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

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Prove that any finitely generated submodule of $R^+$ (the field of quotients) is free of rank $1$

I am working on the following problem: Let $R$ be a principal ideal domain and $R^+$ the field of quotients. Then $R^+$ is an $R$-module. Prove that any finitely generated submodule of $R^+$ is a ...
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38 views

ED,PID and UFD and the relation between them

Let R be a Commutative ring with unity, such that R[x] is UFD. If R[x] is a PID then it is a Eucledian Domain? Is the last statement about being eucledian domain correct?
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24 views

$R/\langle p^k\rangle$ is an associator (i.e. if $\langle a\rangle = \langle b\rangle,$ then $a$ and $b$ are associates) when $R$ is a PID.

As the title says, I want to show that when two principal ideals are equal in $R/\langle p^k\rangle,$ where $R$ is a principal ideal domain and $p\in R$ is a prime element, then their generators are ...
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1answer
63 views

Any unit has an irreducible decomposition

The proposition is the following: Let $R$ be a principal ideal domain. Then every $a \in R$ with $a \neq 0$ has an irreducible decomposition, that is, there is a unit $u$ and irreducible elements $...
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142 views

Can we characterize all infinite Euclidean-domains having exactly one invertible element?

$\mathbb Z_2$ and $\mathbb Z_2[x]$ are two euclidean-domains having exactly one invertible element ; my question is ; Can we characterize all euclidean domains $D$ having exactly one invertible ...
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1answer
98 views

Luröth's Theorem

I've been struggling trying to understand the Jacobson's Basic Algebra vol. II proof of the Luröth's theorem. Let $K$ be a field, $K(X)$ the field of rational fonctions and take $L$ to be a sub-...
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53 views

Finding the structure of an $F_p[X]$module

$p$ is a prime and $M$ is an $F_p[X]-$ module. Given $(X-1)^3M=0, \vert (X-1)^2M\vert=p, \vert (X-1)M\vert=p^3$ and $\vert M\vert =p^7$, determine $M$ as an $F_p[X]-$module, up to isomorphism. Using ...
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2answers
46 views

Cardinal of quotient rings of gaussian integers. [duplicate]

It is known that $\mathbb{Z}[i]$ is a PID and that $\mathbb{Z}[i]/(a+bi)\mathbb{Z}[i]$ is finite for all $(a,b) \in \mathbb{Z}^2\backslash \{(0,0)\}$. My question : Is there any result on the ...
2
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4answers
125 views

In a P.I.D., if $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$ with $\gcd(m,n) = 1$, then $a=b$

Let $R$ be a principal ideal domain, and $a, b \in R$ with $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$, and $\gcd(m, n) = 1$. I now want to show that we then already have $a = b$. I think ...
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4answers
74 views

The principal ideal $(1+i)$ in $\mathbb{Z[i]}$ is maximal.

I'm trying to prove that the principal ideal $(1+i)$ in $\mathbb{Z[i]}$ is maximal. My approach: $m+in \equiv (m-n)$ mod $(1+i), \forall m,n \in \mathbb{Z}$ and $2 \equiv 0$ mod $(1+i)$ then what?
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1answer
27 views

For $\mathbb{Z}[\sqrt{-10}]$ and an ideal $I=(2,\sqrt{-10})$, find $n>0$ s.t. $I^n$ is a principal ideal.

I can show that $I$ itself is not a principal ideal by using the multiplicative norm $N(a+b\sqrt{-10}) = a^2 + 10 b^2$ using the following: Suppose that $I=(\alpha )$. Then $N(\alpha) | 2$ and $N(\...
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1answer
72 views

Example of non-principal ideal of the quotient ring $R[x,y,z]/(xyz-1)$

Suppose $R$ is an integral domain, $R[x,y,z]$ is polynomial ring over $R$, and $$Q=R[x,y,z]/(xyz-1)$$ is a quotient ring. How to prove, that $Q$ is not principal ideal ring? I was trying to compose an ...
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1answer
31 views

quotient of P.I.D by a prime power a P.I.D? [closed]

If $R$ is a P.I.D. and $p\in R$ is prime, is it the case that $R/<p^k>$ will be a P.I.D for all k? If so how would one show this?
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1answer
10 views

if an element $q_1$ of a principal ideal domain is irreducible, then the ideal $q_1 A$ is a maximal ideal

In my lecture, my professor wrote within a proof: $q_1$ in the principal ideal domain $A$ is irreducible, therefore, the ideal $q_1A$ is a maximal ideal. I don't understand how that's true. I tried ...
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56 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
93 views

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

Prove that $\mathbb{Z}[x]$ doesn't have principal maximal ideals. Please, I need help with this problem. Thanks!
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1answer
39 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 \...
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20 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 $f=...
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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
38 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: $\gcd(...
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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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72 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 finitely generated case. Let $M$ be a free abelian group and $N$ a ...
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2answers
49 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
23 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
22 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 ...
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2answers
191 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
43 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 R\...
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3answers
126 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?
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1answer
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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
17 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 ...
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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?
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1answer
50 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 ...
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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
70 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 so ...
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1answer
54 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 $\gcd(...
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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 ...
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1answer
86 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 $\mathbb{Z}[x]$ so I ...
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1answer
78 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
29 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 $q_1,...,q_s \in R\setminus \{0_R\}$...
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0answers
139 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
26 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 ...
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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''?
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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 problem. ...
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2answers
52 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 principal ...
2
votes
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 $A$ ...
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
32 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 ...