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

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37 views

$R$ 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.
6
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2answers
134 views

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 ...
2
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1answer
91 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 ...
2
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1answer
22 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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1answer
38 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 ...
4
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1answer
64 views

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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1answer
55 views

Prime ideals in $R[x]$, $R$ a PID

Let $R$ be a PID. Show that if $r \in R$ and $$p = (r, \underline{f}(x), \underline{g}(x))$$ is prime, where $\underline{f}(x), \underline{g}(x) \in R[x]$ are nonconstant irreducible polynomials, ...
3
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2answers
107 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 ...
2
votes
1answer
249 views

Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PIR.

Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PIR. I think I have proved this: Let $J$ be an ideal of $S$. Then $f^{-1}(J)=(a)$ is a principal ideal of ...
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5answers
4k views

Show that every ideal of the ring $\mathbb Z$ is principal

Let $\mathbb Z$ be the ring of integers. The question asks to show that every ideal of $\mathbb Z$ is principal. I beg someone to help me because it is a new concept to me.
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2answers
45 views

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 ...
1
vote
1answer
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 ...
2
votes
1answer
51 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 ...
2
votes
2answers
27 views

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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0answers
26 views

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 ...
3
votes
2answers
48 views

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*} ...
4
votes
1answer
119 views

PID modulo a non-zero ideal is a semilocal ring

Let $R$ be a commutative ring, $\mathfrak{m}\subset R$ a maximal ideal and $f$ a monic polynomial in $R[x]$. I want to show that $A:=\frac{R[x]}{\mathfrak{m}[x]+(f)}$ is a semilocal ring, where ...
0
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1answer
28 views

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 = ...
0
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1answer
34 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 ...
2
votes
1answer
126 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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0answers
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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0answers
27 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 ...
0
votes
1answer
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 ...
2
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2answers
151 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} ...
0
votes
1answer
39 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 ...
2
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1answer
50 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 ...
2
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0answers
45 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 ...
2
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4answers
48 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
49 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
24 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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0answers
36 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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2answers
41 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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2answers
38 views

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
29 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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1answer
36 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 ...
2
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1answer
25 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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1answer
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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votes
1answer
31 views

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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2answers
46 views

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 $J$ ...
3
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1answer
110 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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votes
4answers
160 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 ...
2
votes
1answer
74 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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0answers
47 views

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 $x,y \in R$.
0
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1answer
99 views

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
68 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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2answers
35 views

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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votes
1answer
76 views

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 ...
4
votes
0answers
78 views

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 $R=M=K[x,y]$ , $L=<x>$ , ...
3
votes
2answers
179 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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vote
2answers
59 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 ...