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

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17 views

What exactly does it mean for a maximal ideal to be unique in a principal ideal domain?

I'm currently reading about PIDs and have come across a question involving maximal ideals which at one point reads "Suppose that a Euclidean domain $R$ had a unique maxima ideal $P$". Does this mean ...
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1answer
42 views

Suppose that $K$ is a field and that $f$ and $g$ are relatively prime in $K[x]$. Show that $f - Yg$ is irreducible in $K(y)[x]$.

I'm a bit confused of the notation $K(y)[x]$, is that simply $K[y][x]$ so... $K[y,x]?$ Anyways, here's my attempt at trying this before I get stuck. Since $f$ and $g$ are relatively prime, that ...
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0answers
24 views

Domain of reciprocal of log in complex plane

Ok so what i already know is that the function f(z) = log(z) is undefined when z = negative or zero which is quite a basic concept. Can someone explain the mystery ...
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1answer
44 views

Principal Ideal Domains with Many Primes of Index 2

By the index of a prime $p$ in a principal ideal domain $R$, I mean the number of cosets of the ideal $(p)$ in $R$ (i.e. $\vert R/(p)\vert$). If I am given a positive integer $m$, I am wondering ...
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1answer
48 views

$\Bbb Z[x]$ is not a principal domain

I know this is already answered here but I am wondering that if the following way to prove is also correct - let $$f(x) = 4x^2+4x+1$$ $$g(x)=4x^2-1$$ Since this is a UFD so a unique gcd will ...
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1answer
104 views

Problem with Smith normal form over a PID that is not an Euclidean domain

This is an homework exercise of the Algebra lecture. I need to evaluate the Smith normal form of the following matrix $$A:=\begin{pmatrix}1 & -\xi & \xi-1\\2 ...
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23 views

algebraic poset

I learn domain theory and stack in definition of algebraic poset. Recall $P$ is algebraic if for every $x\in P$,the set of compact element $y$ below $x$ is directed and has $x$ as least upper bound. ...
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28 views

Principal Ideal Domain Modulo a Power of a Prime Ideal

If $R$ is a PID, $p$ is a prime element of $R$, $R/(p)$ is finite, and $\alpha$ is a positive integer, is it true that $\vert R/(p^{\alpha})\vert=\vert R/(p)\vert^{\alpha}$? I seem to recall seeing ...
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1answer
36 views

The number 2 in a PID

Let $R$ be a PID. Then $R$ is a commutative ring with multiplicative identity $1$. We can then define $2=1+1$. From here, what is known about $2$ and its prime factors? I suppose this breaks into two ...
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38 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 don't ...
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1answer
46 views

Difference between PID and principal ideal ring

All rings are commutative, associative and with 1. Wikipedia states that the difference between PID and Principal Ideal Ring is that the former has to be integral domain while the latter does not. ...
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1answer
25 views

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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2answers
43 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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149 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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72 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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32 views

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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3answers
72 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
356 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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330 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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25 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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57 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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1answer
21 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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87 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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148 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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56 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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25 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
21 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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33 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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48 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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10 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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42 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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1answer
81 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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21 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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1answer
87 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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66 views

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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71 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
64 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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13 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
78 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
46 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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398 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
44 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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25 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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111 views

Homology out of Smith normal form: simultaneous or independent diagonalization?

Let $R$ be a PID and $R^m\overset{A}{\longrightarrow} R^n\overset{B}{\longrightarrow} R^o$ matrices with $BA=0$ and Smith normal forms ...
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1answer
215 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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125 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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58 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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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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149 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 ...