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

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Help understanding statement relating to structure of modules over PIDs

Lemma IV.6.11 of Hungerford's Algebra is Lemma 6.11. Let $R$ be a principal ideal domain. If $r \in R$ factors as $r = p_1^{n_1} \cdots p_k^{n_k}$ with $p_1,\ldots,p_k \in R$ distinct primes and ...
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1k views

If F is a field, then $F[x,y]$ is a Principal Ideal Domain?

Let $F$ be a field, and $F[x,y]$ be a ring of polynomials in two variables. Is $F[x,y]$ a Principal Ideal Domain? Also show that $F[x,y]/(y^2-x)$ and $F[x,y]/(y^2-x^2)$ are not isomorphic for any ...
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1answer
65 views

Finitely generated graded modules over $K[x]$

I need some help on this exercise from A Course in Ring Theory by Donald S. Passman The result is supposely similar to the well-known structure theorem in the non-graded case. So let $M$ be a ...
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2answers
41 views

Is the ideal $(2,x^4+x^2+1)<\mathbb{Z}[x]$ maximal?, principal?

I'm trying to solve the following problem: Let I=$(2,x^4+x^2+1)<\mathbb{Z}[x]$ be an ideal. Is $I$ maximal? Is $I$ principal? Any help would be appreciated.
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1answer
563 views

Proofs of the structure theorem for finitely generated modules over a PID

I'm looking for different proofs (references or sketch of main ideas) of the structure theorem for finitely generated modules over a PID. If possible, a comparison in terms of clarity, elegance or ...
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2answers
135 views

Is $\mathbb{C}[x,y] / (y^2-x^3)$ a PID?

First, I'd like to show $\mathbb{C}[x,y] / (y^2-x^3)$ is an integral domain. Then I need to find out whether or not it is a PID. For the first part, I want to show $y^2-x^3 \: | \: fg \implies ...
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57 views

module over a quotient of a principal ideal domain

The Statement I suspect the following proposition is well known, but I found no reference. Proposition If $A$ is a principal ideal domain, if $I$ is a nonzero ideal of $A$, and if $M$ is an ...
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R is a PID, and a is a nonzero nonunit in R. How can we show R/Ra is an injective module over R?

If we use Baer's criterion then it suffices to show that if there exist a map from an ideal $I$ to $R/Ra$ we must find a map $g$ such that $g\circ i=f$.
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2answers
277 views

Quotient of polynomials, PID but not Euclidean domain?

While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the Euclidean domain page of Wikipedia that $$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$ is ...
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3answers
51 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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2answers
53 views

If $N\cap rM=rN$ for all $r\in R$, then is $M=N\oplus K$ for some $K$?

Suppose $M$ is a finitely generated free module over a principal ideal domain $R$, and $N$ a submodule. Why does the condition $N\cap rM=rN$ for all $r\in R$ implies that $M=N\oplus K$ for some ...
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275 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
291 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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24 views

Smith normal forms and a math program

I am interested to know the Smith normal form of $4 \times 2$ matrices $M$: The two cases of my interests are: (1). $$M_1= \begin{pmatrix} 3 & 0\\ -5 & 4\\ 4 & -5\\ 0 & 3 ...
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2answers
56 views

Intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ is not finitely generated.

Consider the subring $\mathbb{Z}[2x,2x^2,2x^3,\dots]\subset \mathbb{Z}[x]$. Then show that the intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ i.e., $I\cap ...
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146 views

$M \oplus M \simeq N \oplus N$ then $M \simeq N.$

Let $M$ and $N$ be finitely generated $R$-modules where $R$ principal domain. Show that if $M \oplus M \simeq N \oplus N$ then $M \simeq N.$
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1answer
20 views

Show that if $R$ is principal, $N $ is pure and $Ann(x+N)= Rd$ then there exists $y \in M$ such that $x+N=y+N$ and $Ann(y)=Rd$

Let $R$ a integral domain and $M$ a $R$-module. A submodule $N$ of $M$ is pure if for all $x \in M$ and $a \in R$ such that $ax \in N$, there exists $y \in N$ such that $ax=ay$. Show that if $R$ ...
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84 views

Ring Sandwiched between PIDs

If I have three commutative rings $R \subset S \subset T$, such that $R$ and $T$ are principal ideal domains, will this imply that $S$ itself is a principal ideal domain?
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51 views

Show that an integral domain $R$ is principal if and only if every submodule of a cyclic $R$-module is cyclic.

Good morning, I have difficulty with this problem: Show that an integral domain $R$ is principal if and only if every submodule a cyclic $R$-module is also cyclic.
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1answer
1k views

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 ...
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24 views

$Hom(F,T)$, where $F$ is torsion-free and $T$ is torsion module [closed]

If the underlying ring is a PID, then is it true that $Hom(F,T)=0$?
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1answer
20 views

Quick question: SES where base ring is a PID

Let $M$ be an $R$-module, where $R$ is a PID. Assume there is a free $R$-module $F$ and a surjective map $\phi :F\rightarrow M$. Then why is $\ker(\phi)$ also free? Thank you.
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31 views

principal ideals, integral domains, ideals,?

I am stuck trying to grasp this concept. I know that $\Bbb{Z}$ is a PID, $R=\Bbb{Z}[X]$ is not a PID, $\Bbb{Z}[i]$ is a PID. If someone could help me grasp these concepts it would be helpful. ...
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1answer
45 views

A submodule of a free module over a PID

$R$ is a principal ideal domain and $F$ is a free module over $R$ of infinite rank with basis $\{e_1,...,e_n,...\}$. Is it true that the $R$-submodule of $F$ spanned by $\{e_1,...,e_n\}$ has a ...
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7 views

Number of solutions to a congruence in a PID

This is the proof that there are $(a,m)$ solutions to $ax\equiv b \mod m$ for the ring $\mathbb{Z}/m\mathbb{Z}$. Where does this proof not hold for a general PID, allowing for (for instance) an ...
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39 views

Examples of PIDs and prime ideals

(a) Give a specific example of a PID with exactly two prime ideals. Give a brief proof of your answer. (b) Give an specific example of a PID with infinitely many prime ideals. Give a brief proof of ...
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385 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 ...
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1answer
55 views

Irreducible elements in a PID are prime

How can I see that all irreducible elements in a principal ideal domain are prime? $u$ is irreducible when $u_1 u_2 = u \implies u_1 $ or $u_2$ is a unit. $u$ is prime when $u | ab \implies u|a$ or ...
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1answer
19 views

Are rank and determinantal rank the same over a PID?

Are the notions of rank and determinantal rank equivalent for an $m\times n$ matrix $A$ with entries in a principal ideal domain $D$? I'm specifically interested in the case $D=\mathbb{Z}$.
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772 views

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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Proof for Unique Factorization Domain

Prove that the quotient ring $\mathbb{C}[x,y]/(x^2+y^2-1)$ is a unique factorization domain. I am trying to prove first it is a principal ideal domain. However I am really stuck on this problem
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62 views

Proof for maximal ideals in $\mathbb{Z}[x]$

I have been trying to prove the following theorem: Every maximal ideal in $\mathbb{Z}[x]$ has the form $(p, f(x))$ where p is prime integer and f is primitive integer polynomial that is irreducible ...
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1answer
62 views

Abstract Algebra: integral domain and principal ideal domain

I am studying by myself and I needed help for few question which I am confused how give proof of that. Let $\varphi : J \to K$ be a ring epimorphism with $\varphi(1) = 1$, where $J$ and $K$ are ...
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11 views

Domain GCD Property

Let D be a domain and $\emptyset \subset A \subseteq D^*$ $d \in GCD(A)$ if and only if (d) is a minimum among the principal ideals containing (A) If $d \in GCD(A)$ then d|a for all $a \in A$ and ...
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1answer
73 views

Structure Theorem For PIDs

So, I'm a biologist at KCL, but I quite like mathematics and so am going through a book of exercises in algebra. Unfortunately, I've run into a problem in trying to answer some of the questions. I've ...
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1answer
45 views

Projective Modules on PIDs

We are given a PID. Then we have to show that if F is a finitely generated free module of rank n, then a submodule M of F is free. This is fine. However, it then says to deduce that every finitely ...
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Ring Theory Domain Proof

Let D be a domain. Show that $D[X]^x$=$D^x$. Because D is a domain it means that it is cancellative and D has no nonzero zero divisors. The only units in $D[X]^x$ are the units in $D^x$ so it's ...
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1answer
191 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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40 views

$IJ =(I\cap J)(I+J)$ holds in a PID? $I,J$ Ideals of a PID

One inequality is obvious, but the other one im not sure if holds. Any idea?
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24 views

Show $M(\mathfrak{p})$ is an $R$-submodule of $M$

Let $R$ be a PID, $\mathfrak{p}$ be a prime ideal in $R$, and $M$ an $R$-module. Then $$M(\mathfrak{p})=\{m\in M \mid \operatorname{Ann}_Rm=\mathfrak{p}^j, \, \text{ for some } j\geq 0 \}$$ is ...
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94 views

Homology out of Smith normal form

Let $R$ be a PID and $A: R^m\rightarrow R^n$ and $B:R^n\rightarrow R^o$ with $BA=0$ and Smith normal forms $A=P\mathrm{diag}(a_1,\ldots,a_r,0,\ldots,0)Q^{-1}$ and ...
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2answers
56 views

Nonprincipal prime ideals contain two relatively prime elements

Let $R$ be a principal ideal domain and let $P$ be a nonprincipal prime ideal of $R[x]$. I'm having trouble seeing why $P$ must contain two elements with no common divisor. Can anyone help me? ...
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123 views

An ideal in a domain is free if and only if it is principal [closed]

Let $R$ be a domain. Show that every ideal in $R$ is $R$-torsion-free, but is free if and only if it is principal. In particular, show that $R$ is a PID if every submodule of a free module is free. ...
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145 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 ...
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1answer
21 views

Help decomposing an ideal.

Let $R$ be a PID and $A$ an ideal of $R$ such that for any $r^2 \in A$ where $r \in R$, then $r \in A$. I'm trying to show that $R / A$ is isomorphic to a finite direct product of fields. But I'm a ...
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1answer
619 views

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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2answers
996 views

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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1answer
238 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 ...
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2answers
131 views

Prove that all ideals in Q[x] are principal

Prove that all ideals in the polynomial ring $\mathbb{Q}[x]$ are principal. There is probably some elegant shortcut one can use for this proof, but I am only just beginning to study ring theory and ...
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1answer
102 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 ...