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

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44 views

Shortest proof for showing $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID.

I'm looking for an easy proof for that $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID. One proof I know is to show that the field norm is a Dedekind-Hasse norm, but this proof is quite dirty( that it ...
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1answer
83 views

$\mathbb{Z}[x]$ doesn't have principal maximal ideals [on hold]

Prove that $\mathbb{Z}[x]$ doesn't have principal maximal ideals. Please, I need help with this problem. Thanks!
1
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1answer
31 views

What's the importance of invariant factors?

I understand why the following theorem holds: If $R$ is a PID and $M$ is a finitely-generated torsion $R$-module, then there exist $q_1,...,q_s$, non-invertible elements of $R$, such that $q_i ...
3
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0answers
19 views

Existence of alternative basis element in free module over a PID

first question on stack exchange, please let me know if I have made any errors with formatting or in general! :) Let $f_1,f_2, ...,f_s$ be a basis of a free module $V$ over a PID $R$. Suppose that ...
2
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1answer
48 views

Prove that $f=g$

Let $A=K[x]$ be the ring of polynomials over the field $K$. Let $m\in A$ with $\deg(m)\ge1$, and let $I=(m)$ the principal ideal generated by $m$. Let $f,g \in A$ so that $\deg(f)<\deg(m)$ and ...
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1answer
37 views

Show that $(a) + (b)= R$ for $\gcd(a,b) = 1$

The question I am trying to solve it: Let $R$ be a principal ideal domain, $a,b\in R$. Suppose $\gcd(a,b) = 1$. Show that $(a)+(b)=R$. First I have tried to show that $(a)+(b)$ is in R: ...
2
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1answer
48 views

Each proper ideal is a product of prime ideals

$R$ is a commutative ring with unity. If $R$ is P.I.D. I want to show that each of its proper ideal is written as a product of prime ideals. $$$$ Since $R$ is a P.I.D. every ideal is a prime ...
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0answers
32 views

On describing a sort of “well-behaved” subgroups of a free abelian group.

I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finite generated case. Let $M$ be an free abelian group, $N$ a subgroup ...
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2answers
48 views

Principal Ideal using coordinates?

I thought I understood principal ideals but now im stuck... I want to find the elements of the principal ideal $\langle(1,0)\rangle$ in the ring $\mathbb Z_3\times \mathbb Z_3$ with $+_3$ and $*_3$ in ...
2
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1answer
21 views

Let $A$ be a principal ideal domain, and $a,b,d$ elements of $A$. Prove that $d$ is a gcd of $a$ and $b$ if and only if $aA+bA=dA$.

I can prove that $aA+bA=dA$ implies that $d$ is a gcd of $a$ and $b$. I can also prove that $d$ being a gcd of $a$ and $b$ implies that $aA+bA\subset dA$, since $a+b$ is a multiple of $d$. What im ...
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1answer
21 views

Examples of PID's with finitely many maximal ideals

Do you know some examples of principal ideal domains which have finitely many maximal ideals? More generally, do you know how to build such domains? I don't look for fields and discrete valuation ...
3
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2answers
135 views

A Bézout UFD is a PID. [duplicate]

Let $R$ be an integral domain and a Noetherian U.F.D. with the following property: for each couple $a,b\in R$ that are not both $0$, and that have no common prime divisor, there are elements $u,v\in ...
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0answers
42 views

Show that there are finitely many different principal ideals [duplicate]

Let $R$ be a U.F.D. and $0\neq d\in R$. I want to show that there are finitely many different principal ideals that contain the ideal $(d)$. $$$$ We have that $R$ is a U.F.D. iff $\forall r\in ...
2
votes
3answers
121 views

Is $\mathbb{Q}[X,Y]/\left<X^2+Y^2-1\right>$ a PID?

I know that $\mathbb{Q}[X,Y]$ is not a PID. Does this imply that the quotient ring $\mathbb{Q}[X,Y]/\left<X^2+Y^2-1\right>$ is not a PID?
4
votes
1answer
45 views

Is a matrix over a PID similar to its transpose?

We say that two matrices $A,\,B\in M_n(R)$ are similar if there is some invertible matrix $P$ such that $P^{-1}AP=B$. Now, if $R$ was a field (or certainly an algebraically closed field) then it is ...
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0answers
16 views

Showing that the integers localized at a prime, p, is a Euclidean Domain

I want to show that the integers localized at some prime natural number $p$: $$R=\Biggl\{\frac mn \in \Bbb Q ~\Bigg\vert~ m,n \in\Bbb Z,\ n\notin p\Bbb Z\Biggr\}$$ is a Euclidean Domain, but I can't ...
0
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1answer
50 views

Domain of $\ln\left(\frac{6}{6+x-x^2}-1\right)+\arcsin\left(\frac{x+1}{3}\right)$

blob:https%3A//mail.google.com/ea67134d-45a0-4cc0-9ec7-abf6d5a50852 I believe that my first condition is wrong but I don't understand why. Can somebody please help?
2
votes
1answer
43 views

When is the prime spectrum of a ring the finite complement topology on a set?

I have been studying the prime spectrum of different rings recently, and I have noticed that for many "nice" infinite rings, the prime spectrum is precisely the finite complement topology on some set ...
3
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2answers
47 views

Information about Problem. Let $a_1,\cdots,a_n\in\mathbb{Z}$ with $\gcd(a_1,\cdots,a_n)=1$. Then there exists a $n\times n$ matrix $A$ …

I would like to find some information about the following propositions, and unfortunately I haven't been able to find any. Let $a_1,\cdots,a_n\in\mathbb{Z}$ with $\gcd(a_1,\cdots,a_n)=1$. Then ...
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1answer
69 views

Polynomial ring is not PID

I was trying to think of a proof of the following proposition: Let $K$ be a field. Then $R=K[x_1,...,x_n]$ is not a PID for $n>1$. So here's what I've written so far: Suppose $R$ is a PID ...
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0answers
11 views

Ideal generated by two monic polynomials in $\mathbb{Q}[x]$ and the GCD of their degrees [duplicate]

Let $a$ and $b$ be positive integers with GCD $d > 0$. Show that in $\mathbb{Q}[x]$, the ideal generated by $x^a - 1$ and $x^b - 1$ equals the ideal generated by $x^d - 1$. Here is what I have ...
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1answer
52 views

Show that for a field $F$, the polynomial ring $F[x_1, x_2, \ldots, x_n]$ is not a PID for $n>1$.

I want clarification of the following solution: Let $I=(x_1)+(x_2)$ be an ideal of $F[x_1, x_2, \ldots, x_n]$. Then if $I=(f)$ is principal then we must have $f \in F \backslash \{0\}$ since ...
0
votes
1answer
68 views

Decomposition of Free Module over PID

Let $M$ be a free module with rank $2$ over the PID $R$ having basis $B=\{b_1,b_2\}$. I have a few questions regarding the submodule $Rm$, where $m$ is some element of our module $M$. Suppose $m$ is ...
2
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1answer
83 views

Why is the polynomial ring of more than one variable not a PID?

Let $R$ be a field. Show that the polynomial ring $R[x_{1},...x_{n}]$ is not a PID if $n>1$. How do I show this? So far, I've shown that $(2)+(x)$ wouldn't be principal in $\mathbb{Z}[x]$ ...
4
votes
1answer
75 views

Is $\mathbb{Z}_p[\mathbb{Z}_p]$ a PID?

Is $\mathbb{Z}_{p}[G]$ a PID, where $G=(\mathbb{Z}_{p},+)$ is the additive group of the $p$-adics $\mathbb{Z}_{p}$? I am studying a paper where the authors implicitly use that claim, but it is ...
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1answer
43 views

Is $\mathbb{Z}[\mathbb{Z}/(p)]$ a PID?

As the title suggests, I'm interested whether $\mathbb{Z}[\mathbb{Z}/(p)]$ a PID or not. Assume $p$ is prime. My feeling is that it is a PID, since $\mathbb{Z}/(p)$ is cyclic an morally if an ideal ...
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1answer
28 views

Decomposition in a PID

I am working towards understanding the statement and proof of the following theorem: Let $R$ be a PID and $a_1,...,a_r \in R \setminus \{0_R\}$. Then, there are $q_1,...,q_s \in R\setminus ...
5
votes
0answers
132 views

Applications of the Dedekind-Hasse criterion

It is a fact that an integral domain $R$ is a principal ideal domain if and only if there is a Dedekind-Hasse function $|R|\setminus\{0\}\xrightarrow{\ \ \delta\ \ }\mathbb{N}$ on $R$, i.e. a function ...
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1answer
25 views

$M$ is the (possibly infinite) direct sum of its $p$-primary components as $p$ runs over all primes of $R$.

Let $R$ be PID, let $M$ be a torsion $R$-module and $p$ be prime in $R$. Prove that $M$ is (possibly infinite) direct sum of its $p$-primary components as $p$ runs over all primes of $R$. Take $m \in ...
5
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2answers
64 views

Is $\mathbf{Z}[X]/(2,X^2+1)$ a field/PID?

I've been asked to determined whether the following are fields, PIDs, UFDs, integral domains: $$\mathbf{Z}[X],\quad \mathbf{Z}[X]/(X^2+1),\quad \mathbf{Z}[X]/(2,X^2+1)\quad \mathbf{Z}[X]/(2,X^2+X+1)$$ ...
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2answers
60 views

Is it possible to find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i)=(\alpha)$? [closed]

Is it possible to find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i)=(\alpha)$? Is anyone could give me a full explication in ''Answer the question''?
3
votes
1answer
45 views

Prime ideal in R[x] is either principal or $\mathfrak p = (q,f)$

$R$ is a PID and $\mathfrak p$ a prime ideal of $R[x]$. Show that $\mathfrak p$ is principal or $\mathfrak p = (q,f)$ for some $q\in R$ prime and $f \in R[x]$ monic. I can't figure out this ...
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2answers
51 views

Show that I is a principal ideal

Let $\mathbb{Z}[\sqrt{-13}]$ be the smallest subring of $\mathbb{C}$ containing $\mathbb{Z}$ and $\sqrt{-13}$ and let the ideal $I = \left<2,\sqrt{-13}\right>$. Show that $I$ is a ...
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2answers
59 views

Find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i) = (\alpha)$ [closed]

Let $\mathbb{Z}[i] = \{a+bi : a,b \in \mathbb{Z}\}$ and $\mathbb{Q}(i) = \{a+bi : a,b \in \mathbb{Q}\}$. Find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i) = (\alpha)$ Definition : If ...
3
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3answers
46 views

$R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ , $R/I$ is finite ; then is $R$ a PID or at least Noetherian?

Let $R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ of $R$ , $R/I$ is finite; then is $R$ a PID or at least Noetherian ? I can only prove that $R$ must be an ...
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1answer
31 views

Torsion elements with relatively prime orders

Let $x$ and $y$ be torsion elements in an R-module $M$ having orders $a$ and $b$. With $a,b \in R$, $a$ and $b$ are relatively prime, and $R$ is a PID. I want to show that $x + y$ has order $ab$. So ...
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0answers
83 views

Elementary divisors for chains of submodules

Given free modules $N \le M$ of finite rank over a PID $R$, it's well-known that there is a basis $\{x_1,\ldots,x_n\}$ of $M$ and there are $e_1,\ldots,e_n \in R$ such that $\{e_ix_i\mid e_i \neq ...
3
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1answer
56 views

Suppose A is a principal ideal domain with every ideal of finite index. Must A be a Euclidean domain?

Suppose $A$ is a principal ideal domain with every ideal of finite index (except the zero ideal). Must $A$ be a Euclidean domain? If it's not known, are there any relevant partial results?
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1answer
41 views

Is $\Bbb Z / p$ where $p$ is not prime a PID? [closed]

Is it possible for a finite ring with unity in the form of $\Bbb Z / p$ where $p$ is not prime to be a PID?
6
votes
1answer
128 views

$\mathbb{Q}(\sqrt[3]{17})$ has class number $1$

Let $\alpha:=\mathbb{Q}(\sqrt[3]{17})$ and $K:=\mathbb{Q}(\alpha)$. We know that $$\mathcal{O}_K=\left\{\frac{a+b\alpha+c\alpha^2}{3}:a\equiv c\equiv -b\pmod{3}\right\}.$$ I have to show that $K$ has ...
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1answer
19 views

A problem involving proving cyclic module defined via invertible linear operator has cyclic inverse module

I have met this recently in my abstract algebra course dealing with modules over PIDs and we are dealing with cyclic modules at the moment, the problem I am having difficulty with is as follows: ...
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0answers
33 views

Computing the invariant factors and elementary divisors of finitely generated module over a PID

I have just encountered this question in my abstract algebra class dealing with finitely generated modules over PIDs stating the following: Let $ D = \mathbb{R}[x] $ be the ring of polynomials ...
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2answers
36 views

Coprime elements in a PID satisfy that any of their powers are coprime

I have recently met this problem in my abstract algebra dealing with PID rings and coprimes stating: Let D be a PID ring $ a,b \in D $ two coprime elements. We are to show that for all $ m,n \in ...
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4answers
181 views

Description of ideals of ring $F[x]/(x^n)$?

What is a description of the ideals of the ring $F[x]/(x^n)$, where $F$ is a field?
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3answers
100 views

Ideal of $\mathbb Q[x]$ which contains two polynomials

Suppose $I$ is an ideal of $\mathbb Q[x]$ which contains $x^2 + 2x +4$ and $x^3 - 3$. Prove $I =\mathbb Q[x]$. This is an exercise in my abstract algebra text book. I know the definition of an ...
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2answers
49 views

Localization of a PID is a PID

I would like a verification of a proof for the following statement. Let $S$ be a multiplicatively closed subset of a ring $R$. If $R$ is a PID, then $S^{-1}R$ is a PID. Let $I = \left<r_1/s_1, ...
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votes
2answers
22 views

Characteristic and Principal Ideal.

This might be a simple question for some of you, but I am quite confused on the whole concept of principal ideals. Question 1: What is the characteristic of $\mathbb{Z}_2[X,Y]$ where it is the ring ...
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1answer
35 views

A commutative ring with unity which is not a PIR has a non-trivial ideal generated by two elements which is not a principal ideal?

Let $R$ be a commutative ring with unity which is not a principal ideal ring . Then is it true that $\exists 0\ne x,y \in R$ such that the ideal $\langle x, y \rangle$ is not a principal ideal ?
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2answers
66 views

Ring of polynomial functions on unit hyperbola is PID

Let $R=\mathbb{R}[X,Y]/(XY-1)$ be the ring of polynomial functions pn the unit hyperbola. How do I prove that $R$ is a principal ideal domain with unit group ...
2
votes
1answer
55 views

Is the polynomial ring over a PID also a PID?

As stated in the title, given a principal ideal domain $R$, is the polynomial ring $R[x]$ necessarily a principal ideal domain? In particular, is the polynomial ring $(\mathbb{Z}[i])[x]$ over the ...