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

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Finitely generated modules over principal ideal domain

Let $A$ be principal ideal domain with field of fractions $K$. $L$ is finite separable extension of $K$ and $B$ is the integral closure of $A$ in $L$. It is obvious that there exists a constant $d$ in ...
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Is any Direct Summand of a Free Module over a PID Also Free?

Let $R$ be a PID and $F$ be a free module over $R$. Suppose we have $F=A\oplus B$ for some $R$-modules $A$ and $B$. Then are $A$ and $B$ necessarily free? If $F$ is finitely generated, then I ...
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28 views

If $A \in M_n(R)$, with $R$ a P.I.D., can $A$ be put in Jordan form iff all the roots of the characteristic polynomial are in $R$?

If $A \in M_n(R)$, with $R$ a P.I.D., can $A$ be put in Jordan form iff all the roots of the characteristic polynomial are in $R$? If this is false in general, is it possibly true for nilpotent ...
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Is this ring a PID? [closed]

Let $R$ be the $k$-subalgebra of $k(t)$ generated by the set $k[t]$, of all polynomials, and a pair of rational functions: ${1\over{t-1}}$ and ${1\over{t-2}}$. Is the ring $R$ a PID?
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Are the ring of integers of the constructible numbers a Euclidean domain?

I suspect that since Euclid uses the Euclidean Algorithm to perform division on constructible numbers in Elements, the ring of integers of the constructible numbers are a Euclidean Domain, but I have ...
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25 views

A direct sum of cyclic modules over a PID

Let $R$ be a PID, let $p$ be a prime of $R$, and let $M$ be the $R$-module $R/Rp^{e_1}\oplus \cdots \oplus R/Rp^{e_n}$ where the $e_i$ and $n$ are positive integers. Define $M(p)=\{m: pm=0\}$ and ...
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27 views

Prove that $I^2$ is principal.

Consider the ideal $I=(2,\sqrt{-10})$ of $\mathbb{Z}[\sqrt{-10}]$. Prove that $I^2$ is principal. My Try: $I^2=(4,-10,2\sqrt{-10})$. I tried to prove that $I^2=(\sqrt{-10})$. But failed. Is my ...
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$R$ a local ring, also a PID. $I,J$ ideals from $R$. Show that $I \subseteq J$ or $J \subseteq I$

$R$ a local ring, also a PID. $I,J$ ideals from $R$. Show that $I \subseteq J$ or $J \subseteq I$ My brief attempt to try use Bezout theorem at a PID. but unsuccess.. Thanks any help.
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111 views

Proof: $\mathbb{Z}[\zeta_6]$ is a PID.

I am reading through A First Course in Modular Forms. In Proposition 2.2.3 they claim that $\mathbb{Z}[\zeta_6]$ is known to be a principal ideal domain. Does anyone have a reference for the proof of ...
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24 views

Decomposition of Torsion Module

Let $k$ be a field, $k[X]$ the polynomial ring in one variable and $M$ a torsion $k[X]$-module (not necessarily finitely generated). Consider the submodules \begin{equation*} M_1 = \{a \in M \mid ...
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Does there exist such an invertible matrix?

Let $n \geq 1$ and $A = \mathbb{k}[x]$, where $\mathbb{k}$ is a field. Let $a_1, \dots, a_n \in A$ be such that $$Aa_1 + \dots + Aa_n = A.$$ Does there exist an invertible matrix $\|r_{ij}\| \in ...
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41 views

Finite (cardinality) modules over a PID

Let $R$ be a principal ideal domain with $|R|=\infty$. Suppose $M$ is an $R$-module such that $2 \le |M| < \infty$. What properties does this imply about $R$? Background: I was hoping that $R\cong ...
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Principal ideal domains that are not integral domains

In the usual definition, a principal ideal domain $R$ is also assumed to be an integral domain? However, the property that every ideal is generated by a single element does not seem to immediately ...
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Prime ideals in $R[x]$, $R$ a PID

Let $R$ be a PID. Show that if $\mathfrak p \subset R[x]$ is a prime ideal, $(r) = \left\{h(0) \colon h(x) \in \mathfrak p \right\}$, and $$\mathfrak p = (r, f(x), g(x)),$$ where $f(x), g(x) \in ...
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33 views

Why is $(2,x)$ non-principal in $\Bbb Z[x]$? [duplicate]

Why is $(2,x)$ non-principal in $\Bbb Z[x]$? Apparently this is the case, I just read it on wiki, as a counter example to $\Bbb Z[x]$ being a PID. What is $x$ here? I mean $2$ can surely generate ...
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Proof about finitely generated torsion-free R-module M is free, where R is a PID

Proof about finitely generated torsion-free $R$-module $M$ is free, where $R$ is a PID. Can I prove by the following way? Proof by induction on $n$, where $M=\langle v_1,...,v_n\rangle$. If ...
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53 views

Principal ideal domain, $\forall x=(x_1,x_2)^t \in R^2~~\exists G \in SL_2(R) : Gx=(\gcd(x_1,x_2), 0)^t$

Let $R$ be a principal ideal domain. Prove that for every $x=(x_1,x_2)^t \in R^2$ exists a matrix $G \in SL_2(R)$ for which $Gx=(\gcd(x_1,x_2), 0)^t$. I think it's easy, but do not know how to ...
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Distributive law of ideals in $\mathbf Z$ relating $\cap$ & $+$

Let $\mathfrak a, \mathfrak b$ and $\mathfrak c$ be ideals in $\mathbf Z$. Then show that $$ \mathfrak a \cap (\mathfrak b + \mathfrak c) = \mathfrak a \cap \mathfrak b + \mathfrak a \cap \mathfrak c ...
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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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Prove that $(5x^3+9x^2-27x+3)$ is a maximal ideal in $\mathbb{Q}[x]$

We have shown that $Q[x]$ is a Euclidean Domain, and thus is a Principal ideal domain. A principal ideal $(f)$ is maximal $\iff$ $f$ is irreducible in $Q[x]$. But how do I show that \begin{equation*} ...
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all abelian groups with 625 elements with 24 elements of order 5

Let $R$ be a principal ideal domain, $p \in R$ a prime element and $M$ a finitely generated $p$-torsion module of the form $M = R/(p^{e_1}) \oplus \cdots \oplus R/(p^{e_t})$. Let $_pM = \{m \in M: p ...
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vector space constructed through a torsion module

Let $R$ be a principal ideal domain, $p \in R$ a prime element and $M$ a finitely generated $p$-torsion module of the form: $$ M = R/(p^{e_1}) \oplus \dots \oplus R/(p^{e_t}). $$ Let now be $_pM = ...
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free submodules of free modules of a PID

Let $R$ be a principal ideal domain. Now I want to show that each submodule $N \subseteq M$ of a free $R$-module $M = R^{(I)}$ is also free. As a hint, it says I might consider the tripel $(J, ...
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properties of finitely generated torsion-modules and their submodules over a PID

Let $R$ be a principal ideal domain, $M$ a finitely generated $R$-torsion module, and $N \subseteq M$ a submodule. I want to show that there exist free R-modules $F, F', F''$ and module homomorphisms ...
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37 views

subring of a quotient field

Let $R$ be a principal ideal domain, and $S \subseteq Q(R)$ a subring of the quotient field of $R$, so that $R \subseteq S$. I want to show that, for any $x, y \in R$: $$\frac{x}{y} \in S \implies ...
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41 views

How we can show that a group algebra over infinite cyclic group is a principal ideal domain?

How we can show that a group algebra $\mathbb{F}G$ over infinite cyclic group $G= \langle a\rangle$ is a principal ideal domain? I tried to calculate the basis of this in the form of ...
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52 views

Some ring isomorphic to $\mathbb Z[i] / \langle 5 \rangle$

I am trying to find some ring isomorphic to $\mathbb Z[i] / \langle 5 \rangle$ . I know that $\langle 5 \rangle=\langle (2+i)(2-i)\rangle=\langle 2+i\rangle \langle2-i\rangle$ , now if $d$ is the ...
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Domain of the function $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$ will be?

I tried solving this question by $1.$ $-1$ and $4$ will not be in domain because denominator can not be zero . $2.$ Either both denominator and numerator will be positive or negative so that whole ...
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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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Intersection of ideals generated by two relatively prime elements

I am wondering how to prove the following statement: Let $R$ be a PID, $a,b$ are relatively prime. Then $\langle a\rangle \cap \langle b\rangle = \langle ab\rangle$ Progress: I think it ...
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If prime p doesn't divide the class number, then if I is an ideal of $O_K$, and $I ^{p}$ is principal, then I is principal

If a prime p doesn't divide the class number of a number field K, then if I is a non-zero ideal of $O_K$, and $I ^{p}$ is principal, then I is principal.
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Prove that $a$ is a prime element of $R$

Let $R$ be a PID and $P = (a)$ is a prime ideal of $R$. Prove that $a$ is a prime element of $R$. Since $P$ is a prime ideal of $R$, let $x,y \in R$ s.t. $xy \in P = (a).$ (WTS $a \mid x$ or $a\mid ...
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Show that $S$ is a field

I'm trying to prove the following result: Let $R$ be a principal ideal domain, $S$ an integral domain and $f: R\to S$ a surjective morphism. Prove that if $f$ is not an isomorphism, then $S$ is a ...
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Why is a submodule of a free module over a PID is free?

Rotman - Advanced modern algebra p.650 Theorem 9.8 Let $R$ be a PID and $M$ be a free $R$-module and $N$ be an $R$-submodule of $M$. Let $\beta$ be an $R$-basis for $M$ and well-order it. Now, ...
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42 views

Is torsion-free equivalent to free for non-finitely generated modules over a PID?

Maybe this is a trivial question. If $A$ is a PID and $M$ is a finitely generated $A$-module, it's well known that $M$ is torsion-free iff $M$ is free. However, if $M$ is not finitely generated, does ...
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Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements

Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements The field portion I understand. $\mathbb{Z}[i]$ is a PID and because $q$ is ...
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One-dimensional Noetherian UFD is a PID

I am looking for a reference which has a self-contained (elementary, that is, at the "undergraduate algebra level") proof of the the fact that any one-dimensional Noetherian UFD is a PID. Does anyone ...
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How to classify the rings of fractions of a principal ideal domain?

Let $A$ be a principal ideal domain and let $K$ be its field of fractions. I proved a) Every ring $B$ such that $A \subset B \subset K$ is a ring of fractions of $A$, and c) Show that any ring of ...
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Let $R$ be a PID and $I$ a prime ideal of $R$ s.t. $0 \subset I \subset 1_R$

Let $R$ be a PID and $I$ a prime ideal of $R$ s.t. $0 \subset I \subset 1_R$ and let $I = \langle a \rangle$, where $a$ is a prime element of $R$. My question is: is there any other prime ideal $J$ ...
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Let $R$ be a PID. Prove that $\exists c \in R$ such that $c\mid a, c\mid b$ and $c = ax + by$.

Let $R$ be a PID and $a,b \in R$. Prove that $\exists c \in R$ such that $c\mid a, c\mid b$ and $c = ax + by$ for some $x,y \in R$.
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$R/Ra$ is an injective module over itself

Let $R$ be a PID, $a\in R$ be a nonzero nonunit in $R$. Prove that $R/Ra$ is an injective module over itself. If $R$ is a PID, every $R$- divisible module is injective, but the question concerns ...
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Is this module injective? [duplicate]

Here's the problem: (This problem is from Hungerford's Algebra Chapter 5 exercise 6.7 ) Let R be a principal ideal domain and $p$ a prime in $R$ and $n$ a positive integer. Then $R/(p^{n})$ is an ...
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79 views

Show that $R[x,y]$ is not a Principal Ideal Domain [closed]

Let $R$ be a commutative ring with $1$. Prove that a polynomial ring in more than one variable over $R$ is not a P.I.D.. In order to show this is not a P.I.D. what do i need to show.
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A principal ideal domain if and only if proof

Show an ideal $(p)$ in a principal ideal domain in a maximal ideal if and only if $p$ is irreducible. This is a new concept i do not know how to go about this.
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Property of free submodules for a module over a PID

It's possible to produce an example of an integral domain $R$ and a free $R$-module $M$ with free submodules $L, L'$ such that $L+L'$ is not free. We can take $R=M=K[x,y]$ , $L=<x>$ , ...
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Why is F[x] a UFD? [duplicate]

When reading the proof for if $R$ is a UFD, then $R[x]$ is a UFD, the author uses a fact that $F[x]$ is a UFD. I don't quite understand this. Why $F[x]$ is a UFD? ($F$ is the fraction field of ...
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78 views

$F$ is a field iff $F[x]$ is a Principal Ideal Domain

A commutative ring $F$ is a field iff $F[x]$ is a Principal Ideal Domain. I have done the part that if $F$ is a field then $F[x]$ is a PID using the division algorithm and contradicting the ...
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68 views

A question regarding Kummer [closed]

As you know, Ernst Kummer noticed that examples such as $$6 = 2\cdot 3 \text{ or } 6 = 3 \cdot 2 \text{ and, crucially } 6 = (1 + \sqrt{-5}) (1 -\sqrt{-5}) $$ proved the failure of unique ...
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For which values of $d<0$ , is the subring of quadratic integers of $\mathbb Q[\sqrt{d}]$ is a PID?

The "integers" of quadratic field $\mathbb Q[\sqrt{d}]$ , for a squarefree integer $d$ , forms an integral domain . I know that for $d<0$ , the quadratic integers of the quadratic number fields ...
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If R is a PID, is it true that $R/\ker \phi$ is also a PID?

I came across this solution that seeks to prove that any submodule of a cyclic module is cyclic. Proof: Let $M$ be a cyclic module, so that $\phi:R \rightarrow M$ is a surjection under $\phi(r)=r ...