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

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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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20 views

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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1answer
92 views

$R$ is a commutative integral ring, $R[X]$ is a principal ideal domain imply $R$ is a field

I've just read a proof of the statement: Let $R$ be a commutative integral ring. If $R[X]$ is a principal ideal domain, then $R$ is a field. In one part of the proof there is a step which I ...
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16 views

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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24 views

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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35 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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2answers
143 views

Every principal ideal domain satisfies ACCP.

Every principal ideal domain $D$ satisfies the ACCP. Proof. Let $(a_1) ⊆ (a_2) ⊆ (a_3) ⊆ · · ·$ be a chain of principal ideals in $D$. It can be easily verified that $I = \displaystyle{∪_{i∈N} ...
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34 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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1answer
46 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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35 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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4answers
44 views

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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2answers
37 views

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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1answer
20 views

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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24 views

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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38 views

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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2answers
27 views

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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2answers
88 views

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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1answer
28 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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1answer
24 views

Solution of a congruence equation in a PID

If $D$ is a PID and $a,b,m\in D$, then the equation: $$ ax\equiv b \pmod{m} $$ has solution $x \in D$ iff $b$ divides $(a,m)$. I have proved the left to right implication but I'm trying so ...
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53 views

Smith Normal Forms over Fields and PIDs

I need to reduce the following matrices into the Smith Normal form over the field $(\mathbb{Z}/2\mathbb{Z})[x]$: $$M_{1} = \left ( \begin{array}{ccc} x & 1 & 0 \\ 0 & x & 1 \\ 0 ...
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1answer
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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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101 views

When a holomorphy ring is a PID?

I will use the notation and language of Stichtenoth, Algebraic Function Fields and Codes. Let $F$ be a function field over a finite field $\mathbb F_q$, $S$ a non empty set of places (possibly ...
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4answers
155 views

Why is $(2, 1+\sqrt{-5})$ not principal?

Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$? Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb ...
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1answer
60 views

$R/Ra$ 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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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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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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2answers
46 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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Classifying torsion-free injective modules over a PID

So the question is as in the title: Let $R$ be a PID. Classify all torsion-free injective modules. I know that it is going to be divisible, and using torsion free, if we define ...
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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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175 views

Showing a ring is not a principal ideal ring

If I have a ring and suppose that I want to show that it is not a principal ideal ring. How can I construct an ideal (that is not a principal ideal) as a counterexample? For example, I saw this ...
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54 views

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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1answer
66 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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1answer
57 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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25 views

prove that the quotient ring S3/T3 is isomorphic to D3

Could you please help with this question? I've already shown that T_3 is an ideal of S_3. Thanks,
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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 ...
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50 views

Does validity of Bezout identity in integral domain implies the domain is PID ?

Let $D$ be an integral domain such that for any $a,b \in D$ , $Da+Db$ is a principal ideal , then must $D$ necessarily be a principal ideal domain i.e. should all the ideals of $D$ be principal ? ...
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Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal

Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal. The forward direction is standard, and the reverse direction is giving me trouble. In particular, I can prove that ...
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1answer
34 views

Trapezoidal Motion Profile Using Discrete Method

I'm trying to program an arduino to generate a Trapezoidal Motion Profile to control a DC motor with a quadrature encoder. Essentially, the user will input the desired Target Position, Max Velocity ...
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Under what conditions is a tower of quadratic extensions a UFD, GCD domain, or just an Integral Domain?

I have been studying towers of quadratic extensions to $\mathbb Q$ and have noticed the following: $\mathbb Q[\sqrt 2]$ and $\mathbb Q[\sqrt 2][\sqrt 3]$ are unique factorization domains(UFDs), but ...
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What is the dimension of $\mathbb R[x] / \langle x^3-x\rangle$ as a vector space over $\mathbb R$ ?

What is the dimension of $\mathbb R[x] / \langle x^3-x\rangle$ as a vector space over $\mathbb R$ ? Can someone please give some links , articles where I can study about polynomila rings and its ...
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The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$)

The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$) Can I get walked through one of them so that I ...
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6answers
2k views

Proving the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain [closed]

Please help me prove that the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain. This was from Abstract Algebra
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Another question about free modules of finite rank over a PID

I am now trying to prove the following: Let $R$ be a PID and let $M$ be a free $R$-module of finite rank. If $N$ is a submodule of $M$ and $M/N$ is finite, then rank$(N)$ = rank$(M).$ Attempt at a ...
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Bases for submodules of free modules over a PID

I have proved the following: If $G$ is a free abelian group of rank $n$ and $H$ is a subgroup of $G$, then $H$ is free of rank $m\leq n$. Moreover, there exists a $\mathbb{Z}$-basis $x_1,\ldots,x_n$ ...
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108 views

Is $\mathbb Z\left[\frac{1+\sqrt{-15}}{2}\right]$ a PID?

$1)$ Let $R:=\mathbb Z[w]$, where $w=\frac{1+\sqrt{-15}}{2}$. What is the norm $N_{R/\mathbb Z}(x+yw)$ in terms of $x,y\in\mathbb Z$? Which of the integers $1,\dots,10$ occur as the norm of some ...
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14 views

Inclusion with some principal ideals

Let $A$ be a principal ideal domain, $M$ a free $A$-module of rank $n$, $M'$ a submodule of $M$ with $M' \ne (0)$, and $L(M,A)$ the set of linear forms on $M$. For $v \in L(M,A)$, we can write ...