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

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Principal Ideal Domain and Factorization

If $A$ is a local domain such that each non-trivial ideal factors uniquely into primes then does it follow that $A$ must be a principal ideal domain?
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Application of Structure Theorem to Prove Simultaneous Diagonalizability and Group of Units of Cyclic Groups

I am reading these notes on Modules over PID. Exercise 67 (pg 24) asks to prove that: Problem. Let $A$ and $B$ be $n\times n$ matrices with complex entries. Then $A$ and $B$ are simultaneously ...
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39 views

$\Bbb Z[\sqrt{-5}]$ is not a PID [duplicate]

I want to show: In a PID $R$ two elements $a,b\in R$ always have a greatest common divisor. Therefore $\Bbb Z[\sqrt{-5}]$ is not a PID. For the first part: $I=\{ax+by:x,y\in R\}$ is an ideal, ...
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Is $\Bbb Z[i]$ a Euclidean ring? [duplicate]

Is $\Bbb Z[i]$ a Euclidean ring? If not, what would be the simplest way of seeing that $\Bbb Z[i]$ is a PID?
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$R/Rg$ is a field iff $g\in R$ is irreducible.

Let $R$ be a PID and $g\in R$. I want to show: $R/Rg$ is a field iff $g\in R$ is irreducible. I.e. I want to show that all $a\notin Rg$ are invertible modulo $g$ iff $g$ is irreducible. So if I ...
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In a local ring, the maximal ideal $\mathfrak{m}$ is principal $\implies \dim_k(\mathfrak{m}/\mathfrak{m}^2)\leq 1$

This is Proposition 8.8, $ii)\implies iii)$ in Atiyah and Macdonald and it says there that this is clear. It isn't for me though. How would I prove this?
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Why is a discrete valuation ring's unique prime ideal generated by a non-nilpotent element?

Serre gives first the definition of a discrete valuation ring as a "principal ideal domain that has a unique non-zero prime ideal $m(A)$." Next, he says: Let $A$ be a commutative ring. In order that ...
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Is it true that all modules over $\mathbb{Q}$ are of form $\mathbb{Q}^n$ where $n\geq 0$

Is it true that all modules over $\mathbb{Q}$ are of form $\mathbb{Q}^n$ where $n\geq 0$ I feel like this is true because $\mathbb{Q}$ is a principal ideal domain
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Generator of intersection of ideals in a PID via adjunction?

In a PID we have the formulas $ \left\langle f\right\rangle + \left\langle g \right\rangle = \left\langle \gcd(f,g) \right\rangle $ and $ \left\langle f\right\rangle \cap \left\langle g \right\rangle =...
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If ring $R$ is a principal ideal domain, then $R[x]$ is principal ideal domain. [duplicate]

If ring $R$ is a principal ideal domain, then $R[x]$ is principal ideal domain. I think that it is false. So I try to find some PID not satisfying above hypothesis. Could you help me?
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I wonder if $K$ is a field then $K[X]$ is a PID then the ideals of it have formula that is $<f(X)>$ when $f(X)$ is a polynomial with coefficent in $K$ [closed]

I see the solution of proving if K is a field then $K[X]$ is PID that is taking the minimal polynomial of ideal but I do not see they use the hypothesis K is field? Especially, I do not see the ...
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32 views

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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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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$R/\langle p^k\rangle$ 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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64 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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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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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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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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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 ...
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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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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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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
74 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
36 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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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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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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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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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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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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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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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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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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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 ...
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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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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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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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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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Is $\mathbb{Q}[X,Y]/\langle X^2+Y^2-1\rangle$ a PID?

I know that $\mathbb{Q}[X,Y]$ is not a PID. Does this imply that the quotient ring $\mathbb{Q}[X,Y]/\langle X^2+Y^2-1 \rangle$ is not a PID?
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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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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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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
52 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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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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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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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
56 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
90 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
80 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 ...