Questions tagged [principal-ideal-domains]

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

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GTM73 Hungerford's Algebra. Problem IV.6.4(ii)

Problem IV.6.4 If $R$ is a principal ideal domain and $A$ is a cyclic $R$-module of order $r$, then (i) every submodule of $A$ is cyclic, with order dividing $r$; (ii) for every ideal $(s)$ containing ...
2 votes
1 answer
43 views

What does it mean for a submodule of a module over a PID to have invariant factors $1$ or $0$?

I will take a particular example for simplicity: suppose $D$ is a PID and $M$ is finitely generated submodule of $D^5$, say, with set of generators $x_i, i=1,2,3,4,5$. Suppose also that the smith ...
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There exists $x\in M$ such that $Ann(x)=\langle lcm(x_1,x_2) \rangle$

Let $R$ be a commutative ring with unit which is also a PID. Let $M$ be an $R$-module. For $m\in M$, we define $$Ann(m)=\{r\in R:r\cdot m=0\}.$$ Also, if $Ann(m)=\langle c \rangle$ for some $c\in R\...
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Let $I$ be a non null ideal in $A$ P.I.D, if $A/I$ is an integer domain it is also a field? [duplicate]

I have no work to show. This fact is presented in my book as an obvious corollary of the following theorem: Theorem: If I=$(p)$ is an non null ideal in A P.I.D, $I$ is prime iff it is maximal. Thanks ...
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4 votes
2 answers
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A module over PID which is not direct sum of cyclic mdules [duplicate]

I am trying to solve the following qualifying exam problem: “Give an example of a module over a PID that is not isomorphic to a direct sum of cyclic modules. Justify your example”. (Carnegie Mellon, ...
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4 votes
1 answer
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modules over C[x]

$M$ be a module over $\mathbb C[x]$ and let $T=$ {$m\in M: x^n\cdot m=0\ \text{for some}\ n>0$}. Is it true that $$dim_{\ \mathbb C(x)} M\otimes_{\mathbb C[x]}\mathbb C(x)=dim_{\ \mathbb C} (M/T)\...
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Modules over PIDs

Let $R$ be a PID, let $r \in R$ generate a maximal idal $(r)$ in $R$, and let $k = R/(r).$ Let $M$ be a module over $R$ and let $T = \{ m \in M : \text{$r^n \cdot m = 0$ for some $n > 0$}\}.$ Is ...
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Existence of an 'almost euclidean function' on $\mathbb{Z}[\frac{1+i\sqrt{19}}{2}]$ making it a PID

I'm working out in detail the well known example of a PID which is not euclidean, $A=\mathbb{Z}\left[\frac{1+i\sqrt{19}}{2}\right]$. I've already shown it is not euclidean, and now I'm trying to show ...
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2 answers
68 views

Localization of a PID on a prime ideal is a DVR?

I’d like to find a proof that the localization of a principal ideal domain on a prime ideal is a discrete valuation ring. Apparently this is a well-known result (according to some question on math ...
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1 answer
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Free $A$-module and $I$ is a principal ideal generated by a non zero divisor

This question was asked in my assignment of commutative algebra and I was not able to solve this question. So, I am looking for help here. Question: Let $I$ be a non-zero ideal in a ring $A$. Suppose ...
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1 vote
1 answer
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Give an example of a ring $A$ and a multiplicative set $S$of $A$ such that $S^{-1}A$ is a PID (but not a field), but $A$ is not a PID.

I know that if $A$ is a PID and $S\subset A^*$ is a multiplicative set. Then $S^{-1}A$ is a PID since ideals of $S^{-1}A$ is of the form $S^{-1}I$ where $I$ is an ideal of $A$. We have to find a ...
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1 answer
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Showing $\mathbb{Z}[\alpha] / (2) $ and $\mathbb{Z}[\alpha] / (3) $ are fields [duplicate]

I'm doing an elementary commutative algebra / number theory exercise, the goal is to show that $A=\mathbb{Z}[\alpha]$ ($\alpha = \frac{1+i\sqrt{19}}{2}$) is a principal ideal domain which is non ...
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22 views

Modules over a PID with tensor relation

I am trying to work through the following problem: Let $R$ be a principal ideal domain and let $K$ be the field of fractions of $R$. Let $M$ and $N$ be finitely generated $R$-modules. Let $Hom_R (M,N)$...
1 vote
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22 views

Algorithm to Compute Greatest Common Divisors in PID

In a PID, given two elements $a,b$ we can define their greatest common divisor, $d$, as the generator of the principal ideal $$(a,b) = (d)$$ Is there an algorithm to compute this GCD or would this ...
1 vote
0 answers
36 views

Ideals of a PID and Localization

Let $R$ be a PID and let $S \subseteq R$ by any multiplicatively closed set such that $0 \not \in S$. Assume that for any collection $\left\{s_{i}\right\}_{i \in \mathbb{N}} \subseteq S$ that $$\...
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Find representatives for all the conjugacy classes of elements of order dividing 8

Problem: Find representatives for all the conjugacy classes of elements of order dividing 8 in $\text{GL}_4(\mathbf{F}_2)$ and give the orders of the representatives. My attempt: Firstly, $\vert \text{...
0 votes
1 answer
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If $R$ is a principal ideal domain and $P \in Spec(R)$, why is $P^mR/P^n=0$ for $m \geq n$?

If $R$ is a principal ideal domain and $P \in Spec(R)$, why is $P^mR/P^n=0$ for $m \geq n$? This is a step on a proof I’m trying to understand. Answering to the question in the comments: The context ...
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How to show the quadratic integer ring O is not a UFD.

Let $R=\mathbb{Z}[\sqrt{−n}]$ where $n$ is a squarefree integer greater than 3. Prove that $R$ is not a UFD. Conclude that the quadratic integer ring O is not a UFD for $D\equiv 2, 3$ mod $4$, $D < ...
-3 votes
1 answer
33 views

If $A$ is a principal ideal domain, then $_A A$ is uniform

An $A-$module $M$ is called a uniform module if the intersection of any two nonzero submodules of $M$ is nonzero. I don't know how to relate $A$ being a pid with $_AA$ being uniform. Any hint, please?
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Question about pure submodules over a P.I.D.

This question comes from the following proof of the theorem. Definition: Let $R$ be a P.I.D and $M$ be a left $R$-module.$N$ is a submodule of $M$. N is a pure submodule if whenever $y\in N$ and $a\in ...
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2 votes
2 answers
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Torsion is "multiplicative"

Let $A$ be a PID and let $M$ be a finitely generated $A$-module. Then there exist $m\in \mathbb{N}$ and $d_1, \dotsc,d_n\in A$ such that $M\cong A^m\times (A/d_1A)\times\cdots\times (A/d_nA)$, with $...
-1 votes
1 answer
37 views

In which case is the integral domain isomorphic to any proper nonzero ideal? [closed]

In which case is the integral domain isomorphic to any proper nonzero ideal? This should be the case if and only if the integral domain is a PID. But i can't understand, how to prove this statement. ...
0 votes
1 answer
34 views

Noetherian ring with extra condition is a PID

Let $R$ be a noetherian integral domain. Suppose that every finitely generated $R$-module is a direct sum of cyclic $R$-modules. Show that $R$ is a PID. Here's my attempt: Let $I$ be an ideal. Since $...
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1 vote
1 answer
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Sage: Constructing maximal ideal of a local ring, $M = \langle x - P_x \rangle$ gives invalid result, but $M = \langle x - P_x, y - P_y \rangle$ works

Anothony Knapp's book on Elliptic Curves, page 350 states: $M$ is principal given by $M = \langle t \rangle$ for any element $t$ of $M$ not in $M^2$. He later writes on that page, that $t$ is the ...
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58 views

When does a common divisor of a and b generate the same ideal of a and b?

Let R be a unitary commutative ring: let $a,b,d\in R$. I'm trying to prove that $\langle d \rangle = \langle a,b \rangle \Leftrightarrow d = \gcd(a,b)$ I can prove the $\Rightarrow$ part. The $\...
5 votes
3 answers
128 views

Neat way to show that $\mathbb{Z}_p$ is a principal ideal domain via inverse limit definition

I have encountered the ring of $p$-adic integers $\mathbb{Z}_p$ defined purely algebraically as the inverse limit of the rings $\mathbb{Z}/p^k\mathbb{Z}$ for $k\geq1$, and am curious to see if there ...
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5 votes
0 answers
57 views

Different approaches to Jordan Canonical Form

I know two different proofs of the existence of JCF. Let $V$ be a finite-dimensional vector space over base field $\mathbb{C}$ and $\alpha \in \mathsf{End}_{\mathbb{C}}(V)$. Given transformation $\...
0 votes
1 answer
39 views

Let $R$ be a PID and $a \in R$ is non-invertible,then $\exists$ some prime element $p$ such that $p|a.$

I tried this and i don't know , if there is any fallacies in my arguement. If $a$ is prime then it is easy to show. If $a$ is not prime the $<a>$ is cannot be a maximal ideal of $R$, since $R$ ...
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0 answers
32 views

Prove that in a PID if $a$ is not writable as product of finitely many prime elements then $a$ is not prime

The statement I want to prove: Suppose $I$ is a principal ideal domain. An element $a \in I$ can not written as a product of finite number of prime elements. Then $a$ is not a prime element in $I$? My ...
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0 answers
22 views

Is it true that an uniformizing element of a discrete valuation ring $R$ is either zero or prime?

Let $R$ be a DVR. This means that it is a PID with a unique maximal ideal. Let us denote $m$ to be the maximal ideal. Then there is a unique irreducible element $\pi$ up to multiplication my a unit ...
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1 vote
1 answer
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If a ring is not an integral domain does it then implies that it is not a PID?

I have some problems in understanding the relation between domains. As I understood it I know that we have the following implications PID$\Rightarrow$UFD$\Rightarrow$integrally closed domain$\...
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1 vote
1 answer
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How to obtain generator of principal ideal

Let $f(x) = x^3-x^2+2x+8$, and $K = \mathbb{Q}[x]/(f(x))$. Using PARI/GP, I knew that the prime $(2)$ is factored into $3$ prime ideals in $K$ as $(2) = P_A P_B P_C$, where $P_A=(2,x^2/2-x/2)$ $P_B=(2,...
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1 vote
1 answer
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Is the ring of formal power series $\mathbb K[[x_1, \dots, X_n]]$ in $n$ indeterminates over a field a PID?

Is the local ring of formal power series $\mathbb K[[x_1, \dots, X_n]]$ in $n$ indeterminates $x_1, \dots, x_n$, $n>1$, over a field $\mathbb K$ a principal ideal domain (PID)? This is true when $n=...
0 votes
1 answer
36 views

Localization of the ring of Laurent polynomial

Let $\mathbb{Q}[t,t^{-1}]$ be the ring of Laurent polynomials. I view this as the localization of $\mathbb{Q}[t]$ by $S= \{ t^i \mid i \in \mathbb{N} \cup \{0\} \}$. I want to understand the ...
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1 vote
1 answer
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Quotient, regular modules and submodules PID.

Let $R$ be a Principal Ideal Domain and $I\neq \{0\}$ an ideal of $R$. Are the $R$-submodules of $R/I$ the same as the $R/I$-submodules of $R/I$? If the action of $R/I$ as $R$-module is $r(r'+I) = rr' ...
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Bezout Lemma in a PID [duplicate]

The Bezout Lemma in the integers states that For any $a, b \in\mathbb Z$, let $g = \gcd(a, b)$, There exists $x, y$ such that $ax+by = g$. This can be generalized to a commutative ring that is a ...
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0 answers
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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
117 views

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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1 vote
1 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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1 vote
1 answer
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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
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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 ...
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1 vote
1 answer
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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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1 answer
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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 ...
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-3 votes
2 answers
76 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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1 answer
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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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0 votes
1 answer
61 views

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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0 votes
1 answer
24 views

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 ...
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0 votes
0 answers
44 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 votes
0 answers
126 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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