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

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Special subrings of a PID is PID

Suppose you have $A\subseteq B$ domains, and $B$ is PID. Suppose that the extension is 'purely separable', i.e. for every $I$ ideal in $B$ such that $I \cap A \neq (0)$ we have $I \subseteq A$. Is ...
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34 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: ...
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1answer
28 views

Modules over rings which are NOT a PID, or NOT a UFD [closed]

I am interested in studying the properties of modules over rings which are not Principal Ideal Domains or are not Unique Factorization Domains, but I am finding it very difficult to find any material ...
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1answer
253 views

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

Let $R$ be a PID (such as $\mathbb{Z}$) and $R^m\overset{A}{\longrightarrow} R^n\overset{B}{\longrightarrow} R^o$ matrices with $BA=0$ and Smith normal forms ...
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1answer
44 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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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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47 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 ...
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1answer
43 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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14 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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3answers
120 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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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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2answers
134 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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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 ...
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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 ...
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42 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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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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2answers
114 views

Let D be a principal ideal domain. Prove that every non-zero prime ideal in D is a maximal ideal in D.

Let D be a principal ideal domain. Prove that every non-zero prime ideal in D is a maximal ideal in D. So I'm think I need to use the fact that all PID's are UFD's. If it is a UFD I can infer ...
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817 views

Prove that $n^2+n+41$ is prime for $n<40$

Here's a problem that showed up on an exam I took, I'm interested in seeing if there are other ways to approach it. Let $n\in\{0,1,...,39\}$. Prove that $n^2+n+41$ is prime. I shall provide my ...
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1answer
41 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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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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125 views

Is a localization of a UFD at a prime ideal a PID? [duplicate]

I have this problem: Let $R$ be an UFD and $p$ a prime element. Prove $R_{(p)}$ is a PID. I think I'm just struggling to get started. I know that $R_{(p)}=\{\frac{a}{b}: a \in R, b \in ...
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68 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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1answer
150 views

If in a UFD every maximal ideal is principal then it is a PID

I want to prove that if in a UFD every maximal ideal is principal then it is a PID. My line of attack is: If it is a field i.e. it has no non-zero proper ideal, then we are done. Otherwise ...
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1answer
51 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 ...
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1answer
81 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 $\mathbb{Z}[x]$ ...
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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 ...
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1answer
67 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
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
27 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 ...
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128 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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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 ...
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62 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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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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167 views
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49 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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57 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 ...
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42 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
30 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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81 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 ...
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1answer
400 views

Contraction of maximal ideals in polynomial rings over PIDs

Let $R$ be a principal ideal domain which is not a field, and let $M$ be a maximal ideal of the polynomial ring $R[X_1,\dots,X_n]$. If $n=1$ it is very easy to see that $M \cap R \neq 0$. Is this also ...
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1answer
48 views

Proving that for an integral domain $R$, $y\in (x)\iff (y)\subseteq (x)$.

I am trying to prove the following statement. Let $R$ be a integral domain. Then for all $x,y\in R$ we have $$x\mid y\iff y\in(x)\iff (y)\subseteq (x).$$ Note that $(x)$ denotes the principal ...
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1answer
55 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
210 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 Find all finitely generated graded $K[x]$-modules up to abstract isomorphism. Remember, $K[x]$ is a principal ...
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1answer
127 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
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?
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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
29 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
103 views

$R=\{a/b: a,b \in \mathbb{Z}$ and $b$ is odd}. Show that the ring $R$ is a PID.

Let us have a ring $R$ defined as $R=\{a/b: a,b \in \mathbb{Z}$ and $b$ is odd}. I want to show that $R$ is a PID. I think I should start with that $I\cap Z = n \mathbb{Z}$ for some $n \in ...
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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 ...