2
votes
0answers
50 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
1
vote
0answers
30 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 ...
0
votes
0answers
10 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 ...
3
votes
1answer
62 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 ...
0
votes
0answers
17 views

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 ...
0
votes
1answer
34 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?
2
votes
1answer
157 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 ...
1
vote
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 ...
4
votes
1answer
123 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 ...
1
vote
1answer
79 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 ...
2
votes
2answers
43 views

proper ideals in the principal ideal domain

I'm to prove that every proper ideal is a product of maximal ideals which are uniquely determined up to order. I have no idea even how to start in the proof to solve this question :( May anybody help ...
1
vote
1answer
88 views

Infinite direct product of rings free.

Let $A$ be a commutative ring (viewed as an $A$-module over itself) that is not a field. Are there some conditions that guarantee that $\prod_{k=0}^\infty A$ is free? What if $A=\mathbf{Z}$ or more ...
1
vote
1answer
36 views

irreducible elements of polynomial rings

Let $p$ be a prime integer. For $x\in\mathbb{Z}$, let $x'$ be the remainder of $x$ when divided by $p$. Let $\sum_{i=0}^{n}a_iX^i\in \mathbb{Z}[X]$ with $p$ does not divide $a_n$ in $\mathbb{Z}$. Then ...
0
votes
1answer
50 views

to show that a ring is a principal ideal ring

I'm asked to show that $\mathbb{Z}_m$ (the integers mod $m$) is a principal ideal ring for every $m > 0$ I see that it is the same discussion used in verifying that $\mathbb{Z}$ (the set of all ...
0
votes
0answers
95 views

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

Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PID. I think I have proved this: Let $J$ be an ideal of $S$. Then $f^{-1}(J)=(a)$ is a principal ideal of ...
0
votes
1answer
55 views

Let $R$ be an integral domain and $I$ be a prime ideal of $R$. If $R/I$ is a Euclidean domain, will $R$ be a unique factorization domain?

Let $R$ be an integral domain and $I$ be a prime ideal of $R$. If $R/I$ is a Euclidean domain, will $R$ be a unique factorization domain? I have no idea to prove or disprove this... should I prove ...
0
votes
1answer
62 views

Prove that $M$ is a free module if and only if $M$ is a projective module over $PID$.

Let $R$ be a principal ideal domain and $M$ a finitely generated $R$ module. Prove that $M$ is a free $R$-module if and only if $M$ is a projective $R$-module. I am quite confused and totally not ...
2
votes
2answers
48 views

Inverse image of a PID is a PID

Let $f : R \to S$ be a ring homomorphism from $R$ onto $S$. If $S$ is a PID, is $R$ then a PID? If this is not possbile, is there an example to contradict it?
1
vote
2answers
199 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 ...
3
votes
3answers
56 views

Showing that an integral domain is a PID if it satisfies two conditions

This is just a textbook problem from Dummit and Foote, but the issue is that our class barely touched on PIDs and the preceding material, so I don't really know or understand much. Anyway, Let ...
0
votes
2answers
54 views

Why $\langle I, J\rangle =R$ for distinct prime ideals $I$, $J$ of a principal ideal domain $R$?

Let $R$ be a principal ideal domain with identity and $I$, $J$ be distinct prime ideals of $R$. Prove that $1 \in \langle I, J\rangle$ hence $\langle I, J\rangle = R$. How to prove?
0
votes
0answers
88 views

Subring of the field of rational numbers

Let $R=\{a\cdot2^n\mid a,n \in \mathbb{Z}\}$ be a subring or the field of rational numbers $\mathbb Q$. i) What kind of elements are invertible in $R$? ii) Prove that $R$ is a principal ...
2
votes
1answer
35 views

Generator for the ideal $I + J$ where $I = (2 + 3i)$ and $J = (1 - i)$

On a related question I calculated the GCD of $I = (2 + 3i)$ and $J = (1 - i)$ to be $1$. Now I know that $\mathbb{Z}[i]$ is a principal ideal domain. And I also know that the greatest common divisor ...
0
votes
1answer
71 views

Ring homomorphism, maximal ideals

Here's a question from my worksheet, i solved subquestion (1) but can use help with the other 2...And also would appreciate any comments on my answer for subquestion (1). Let $\psi: R->S$ be a ...
1
vote
1answer
58 views

Number of ideals in $\Bbb Z[x]/(x^3+1, 7)$

I am trying to find the number of ideals in $R:=\Bbb Z[x]/(x^3+1, 7)$ and $S:=\Bbb Z[x]/(x^3+1, 3)$. I started with $R$ and tried to write it in terms of familiar rings, by using fundamental ...
3
votes
1answer
93 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 ...
2
votes
1answer
63 views

Counting the ideals of $\frac{\mathbb{R}[X]}{(X^2)}$

I want to ask you guys if I'm on the right track: Here's the question: Suppose $a \in \mathbb{R}$. Count the ideals of $\frac{\mathbb{R}[X]}{(X^2-a)}$. Give an example of a ring with exactly 3 prime ...
3
votes
1answer
95 views

number of element in a principal ideal domain can be $25/36/35/15$?

Could any one tell me number of element in a principal ideal domain can be $25/36/35/15$ ? I just know a principal ideal domain is generated by a single element. what the knowledge I need to find ...
0
votes
1answer
101 views

Principal Ideal Domain Basics.

Let $R$ be a Principal Ideal Domain and $a,b,c,d$ elements in $R$, such that $ab-cd=1$. I am trying to figure out why $Rb \cap Rd=Rdab+Rbcd$. In case this is true, I am wondering weather it is enough ...
1
vote
1answer
45 views

Ideals of nested PID's

Let $R\subset K$ be principal ideal domains. If $a,b$ are nonzero elements of $R$, prove that $I=J\cap R$, where $I$ and $J$ denote the ideals generated by $a,b$ in $R$ and $K$, respectively. Showing ...
7
votes
3answers
226 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 ...
5
votes
3answers
135 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 ...
2
votes
1answer
110 views

which of the following statements are true and why?

which of the following statements are true and why? Any two irreducibles in any UFD are associates. If $D$ is a PID, then $D[x]$ is a PID. In any UFD, if $p|a$ for an irreducible $p$, then $p$ ...
0
votes
2answers
370 views

Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain.

I'm having difficulties with a homework problem from Algebra by Hungerford. Let $R$ be the following subring of the complex numbers: $R = \{a + b(1 + \sqrt{19}i)/2 \mid a, b \in \mathbb{Z}\}$. ...
4
votes
3answers
689 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 ...
3
votes
1answer
149 views

Units in $\mathbb{Z}[\sqrt[3]{2}]$ : $\pm(1+\sqrt[3]{2}+(\sqrt[3]{2})^2)^n$?

The subring $\mathbb{Z}[\sqrt[3]{2}]\subset\mathbb{C}$ is a PID. I remember reading somewhere that the units in $\mathbb{Z}[\sqrt[3]{2}]$ are precisely the elements ...
4
votes
1answer
110 views

Is $(x^3-x^2+2x-1)$ prime in $\mathbb{Z}/(3)[x]$?

This is somewhat of a follow up on this question: Why is $(3,x^3-x^2+2x-1)$ not principal in $\mathbb{Z}[x]$? I'm curious, is $\mathbb{Z}[x]/I$ a domain, with $I=(3,x^3-x^2+2x-1)$? I know $I$ is not ...
2
votes
1answer
180 views

Ring of analytic functions on the circle

Let $A = C^\omega(S^1)$ (resp. $C^\omega_{\mathbb C}(S^1)$) the ring of real-analytic real-valued (resp. complex valued) functions on the circle. These rings have maximal ideals $\mathfrak m_p = ...
2
votes
4answers
178 views

If two elements in a ED have the same Euclidean norm, they are associates?

Is it very obvious that on a Euclidean Domain, two elements $x$ and $y$ have the same Euclidean norm $\nu(x) = \nu(y)$ then they are associates? Can someone give me a proof of this?
13
votes
2answers
907 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 ...
1
vote
1answer
228 views

Rings such that $A[x]$ is a principal ideal domain

Let $A$ be a commutative ring. Then the following assertions are equivalent. $A$ is a field; $A[x]$ is a Euclidean domain; $A[x]$ is a principal ideal domain; $A[x]$ is a unique factorization ...
1
vote
1answer
136 views

Queries on proof that every PID is a factorisation domain

I'm reading a proof from C. Musili's Rings and Modules that every PID is a factorisation domain. The author defines a factorisation domain as a commutative integral domain $R$ with a unit such that ...
10
votes
1answer
356 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 ...
6
votes
3answers
759 views

Dedekind domain with a finite number of prime ideals is principal

I am reading a proof of this result that uses the Chinese Remainder Theorem on (the finite number of) prime ideals $P_i$. In order to apply CRT we should assume that the prime ideals are coprime, i.e. ...
8
votes
1answer
475 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 ...
2
votes
1answer
183 views

Number of ideals of a PID modulo an ideal

Let $R$ be a Principal Ideal Domain and $(a)\neq(0)$ an ideal of $R$. Prove $R/(a)$ has a finite number of ideals.
1
vote
1answer
264 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 ...
6
votes
2answers
581 views

How to show that $R/I$ is Artinian when R is PID

I'm working through some of Hungerfords "Algebra", and having trouble with Excercise VIII 1.2.: Show that if $I$ is a non-zero ideal in a principal ideal domain (PID) $R$, then the ring $R/I$ is ...
2
votes
1answer
671 views

Norm-Euclidean rings?

For which integer $d$ is the ring $\mathbb{Z}[\sqrt{d}]$ norm-Euclidean? Here I'm referring to $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a,b \in \mathbb{Z}\}$, not the ring of integers of ...
1
vote
1answer
180 views

Can we consider set of all composite integers as an ideal? And if yes, why then Z is a PID?

In this wikipedia article it is said that set Z is a principal ideal domain, i.e. each one of its ideals can be generated by a single element. But if we consider set C of all composite integer numbers ...