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

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14
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2answers
983 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 ...
14
votes
2answers
253 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 ...
12
votes
5answers
521 views

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 ...
11
votes
3answers
380 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 ...
10
votes
2answers
253 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 ...
10
votes
1answer
371 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 ...
10
votes
1answer
346 views

Finitely generated modules over PID

Let $A$, $B$, $C$, and $D$ be finitely generated modules over a PID $R$ such that $A\oplus $ $B$ $\cong$ $C\oplus $ $D$ and $A\oplus $ $D$ $\cong$ $C\oplus $ $B$ . Prove that $A$ $\cong$ $C$ and $B$ ...
8
votes
4answers
913 views

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).
8
votes
1answer
595 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 ...
7
votes
3answers
268 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 ...
7
votes
2answers
143 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 $M \simeq N.$
6
votes
3answers
218 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 ...
6
votes
3answers
820 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. ...
6
votes
1answer
177 views

Integer extensions, rings $\mathbb{Z}[\sqrt{s}]$

I'm not sure if this type of question is acceptable here, but I'd really appreciate someone's help. I'm about to start writing a semestral work that we need to achieve the Bachelor's degree in our ...
5
votes
3answers
143 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 ...
5
votes
1answer
636 views

How many real quadratic number fields have the class number 1?

I know that in general the number of ideal classes are not 1, and that there are only 9 imaginary quadratic number fields which are principal ideal domains, i.e. $\mathbb(Q(\sqrt{-m}))$ where m is 1, ...
5
votes
2answers
659 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 ...
5
votes
2answers
83 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?
5
votes
1answer
163 views

Quotient of ring of integers

Let $R=\mathcal{O}(K)$ be the ring of the integers of $K=\mathbb{Q}[\zeta_8]$, where $\zeta_8=e^{2\pi i/8}=\sqrt{2}/2(1+i)$ is a primitive eighth root of unity in $\mathbb{C}$. It can be shown that ...
4
votes
3answers
739 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 ...
4
votes
1answer
117 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 ...
4
votes
1answer
118 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 ...
4
votes
2answers
54 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 ...
4
votes
1answer
227 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 ...
4
votes
1answer
143 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 ...
4
votes
1answer
187 views

Question about torsion submodules and Decomposition theorem

Let $A$ be an principal ideal domain, and $M$ an $A$-module. If $p$ is irreducible in $A$, let's define $$\mathrm{Tor}_p(M):=\{m\in M\mid p^km=0\text{ for some }k\in\mathbb{N}\}.$$ I need to show ...
3
votes
6answers
2k views

Proving the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain

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
3
votes
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 ...
3
votes
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 ...
3
votes
2answers
252 views

A question about Euclidean Domain

This is a problem from Aluffi's book, chapter V 2.17. "Let $R$ be a Euclidean Domain that is not a field. Prove that there exists a nonzero, nonunit element $c$ in $R$ such that $\forall a \in R$, ...
3
votes
1answer
61 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 ...
3
votes
2answers
129 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 ...
3
votes
3answers
68 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 ...
3
votes
1answer
132 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? ...
3
votes
1answer
106 views

Quotient ring of $\Bbb Z[x]$ by an irreducible polynomial is a PID

I don't know what can I do with this problem. How can I prove that $\mathbb{Z}[x]/(x^{3}-4x+2)$ is PID?
3
votes
1answer
98 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 ...
3
votes
1answer
154 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 ...
3
votes
1answer
813 views

How do I get this matrix in Smith Normal Form? And, is Smith Normal Form unique?

As part of a larger problem, I want to compute the Smith Normal Form of $xI-B$ over $\mathbb{Q}[x]$ where $$ B=\begin{pmatrix} 5 & 2 & -8 & -8 \\ -6 & -3 & 8 & 8 \\ -3 & ...
3
votes
1answer
72 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 ...
3
votes
0answers
50 views

Is quantum torus a principal ideal domain?

For a quantum torus $C_q[x_1^{\pm1}, ...,x_n^{\pm1}]$ satisfying $x_ix_j=q_{ij}x_jx_i$. Question: Is this quantum torus a principal ideal domain?
3
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0answers
341 views

Exercise on modules over PID involving injective modules, Baer's criterion.

I'm interested if I solved this somewhat correctly, and would like to be set straight if it is wrong. This is an exercise from an introductory text. Let $A$ be a module over a principal ideal ...
2
votes
4answers
196 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?
2
votes
2answers
126 views

Sub-module over the Fraction field of a PID

R is a PID with field of fractions K. $M \subseteq K$ is a fin. generated R-Submodule. I am trying to show M is in fact generated by one element.
2
votes
1answer
191 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.
2
votes
2answers
282 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 ...
2
votes
2answers
51 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?
2
votes
2answers
1k views

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: ...
2
votes
1answer
754 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 ...
2
votes
1answer
127 views

Why does the result hold for PIDs but not for UFDs?

Let $R$ be a subring of an integral domain $S$, and suppose $R$ is a PID. Then it follows that if $r\in R$ is a gcd of $r_1$ and $r_2$ in $R$, where $r_1$ and $r_2$ are not both zero, then $r$ is a ...
2
votes
2answers
1k views

Every prime ideal is either zero or maximal in a PID.

$(1)$ Let $R$ be a commutative ring with $1\neq 0.$ If $R$ is a PID, show that every prime ideal is either zero or maximal. In many books I have found the proof of the above statement where they ...