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Questions tagged [principal-ideal-domains]

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

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R is UFD. R is PID if every prime ideal is principal. [duplicate]

Suppose not. We consider the set of all non-principal ideals, $S$. Order $S$ by inclusion. We show S satisfies all the conditions in Zorn's Lemma. So it has a maximal element. If we show the maximal ...
Dwaipayan Sharma's user avatar
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Why Smith normal form gives isomorphic modules?

I have an answer to the problem but I use some (trivial) diagram chasing by $5$-Lemma. Consider a principle ideal domain $A$ and a finitely generated module $M$ over $A$. Since $A$ is Noetherian, we ...
user108580's user avatar
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2 answers
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Prove that $R / J$ is a field and determine $|R / J|$ with $J=(1-2 i) \subseteq \mathbb{Z}[i] .$

Exam question: In the ring of Gaussian integers $\mathbb{Z}[i]$, consider the ideal $J=(1-2 i) \subseteq \mathbb{Z}[i] .$ (a) Prove that $R / J$ is a field and determine $|R / J|$. (Check: The order $|...
Isaac16726's user avatar
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1 answer
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Decomposition of $A$ torsion $R$-module over a PID $R$, where every element is of the order of some power of a prime $p\in R$

Let $A$ be a module over a PID $R$ such that $p^{n}A = 0$ and $p^{n-1}A \neq 0$ for some prime $p \in R$ and positive integer $n$. Let $a$ be an element of $A$ of order $p^{n}$. Then need to show ...
Dwaipayan Sharma's user avatar
2 votes
2 answers
190 views

An easier example of a non-PID where every finitely generated ideal is principal

Say that an integral domain $\mathcal{X}$ is an almost-PID if $\mathcal{X}$ is not a PID but every finitely generated ideal of $\mathcal{X}$ is principal. The question of whether almost-PIDs exist ...
Noah Schweber's user avatar
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Under what conditions can Noetherian /artinian rings be considered principal ideal domain.

Background Definition 1: A ring $R$ is said to satisfy the $\textbf{ascending chain condition}$ ($\textbf{ACC}$) for left (right) ideals if for each sequence of left (right) ideals $A_1, A_2, \ldots$ ...
Seth's user avatar
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For a PID $R$ prove that restriction-of-scalars induces equivalence of categories $\coprod_p (R_{(p)}-\mathrm{mod})_{tors}\to (R-\mathrm{mod})_{tors}$

Let $R$ be a PID. Prove that restriction-of-scalars induces an equivalence of categories of finitely generated modules $$\coprod_p (R_{(p)}-\mathrm{mod})_{tors}\to (R-\mathrm{mod})_{tors},$$ where the ...
Squirrel-Power's user avatar
1 vote
2 answers
39 views

The polynomial ideal associated to a prime ideal is prime

Let $A$ be a ring, $I \subset A$ a prime ideal and $\Phi : A \rightarrow A[x]$ the canonical inclusion homomorphism. Is $I[x]= \{ \sum_{k=0}^n a_k x^k : n\in \Bbb{N}, (a_k)_{k=0}^n \in I^n \}$ also ...
Superdivinidad's user avatar
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Determine the domain-sets of $ Log(e^z-e^{-z})$

Determine the domain-sets of $$ Log(e^z-e^{-z})$$ I only know that $e^z-e^{-z} > 0$ and so z must be positive. Also, $e^z = e^{x+iy} $ The solution is $\Bbb{C}$ \ { $k \pi i | k \in \Bbb{Z}$ } I ...
Confused's user avatar
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Determine the domain-sets of $f(z)=z\bar{z}/(z^2+\bar{z}^2)$

Determine the domain-sets of $$f(z)=\frac{z\bar{z}}{z^2+\bar{z}^2}$$ Here's my attempt: Consider z = x+iy, $z^2+\bar{z}^2 = (x+iy)^2 +(x-iy)^2 = 2x^2$ we know that $z^2+\bar{z}^2 \neq 0$ and so $x \...
Confused's user avatar
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Prove every ideal in ring of integers is principal [duplicate]

I wrote a proof and would like to see how sturdy it is. I'm very new to this subject, and am curious. I wrote: We have $I \vartriangleleft Z$. Then for $x,y\in I$ and $z\in \mathbb{Z}$, $x - y \in I$, ...
rlaivsezlt's user avatar
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If $I$ is an ideal of $R=\{\frac{a}{b}\in \mathbb{Q}\mid p\not\mid b\}$, for some fixed prime $p$, show that $(p^t)\subset I.$

Background: Exercise 14: Let $p$ be a fixed prime integer and let $R$ be the set of all rational numbers that can be written in the form $\frac{a}{b}$ with $b$ not divisible by $p$. Prove that (a) $R$ ...
Seth's user avatar
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Where does $p^t$ come from in $R=\{\frac{a}{b}\in \mathbb{Q}\mid p\not\mid b, \text{ for some prime-}p \}$

Background: Exercise 14: Let $p$ be a fixed prime integer and let $R$ be the set of all rational numbers that can be written in the form $\frac{a}{b}$ with $b$ not divisible by $p. Prove that (a) $R$ ...
Seth's user avatar
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Explicit proof of the fact that a domain which is not a UFD is not a PID

In the same spirit as this question, I would like to prove explicitely that if $R$ is a domain which is not a UFD, then it is not a PID. I am interested in the case where there is an element $a\in R$ ...
GreginGre's user avatar
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Need help understanding the proof showing the existence of gcd in an UFD.

Background: Definition: Let $a_1,a_2,\ldots a_n$ be elements (not all zero) of an integral doain $R.$ A $\textbf{greatest common divisor}$ of $a_1,a_2,\ldots, a_n$ is an element of $d$ of $R$ such ...
Seth's user avatar
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Need help understanding the proof that if $R$ is a PID, every nonzero, nonunit elements of $R$ is a product of irreducibles.

Background: Lemma 10.9: Let $a$ and $b$ be elements of an integral domain $R.$ Then $(a)\subset (b)$ if and only if $b\mid a.$ $(a)=(b)$ if and only if $b\mid a$ and $a\mid b.$ $(a)\subsetneq (b)$...
Seth's user avatar
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Explicit proof of the fact that a non integrally closed domain is not a UFD

Context. Let $R$ be a domain. It is well-known that $R$ is a PID $\Rightarrow$ $R$ is a UFD $\Rightarrow$ $R$ is integrally closed (in its field of fractions). In other words, we have $R$ is not ...
GreginGre's user avatar
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Let $R$ be a P.I.D. and let $a \neq 0$ be an element in $R$. Prove that for a prime element $p$, $p(R/(a)) = ((p) + (a))/(a)$.

Let $R$ be a P.I.D. and let $a \neq 0$ be an element in $R$. Prove that for a prime element $p$, $p(R/(a)) = ((p) + (a))/(a)$. This is something that came up while reading Dummit and Foote's textbook ...
Squirrel-Power's user avatar
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Algorithms to decompose a graded module over R[x], where R is a PID

I have a certain class of objects, which can be thought of either as modules over a ring $R[x]$ or as functors, and I am looking for algorithms to decompose these objects into indecomposable summands. ...
GHPR's user avatar
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ID's, PID's, Noetherian rings and valuation rings: implications amongst them

I am trying to establish some implications between being an ID, a PID, a Noetherian ring and a valuation ring. First of all, I know that PID $\Rightarrow$ Noetherian, because in a PID every ideal is ...
kubo's user avatar
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The multiplicative group of a finite field is cyclic, using modules

I need to prove, specifically using modules over $\Bbb{Z}$, that the multiplicative group of a finite field is cyclic. This is what I've done already: The multiplicative group of a finite field is in ...
soggycornflakes's user avatar
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Trying to define Dedekind-Hasse norm in a PID

I have the following definition: Let D be an integral domain; a function $N:D \to \mathbb{N}$ is a $\textbf{Dedekind-Hasse}$ norm if $$\forall 0 \neq a,b \in D;\quad b|a \quad \lor \quad \exists s,...
J P's user avatar
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A case where submodules, isomorphic as modules, are isomorphic as submodules

Let $R$ be a discrete valuation ring with maximal ideal $m$ and let $M$ be a finitely generated torsion $R$-module. Let $K$ and $K'$ be two submodules of $M$ which are isomorphic as $R$-modules. Can ...
Stabilo's user avatar
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1 answer
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$ (x-a_1) \cap \ldots \cap (x-a_n) = (p) $ in $ \mathbb{C}[x]$ reasoning [duplicate]

Let $ p \in \mathbb{C}[x] $ be of degree $ n $ with pairwise different roots $ a_1, \ldots, a_n \in \mathbb{C}.$ I read in a script that $ (x-a_1) \cap \ldots \cap (x-a_n) = (p) $ because $ \mathbb{C}[...
2GR8's user avatar
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1 answer
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Need help showing that the only submodules of $M$ are the ones in an ascending chain.

$\color{Green}{Background:}$ $\textbf{Assumed facts:}$ $\textbf{Theorem 1:}$ Let $R$ be a ring. Then the following conditions are equivalent: $(1)$ Every ideal of $R$ is finitely generated $(2)$ ...
Seth's user avatar
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0 votes
1 answer
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Let R be a UFD such that each maximal ideal of R is a principal ideal, prove that R is a PID

I need help to the following problem: Problem: Let $R$ be a Unique Factorization Domain such that each maximal ideal of $R$ is a principal ideal. Then $R$ is a Principal Ideal Domain. Solution(my ...
TrItOs's user avatar
  • 101
1 vote
1 answer
66 views

$\mathbb Z[\xi_{2n}]$ is a pid

Let $\xi_{2n} \in \mathbb C$ a primitive $2n^{th}$ root of unity for some integer $n\ge 2 $. Is the inclusion $\mathbb Z[\xi_{2n}] \hookrightarrow \mathbb C$ flat? It is possible to answer this ...
Conjecture's user avatar
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0 answers
32 views

Deduce Jordan Normal Form from PID finitely generated module structure.

Let $k$ be an algebraically closed field. Any $k$-linear map $\phi:k^n\to k^n$ imposes an extra $k[x]$-module structure on $k^n$ by defining $p(x)\cdot v = p(\phi)(v).$ Clearly then $k^n$ is a ...
user108580's user avatar
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1 answer
54 views

Free submodules of an integral $R.$

I was told by the author of the answer here Showing that the rank of $M$ is exactly $1.$ that: Free submodules of an integral domain $R$ are exactly the principal ideals of $R.$ I am wondering which ...
Brain's user avatar
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2 votes
2 answers
138 views

Pure submodule that is a direct summand

I have been recently studying about Pure submodules of a module in a commutative algebra course. I came up with the following two questions, please help me as I am completely stuck. The definition of ...
Ratanjit 's user avatar
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0 answers
80 views

Primary component of a finitely generated module over PID is a direct sum of cyclic modules.

Recently I have been studying the structure theorem of finitely generated modules over a principal ideal domain $R$. I am stuck with the following statement and I am looking for an elegant and nice ...
Ratanjit 's user avatar
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0 answers
51 views

If every irreducible elements in Noetherian ring is prime, it is UFD

I have problem with the proof of theorem 4 in this link It says if $R$ is an Noetherian ring, we construct $\mathbb U$, the set of ideals generated by each element of $R$ that cannot be written as a ...
MrTanorus's user avatar
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Exhibit all the ideals in the ring $F[x]/(p(x))$, where $F$ is a field and $p(x)∈F[x]$ (describe them in terms of the factorization of $p(x)$)

This is Exercise 9.2.5 in Dummit and Foote's Abstract Algebra Exhibit all the ideals in the ring $F[x]/(p(x))$, where $F$ is a field and $p(x)$ is a polynomial in $F[x]$ (describe them in terms of ...
Stanarth's user avatar
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0 answers
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What is meant by "invertible" matrices in the creation of a SNM

I just read up on wikipedia on the Smith Normal Matrix. But what is meant by an invertable matrix. For example if you have a start matrix with only PID values does that mean the other matrices don't ...
IV-301's user avatar
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0 answers
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Are the invertible matrices which are used to find the SNF always part of the ideal principle domain?

Let's say SNF = TAT^-1. Do T and T inverse always only have elements part of the domain? For example, if we have an integer matrix, will T and T inverse only have integer values (not rational numbers)....
IV-301's user avatar
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0 answers
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Epimorphism of torsion module

Let $M$ be a finitely generated torsion module over a PID $D$. Suppose $\varphi:M\to{N}$ is a module epimorphism and suppose that the invariant factor ideals of $M$ are $(d_1)\supseteq{(d_2)}\supseteq....
zinne98's user avatar
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0 answers
89 views

Let $R$ be a factorial ring in which every ideal generated by two elements is a principal ideal. Show that $R$ is a principal ideal ring. [duplicate]

I want to check if my solutions for this problem are right. Let $R$ be a factorial ring in which every ideal generated by two elements is a principal ideal. Show that $R$ is a principal ideal ring. ...
Marco Di Giacomo's user avatar
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0 answers
100 views

Step in proof of Structure theorem over PID - Jacobson, Basic Algebra I

I'm trying to do exercise 8.6 in chapter 3 of Jacobson's Basic algebra I, which provides a proof of the structure theorem that does not require the Smith Normal form. Let $x_1,x_2,...,x_n$ be a set of ...
zinne98's user avatar
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1 vote
0 answers
62 views

Classify finitely generated modules over $\mathbb{Z}[i]$

Classify all finitely generated modules $M$ over $R=\mathbb{Z}[i]$ where $5M=0$ By the prime decomposition theorem, I know that $M\cong R^n\oplus(\bigoplus_{i=1,...,u} R/(p_i^{k_i}))$ where $p_i$ are ...
Irene's user avatar
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1 answer
42 views

Find a basis for an ideal $I$ of $k[X]$ viewed as a $k$-subspace of the vector space $k[X]$.

I am looking for an answer to the following question. If $k$ is a characteristic zero field and $I=(X^3+X^2,X^6)k[X]$, find a basis for the $k$-subspace $I$ of the vector space $k[X]$. It is clear ...
Kishalay Sarkar's user avatar
3 votes
1 answer
157 views

What is the quotient ring $\mathbb{R}[x]/(x-1)(x+1)(x+3)(x-5)$ using Chinese Remainder Theorem?

I know that it is not a PID since $((x-1)(x+1) + I)(x+3)(x-5) + I) = 0$ where $I$ is the ideal we are quotienting by, and this means it is not even an Integral Domain since it has zero divisors ...
user13121312's user avatar
0 votes
1 answer
108 views

Ring that is not a principal ideal domain has at least 6 ideals

I am stuck on the proof for this. This is what I manage to conclude so far: we have two trivial ideals, (0) and (1). Since it is not a PID, the ring R cannot be a field, so it must have some element ...
helen of troy's user avatar
0 votes
0 answers
56 views

Product of matrices with only one nonzero entry on each column

Let $R$ be a full-rank $n \times m$ rectangular matrix with entries in a DVR, with $n \leq m$ (less rows than columns) and we may assume $n < m$ if useful. Is there an $m \times m$ invertible ...
Matteo Casarosa's user avatar
1 vote
1 answer
85 views

A PID is a semisimple ring iff it is a field

I am trying to prove that a PID $R$ is a semisimple ring iff it is a field. Clearly any field is semisimple. I am not sure about the converse. By Artin-Wedderburn, $R$ is a product of matrix rings ...
Margaret's user avatar
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27 views

For compact and connected $K\subset\mathbb{C}$, $H(K)$ is a PID?

For compact and connected $K\subset\mathbb{C}$, let $H(K)$ be the set of functions that extend to a holomorphic function over an open neighborhood of $K$. This answer states that $H(S^1)$ is a UFD (...
Jianing Song's user avatar
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-1 votes
1 answer
111 views

Proving the following question: Z[3i] isn't PID by use of the quotient map Z/(3i) isn't a field.

I was wondering if this methodology is sound. I've checked that for any element in $\mathbb{Z}[3i]$ if we take the quotient map with ideal $(3i)$, its kernel consists of the elements $ \{ 9a+3bi \}$. ...
veganwithabeef's user avatar
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0 answers
35 views

Prime and maximal ideals in principal domain [duplicate]

I was studying Yves Lequain's elements of algebra and this was a question on the second chapter of the book. I have been able to prove two implications but it remains to prove $(i)\Rightarrow (ii)$: ...
Kadmos's user avatar
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If $P\cap B$ is maximal in $B$, then $P$ is maximal in $B[y]$

As part of a much larger proof, I have encountered the following situation. I have that $B$ is a PID and $P$ is a prime ideal of the polynomial ring $B[y]$. I know that the intersection of $P$ and $B$ ...
kubo's user avatar
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2 votes
1 answer
50 views

Principal ideals of $\mathbb Z[x]/(p(x))$

I am given a series of questions of the form: A reducible polynomial $p(x)\in \mathbb Z[x]$ is given and the task is to decide if all ideals in $\mathbb Z[x]/(p(x))$ are principal. For example, $p(x) ...
user2345678's user avatar
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3 votes
2 answers
163 views

$P$ is a prime ideal $\iff$ for $r, s \in R$ such that $rRs \subset P$, then $r \in P$ or $s \in P$.

If $P$ is an ideal in a not necessarily commutative ring $R$, then the following conditions are equivalent: a) $P$ is a prime ideal, b) If $r,s\in R$ such that $rRs\subset P$ then $r\in P$ or $s\in P$...
Remas 's user avatar
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