Questions tagged [principal-ideal-domains]
For questions about principal ideal domains: rings without zero divisors where every ideal is principal.
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How can I finish the proof for the following proposition: every ideal in k[x] is a principal ideal? [duplicate]
Proposition :
Every ideal in k[x] (polynomial ring) is a principal ideal
Proof :
Suppose that $I\subseteq K[x]$. Take $p(x)\in I$ such that $p(x)$ is monic and $deg(p(x))$ is minimal over all ...
0
votes
1
answer
56
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On Showing that $\mathbb{Z}[x]$ is not a Principal Ideal Domain
I'm working on a question that asks to consider the ring $R = \mathbb{Z}[x]$. I need to show that $(x)$ is a prime ideal, show that $(x,7)$ is a prime ideal and use these facts to conclude that $R$ ...
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Question about the relation between P.I.D. and quotient field. [duplicate]
Let $D$ be a P.I.D. and $F$ is a quotient field of $D$. How to prove that any subring $D'$ contained in $D$ in $F$ is a P.I.D., and $D'$ is a ring of fraction of a subset of $D$ closed under ...
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answer
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Question on Definition of Canonical Epimorphism
Let $P$ be a finitely-generated $R$-module, where $R$ is a principal ideal domain. The text that I am reading refers to
$$
f:\bigoplus_{p \in P}R \rightarrow P
$$
as the canonical epimorphism of $R$-...
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43
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How to Prove an Ideal Can Be Generated From 2 Elements [duplicate]
Given a commutative ring with unity, $R$, an $a,b \in R$, and an ideal $I$ such that $I=\{ax+by \mid x,y \in R\}$.
Prove that $I=(a,b)$.
I think I want to show that $R$ is a PID, but I am not quite ...
1
vote
1
answer
63
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What does the correspondence theorem tell us about the ideals of $\Bbb Z[x]$ that contain $x^2+1$?
What does the correspondence theorem tell us about the ideals of $\Bbb Z[x]$ that contain $x^2+1$?
If I define $\varphi : \Bbb Z[x] \to \Bbb Z[i]$ as $x \longmapsto i$, then $\ker \varphi = (x^2+1)$.
...
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1
answer
31
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Why are the direct summands obtained in the structure theorem for finitely generated modules over a PID indecomposable?
The structure theorem tells us that a finitely generated module $M$ over a principal ideal domain $R$ is isomorphic to a direct sum
$
\bigoplus _{i}R/(q_{i})
$ where $
(
q_i
)
≠
R
(q_{i})\neq R$ and ...
- 1,178
1
vote
1
answer
22
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Basis for Module Over PID
Given $M$ is a free module over PID (let's say R).
$$X = \{x_1, x_2, \cdots, x_m\}$$
span $M$.
Can $X$ can be reduced, so that $X$ is a basis for $M$?
Note: $X$ is a basis for module-$M$ if $X$ span $...
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Principal ideal domain - polynomial ring
Let $f(x)=x^2+3x+5,\ \ g(x)=x+1\in\mathbb{R}[x]$. Show that $1\in f(x)\mathbb{R}[x]+g(x)\mathbb{R}[x]$ and hence that $f(x)\mathbb{R}[x]+g(x)\mathbb{R}[x]=\mathbb{R}[x]$.
My idea is to use the ...
user1016931
-3
votes
2
answers
71
views
Is PID with one divisible element a field? [closed]
Let $R$ be a PID and $r\in R$ be the divisible element i.e. for any $a\in R$ there exists $y\in R$ s.t. $r=ay$. Can we imply that $R$ is a field?
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Question regarding divisibility of invariant factors in Smith Normal Form.
I am trying to understand the algorithm presented in Wikipedia on how to calculate the Smith Normal Form of a matrix. I understand how to transform the matrix into a diagonal matrix and that the ...
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1
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Terminology for invariant factors of quotient module over PID
Let $A$ be a PID, $M$ a finitely generated $A$-module and $N$ a submodule. By the structure theorem of finitely generated modules or by Smith normal form, $M/N \cong \prod A/(a_i)$ for certain $a_i \...
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How does this quotient by a maximal ideal looks like?
I have the following situation:
Let $R$ be a PID. Then let $\mathfrak{p}\subset R$ be a maximal ideal. We consider the set $$\left(R/\mathfrak{p}\right)[X]$$ which is clearly a quotient, but I would ...
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1
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23
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Content of a polynomial and maximal ideals.
I have the following problem:
Let $R$ be a PID. Let us denote for $P=a_0+a_1 X+\dots+a_n X^n\in R[X]$ that $c(P)=\gcd(a_0,...,a_n)$. Show that for $P\in R[X]$ nonzero $c(P)=1$ iff for every maximal ...
- 1,029
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answers
17
views
Equivalent condition for uniserial torsion module over a PID
This is a homework problem.
Suppose, M is a finitely generated torsion module over a PID R. Then, M is uniserial if and only if M has only one elementary divisor.
So far I was able to prove the ...
0
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0
answers
108
views
Bounds involving number of invariant factors and elementary divisors
Let $M$ be a finitely generated torsion module over the PID $R$. Suppose that $M$ has $s$ elementary divisors and $t$ invariant factors. Define $$S:=\{n\in\mathbb{N}\,|\,M\text{ is isomorphic to a ...
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104
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If $N \leq M$, there exist free modules $F_N \leq N$, $F_M \leq M$ with $N = F_N \oplus N_{tor}$, $M = F_M \oplus M_{tor}$ with $F_N \leq F_M$
Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module. Furthermore, let $N$ be a submodule of $M$. Prove or disprove: there exist free submodules $F_N \leq N$, $F_M \leq M$ with $...
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Why does this proof generalize for all norms? $\mathbb Z[\sqrt{-5}]$ is not an Euclidean Domain.
Here's a proof from Dummit and Foote that $\mathbb Z[\sqrt{-5}]$ is not an Euclidean Domain (based on the fact that it's not a PID).
I understand all of the proof except the last step. They carried ...
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Linear transformations over PIDs/ EDs
We have the following linear transformation $\mathbb{Z}^2 \rightarrow \mathbb{Z}^2$ represented by the matrix \begin{equation*}
A =
\begin{pmatrix}
5 & -2 \\
-3 & 1
\end{pmatrix}
\end{...
- 47
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0
answers
90
views
Ideal of Polynomial Function on a Circle that Vanish at a Point
Let $R = \mathbb{R}[x,y]/(x^2+y^2-25)$ and $I$ the ideal of functions which vanish at the point $P = (3,4)$. I have proven that $I$ is generated by $(x-3,y-4)$ and that if $I= (f)$ for some $f \in R$, ...
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$\mathbb{Z}[i]$-modules with 101 elements and cyclic $\mathbb{Z}[i]$-torsion modules [duplicate]
Well I am solving old exam tests and I am confused. It is asking to find two non-isomorfic $\mathbb{Z}[i]$-modules with 101 elements each, but I think this can't happen because as abelian groups both ...
1
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0
answers
41
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Let $D$ be an integral domain and let $c\in D$ be irreducible in $D$. Show the ideal $(x,c)$ in $D[x]$ is not principal. [duplicate]
Question: Let $D$ be an integral domain and let $c\in D$ be irreducible in $D$. Show the ideal $(x,c)$ in $D[x]$ is not principal.
Thoughts: Since $c$ is irreducible in $D$, $c$ is noninvertible in $...
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1
vote
2
answers
73
views
Showing that $x+x^2$ belongs to an ideal in $\mathbb{Z}_2[x]$
I want to show that $x+x^2$ belongs to an ideal $I$ of $\mathbb{Z}_2[x]$ which contains both $1+x^2$ and $1+x^3$. Since $\mathbb{Z}_2[x]$ is a principal ideal domain that contains two polynomials ...
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1
vote
1
answer
51
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If any "dividing" chain "terminates" at some point, does that imply an integral domain being P.I.D.?
I'm working on Dummit & Foote's Abstract Algebra. In Section 8.2 of P.I.D.s I found the following question:
If $R$ is an integral domain, prove that $R$ is a P.I.D. if the following two ...
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1
vote
1
answer
82
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A prime ideal of a polynomial ring over a PID can be generated by two elements. [duplicate]
I am studying for a qualifying exam, and have been working on this problem:
Let $D$ be a PID and let $P$ be a prime ideal of the polynomial ring $D[x]$. Suppose that $P$ contains a non-zero constant ...
- 11
1
vote
1
answer
26
views
If $A$ is a Principal Ideal Domain, and $\mathfrak{a}$ its ideal. prove that $\frac{A}{\mathfrak{a}}$ is also a Principal Ideal Domain.
Let $A$ be a Principal Ideal Domain, and $\mathfrak{a}$ its ideal. I have to prove that $\frac{A}{\mathfrak{a}}$ is also a Principal Ideal Domain.
This is what I've done:
As $A$ is a Principal Ideal ...
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votes
1
answer
50
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Let $D$ be a PID. Show that $d\in D$ is a least common multiple of $a$ and $b$ iff $(a)\cap(b)=(d)$. [duplicate]
Question: Let $D$ be a PID. Show that $d\in D$ is a least common multiple of $a$ and $b$ iff $(a)\cap(b)=(d)$.
Attempt/Thoughts:
First, let's define a least common multiple: Suppose $a,b,c,d\in D$. $...
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votes
1
answer
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How $\frac{Z[x,y]}{\langle y+1\rangle}$ is Unique factorization domain? [duplicate]
The question is,
Given MCQ,
Which of the following is true?
(a) $Z[x]$ is principal ideal domain.
(b) $Z[x,y]/\langle y+1\rangle$ is a unique factorization domain.
(c) If $R$ is a principal ideal ...
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3
votes
1
answer
38
views
Rings of integers of number field and free modules over a PID
If $K \subseteq L$ are number fields and $O_{L}, O_{K}$ are their ring of integers, how can I prove that there exists a basis of $O_{L}$ over $\mathbb{Z}$ that contains a basis of $O_{K}$ over $\...
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1
vote
0
answers
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views
Write $(7,\omega -5)$ as principal ideal in $\mathbb{Z}[\omega]$
Let $\mathcal{O}$ be the number ring of $\mathbb{Q}(\sqrt{-3})$, then
$$\mathcal{O}=\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2} \right]=\mathbb{Z}[\omega]$$
I am told that $\mathcal{O}$ is a PID. I have ...
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0
votes
0
answers
18
views
Checking if given polynomials are units in $\mathbb{Z}_7[x]$ [duplicate]
So, I was doing an algebra exercise related to $\gcd$'s and $PID$'s and I need to check if some polynomials in $\mathbb{Z}_7[x]$ are units. The polynomials are the following:
\begin{equation*}
g = x^2+...
- 599
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0
answers
89
views
Chinese Remainder Theorem for Hurwitz quaternions
I know that if we have a noncommutative ring the CRT (Chinese Remainder Theorem) doesn't work and I know that the CRT works for all PIDs (Principal Ideal Domains).
My question is:
In the case of the ...
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votes
1
answer
42
views
Confusion about multiplying two ideals.
I am trying to solve this question:
Let $R$ be the quadratic integer ring $\mathbb Z[\sqrt{-5}].$ Define the ideals $I_2 = (2, 1 + \sqrt{-5}), I_3 = (3, 2 + \sqrt{-5}),$ and $I_3^{'} = (3, 2 - \sqrt{-...
- 1,187
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1
answer
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Rank and Nullity is preserved by multiplication of invertible matrices (PID).
I want to show that the rank and nullity of a matrix $A$ whose entries come from a PID are preserved by when $A$ is multiplied by invertible matrices
i.e If $A=PBQ$, where $P,Q$ are invertible, rank($...
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votes
1
answer
87
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Hypothesis of Structure Theorem for Finitely-Generated Modules over PID's [duplicate]
As a consequence of the Structure Theorem for Finitely Generated Modules over a PID, we know that if $R$ is a PID, and $M$ is an $R$-module, then $M$ is a direct sum of its torsion submodule $T(M)$ ...
0
votes
0
answers
30
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Show that $\mathbb{Z}[i]=\{m+in\colon\; m,n\in\mathbb{Z}\}$ is a P.I.D [duplicate]
I came across the following problem
Show that the ring of the Gaussian integers, defined as the subring of $\mathbb{C}$ given by the set $\mathbb{Z}[i]=\{m+in\colon\; m,n\in\mathbb{Z}\}$, where $i=\...
1
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0
answers
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Writing modules over PIDs as direct sums of cyclics
It seems a typical exercise in module theory is to write a finitely generated module over a PID as a direct sum of cyclic modules. After doing several problems, I have come to the conclusion that ...
- 239
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1
answer
91
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For which $d$, the algebraic integers of $K =\Bbb{Q}(\sqrt d)$ is a PID?
I know that the matter is settled for $d<0$ and an open problem for $d>0$ but I am asking about already known values. The below theorems are the motivation for asking this question.
Let $d$ be ...
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why we can't write that $\pm2=2r(x) +xs(x)$ instead of $x=2v(x)$?
Prove that the ideal $\langle 2, X\rangle$ in $\mathbb{Z}[X]$ is not principal
My attempt : I found the answer here
In the proof of the theorem it is written that
suppose that the ideal $I=\langle2,...
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0
votes
1
answer
76
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Finitely generated torsion module homomorphism in PID local ring
Problem Statement: Let R be a commutative local ring, PID, but not a field. Given two nonzero finitely generated torsion $R$-modules, show that there exists a nonzero $R$-module homomorphism $M \to N$....
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2
votes
0
answers
30
views
Hypothesis - am I doing this right? (Local, principal ideal ring)
Let $R$ a commutative local principal ideal ring with 1 that is not Artinian. So it's Krull dimension is non zero. Let $P\subsetneq M$ a prime ideal and M the maximal ideal of R, since P and M are ...
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2
votes
1
answer
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views
In A a PID, $d=\gcd(a,b) \iff ax+by=d$
I am aware this is Bézout's identity but I'm trying to prove it using ideals, I have the first half and it goes:
Suppose $ax+by=d$
$ax+by=d \implies d \in (a,b) \implies (a,b)=(d) \implies a,b \in (d) ...
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3
votes
1
answer
142
views
showing a ring is a principal ideal domain
Suppose $R$ is a principal ideal domain. Let $S$ be a multiplicatively closed subset of $R$ not containing $0$. Show that $S^{-1}R$, the localization of $R$ by $S$, is a principal ideal domain.
I ...
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votes
0
answers
39
views
sieve of Eratosthenes generalization to Dedekind domains or even PID's
I'm Interested in finding irreducibles in Dedekind domains, (and especially integer rings) in an efficient manner.
I've tried to look around a bit but found no papers on this (admittedly my paper ...
2
votes
0
answers
68
views
Extracting Rank from a module isomorphism over a PID
Notation: For an ID (integral domain) $R$ and $R$-module $C$, denote the torsion submodule: $C_t$. A unital ring $R$ has $\textbf{invariant basis number}$ [IBN] If every basis of a given free $R$-...
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36
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Some problems in Noncommutative Algebra book of Benson Farb and R. Keith Dennis [duplicate]
In this, page 48, Exercies in chapter 1, there is a following exercise.
Exercise 2. Let $R$ be a ring (with $1$) such that the only left ideals of $R$ are $0$ and $R.$ Show that $R$ must be a division ...
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votes
0
answers
68
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constructive proof that $K[X]$ is a PID
The usual proof that $K[X]$ is a PID goes by finding $p\in I$ of the lowest degree for a non-zero ideal $I$ of $K[X]$. Constructively, this amounts to testing $p\in I$ for all $p$ of degree $0$, then ...
- 2,362
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0
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94
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Implications between $K$ field and $K[x]$ PID and ED
I know that if $K$ is a field, then $K[x]$ is an ED (Euclidean Domain), and that if $K[x]$ is a PID (Principal Ideal Domain) then $K$ is a field. Now, I can say that if $K[x]$ is an ED, then it is a ...
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2
answers
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Questions about following proof regarding why $\mathbb{Z}[x]$ is not a principal ideal domain
Prove that $\Bbb{Z[x]}$ is not a principal ideal domain.
Proof: Consider the ideal $I=\langle x,2\rangle$. We’ll show that this ideal is not principal. First note that I is not equal to $\Bbb{Z[x]}$ ...
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1
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60
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How to prove that ring of polynomials in two variables is not a PID?
I am aware that a few questions like this have been asked before, but I am struggling to even understand what an ideal of such a ring would look like. I've been looking at some of the examples here ...
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