5\\ 0 & 3 ...
2
votes
2answers
58 views

Intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ is not finitely generated.

Consider the subring $\mathbb{Z}[2x,2x^2,2x^3,\dots]\subset \mathbb{Z}[x]$. Then show that the intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ i.e., $I\cap ...
7
votes
2answers
151 views

$M \oplus M \simeq N \oplus N$ then $M \simeq N.$

Let $M$ and $N$ be finitely generated $R$-modules where $R$ principal domain. Show that if $M \oplus M \simeq N \oplus N$ then $M \simeq N.$
2
votes
1answer
21 views

Show that if $R$ is principal, $N $ is pure and $Ann(x+N)= Rd$ then there exists $y \in M$ such that $x+N=y+N$ and $Ann(y)=Rd$

Let $R$ a integral domain and $M$ a $R$-module. A submodule $N$ of $M$ is pure if for all $x \in M$ and $a \in R$ such that $ax \in N$, there exists $y \in N$ such that $ax=ay$. Show that if $R$ ...
5
votes
2answers
88 views

Ring Sandwiched between PIDs

If I have three commutative rings $R \subset S \subset T$, such that $R$ and $T$ are principal ideal domains, will this imply that $S$ itself is a principal ideal domain?
0
votes
2answers
61 views

Show that an integral domain $R$ is principal if and only if every submodule of a cyclic $R$-module is cyclic.

Good morning, I have difficulty with this problem: Show that an integral domain $R$ is principal if and only if every submodule a cyclic $R$-module is also cyclic.
3
votes
1answer
2k views

Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...