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

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A subring of the field of fractions of a PID is a PID as well.

Let $A$ be a PID and $R$ a ring such that $A\subset R \subset \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ denotes the field of fractions of $A$. How to show $R$ is also a PID? Any ...
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2answers
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Greatest common divisor in the Gaussian Integers

Let $a$ and $b$ be integers. Prove that their greatest common divisor in the ring of integers is the as their greatest common divisor in the ring of Gaussian Integers. Ring of Gaussian Integers is: ...
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3answers
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Ring of trigonometric functions with real coefficients

Let $R$ be the ring of functions that are polynomials in $\cos t$ and $\sin t$ with real coefficients. Prove that $R$ is isomorphic to $\mathbb R[x,y]/(x^2+y^2-1)$. Prove that $R$ is not a unique ...
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1answer
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Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...
13
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4answers
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An integral domain whose every prime ideal is principal is a PID

Does anyone has a simple proof of the following fact: An integral domain whose every prime ideal is principal is a principal ideal domain (PID).
14
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3answers
537 views

Prove that $n^2+n+41$ is prime for $n<40$

Here's a problem that showed up on an exam I took, I'm interested in seeing if there are other ways to approach it. Let $n\in\{0,1,...,39\}$. Prove that $n^2+n+41$ is prime. I shall provide my own ...
14
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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
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Ring of Polynomials is a Principal Ideal Ring implies Coefficient Ring is a Field?

I read this proof that if $D$ is an integral domain and $D[X]$ is a principal ideal domain, then $D$ is a field. My question is if the requirements can be relaxed a bit, namely: Is it true that ...
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5answers
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Is $\mathbb{Z}[x]$ a principal ideal domain?

Is $ \mathbb{Z}[x] $ a principal ideal domain? Since the standard definition of principal ideal domain is quite difficult to use. Could you give me some equivalent conditions on whether a ring is a ...
2
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1answer
83 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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6answers
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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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2answers
909 views

Ring of integers is a PID but not a Euclidean domain

I have noticed that to prove fields like $\mathbb{Q}(i)$ and $\mathbb{Q}(e^{\frac{2\pi i}{3}})$ have class number one, we show they are Euclidean domains by tessalating the complex plane with the ...
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2answers
930 views

Prime elements in $\mathbb{Z}[\sqrt{2}]$

What are the prime elements in the ring $\mathbb{Z}[\sqrt{2}]$? Note that since the ring is a PID (and thus a UFD) then prime = irreducible. Even more, it is Euclidean with respect to the absolute ...
3
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2answers
116 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 ...
6
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3answers
401 views

$\mathbb Z\times\mathbb Z$ is principal but is not a PID

I need to find an example of a ring that is not a PID but every ideal is principal. I know that $\mathbb Z\times\mathbb Z$ is not an integral domain, so certainly is not a PID, but here every ideal is ...
2
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2answers
152 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 ...
6
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2answers
149 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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2answers
153 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
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1answer
60 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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2answers
566 views

Specific way of showing $\Bbb Z[\sqrt{-d}]$ is not a Euclidean Domain when $d>2$

Is it true that if a ring is not a UFD then it's not a Euclidean Domain? I have a ring $R=\mathbb{Z}[\sqrt{-d}]=\{ a+b\sqrt{-d} \mid a,b \in \mathbb{Z} \}$ where $d$ is a square free integer. I want ...
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1answer
68 views

Describe units and maximal ideals in these two PIDs

If $p$ is a fixed prime integer, let $R$ be the set of all rational numbers that can be written in the form $(a)$ $\frac{a}{b}$ with $b$ not divisible by $p$. $(b)$ $\frac{a}{b}$ with $b=p^k$ for a ...
0
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1answer
58 views

Over a PID, $\text{rank}(F/N)=0 \Longleftrightarrow\text{rank}(F)=\text{rank}(N)$?

Let $D$ a PID, $F$ a free module rank $n$, $N$ a submodule of $F$. I want to prove (or find a counterexample) of: $\text{rank}(F/N)=0 \Longleftrightarrow\text{rank}(F)=\text{rank}(N)$ ...
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5answers
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How does a Class group measure the failure of Unique factorization?

I have been stuck with a severe problem from last few days. I have developed some intuition for my-self in understanding the class group, but I lost the track of it in my brain. So I am now facing a ...
4
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3answers
893 views

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal. What could be wrong in this approach?

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which ...
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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 ...
2
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1answer
378 views

Is any UFD also a PID?

Is there any counterexample that will disprove that every unique factorization domain (UFD) is also a principal ideal domain (PID)? I mean, any PID is a UFD, does the converse hold? Thanks in ...
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2answers
89 views

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 ...
5
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2answers
267 views

Characterization of primary ideals in a principal ideal domain

On the commutative algebra wiki, a table of properties lists that "for a PID, the primary ideals coincide with the powers of prime ideals." I played around with it, couldn't produce a proof, ...
4
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1answer
168 views

A question on valuation overrings of a PID

Let $A$ be a PID and let $K$ be its quotient field. Let $V$ be a valuation ring of $K$ containing $A$ and assume $V\neq K$. Show that $V$ is a local ring $A_{(p)}$ for some prime element $p$. I ...
3
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1answer
78 views

Finite intersection of DVRs

Let $K$ be a field and $R_1,\dots,R_n$ DVRs of $K$ with $m_i$ the maximal ideal of $R_i$ and $R_i \not\subseteq R_j$ for $j\neq i$ . Define $A=\bigcap_{i=1}^n R_i$. Then $A$ is semilocal with maximal ...
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2answers
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Proving that a ring is not a Principal Ideal Domain

This is my first question on StackExchange. I'm taking a second semester course of Abstract Algebra. I have a general understanding of Principal Ideal Domains, but I am a bit confused on proving that ...
3
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1answer
171 views

Do a matrix and its transpose have the same invariant factors over a PID?

I suspect this is true since it holds in the case over a field. But suppose $A\in M_{m\times n}(R)$ where $R$ is a PID. Does it still hold that $A$ and $A^{T}$ have the same invariant factors? ...
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2answers
93 views

Integral domain, UFD and PID related problem

(i) Let $R$ be an integral domain that has irreducible elements. Prove that $R[X]$ is not A PID. (ii) Let $R$ be a UFD and $K$ its field of fractions. Let $f \in R[X]$ be a monic polynomial ...
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2answers
60 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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2answers
77 views

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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1answer
439 views

$R$ is PID, so $R/I$ is PID, and application on $\mathbb{Z}$ and $\mathbb{N}$

I'm supposed to show in a part of an exercise that if we have a ring $R$ that is a principal ideal domain, then for any ideal $I$ in $R$, $R/I$ will also be a PID. So $I=(i)$ for some $i \in R$, and ...
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Show that the ring of all rational numbers, which when written in simplest form has an odd denominator, is a principal ideal domain.

Show that the ring of all rational numbers $m/n$ with $n$ an odd integer is a principal ideal domain. We haven't really discussed principal ideal domains. I've heard that this is easy, but I just ...
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1answer
459 views

Are all subrings of the rationals Euclidean domains?

This is a purely recreational question -- I came up with it when setting an undergraduate example sheet. Let's go with Wikipedia's definition of a Euclidean domain. So an ID $R$ is a Euclidean domain ...
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3answers
470 views

Why define vector spaces over fields instead of a PID?

In my few years of studying abstract algebra I've always seen vector spaces over fields, rather than other weaker structures. What are the differences of having a vector space (or whatever the ...
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3answers
200 views

Show that $\mathbb{Z}[x]=\lbrace \sum_{i=0}^{n}{a_ix^i}:a_i \in \mathbb{Z}, n \geq 0 \rbrace$ is not a principal ideal ring.

Show that $\mathbb{Z}[x]=\lbrace \sum_{i=0}^{n}{a_ix^i}:a_i \in \mathbb{Z}, n \geq 0 \rbrace$ is not a principal ideal ring. I know the definition of principal ideal ring is that every ideal is ...
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0answers
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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 $R=M=K[x,y]$ , $L=<x>$ , ...
3
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1answer
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For which $d$ is $\mathbb Z[\sqrt d]$ a principal ideal domain?

Is there any general idea about for which $d$, $\mathbb Z[\sqrt d]$ a principal ideal domain (PID)? As for example $\mathbb Z[\sqrt{-1}]$ and $\mathbb Z[\sqrt 2] $ are PIDs, but $\mathbb Z[\sqrt{-5}] ...
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2answers
127 views

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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2answers
57 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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2answers
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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 ...