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

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9
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2answers
233 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 ...
0
votes
0answers
23 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 ...
2
votes
2answers
54 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 ...
2
votes
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 $R$ ...
4
votes
2answers
81 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?
6
votes
2answers
139 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 $M \simeq N.$
0
votes
2answers
48 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.
-1
votes
2answers
23 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 $Hom(F,T)=0$?
2
votes
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.
0
votes
1answer
28 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. ...
0
votes
1answer
41 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 ...
0
votes
0answers
6 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 ...
1
vote
0answers
38 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 ...
1
vote
1answer
46 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 ...
0
votes
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 $D=\mathbb{Z}$.
1
vote
0answers
43 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 ...
2
votes
0answers
57 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
48 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 ...
3
votes
1answer
60 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 ...
0
votes
0answers
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 ...
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 ...
1
vote
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 ...
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 ...
11
votes
3answers
378 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 ...
0
votes
1answer
38 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?
0
votes
1answer
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 ...
0
votes
0answers
90 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 ...
2
votes
1answer
180 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
votes
1answer
121 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. ...
0
votes
2answers
53 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? ...
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
139 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
214 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
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
2answers
52 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 ...
1
vote
1answer
100 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 ...
1
vote
1answer
38 views

primary ideals in principal ideals domain

I'm to prove that an ideal $M$ is primary iff for some $n$, $M = (p^n)$ where $p$ is a prime or $p=0$. The second direction is simply proved referring to the definition of the primary ideal, my ...
2
votes
2answers
51 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
98 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
37 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 ...
2
votes
1answer
75 views

Intersection of a PID and a field

There is an exercise in Bourbaki about the intersection $A = k (x,y) [z] \cap k (z, x + yz)$. These are two subrings of $k (x,y,z)$. The first is PID, the second is a field. Bourbaki requests to prove ...
0
votes
1answer
55 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 ...
2
votes
1answer
56 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 ...
0
votes
0answers
135 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
60 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
0answers
24 views

Example of principal v-ideal

I am confuse in constructing counter example that If $A⊆B$ is an extension of domains such that $J$ is a principal $v$-ideal in $A$, then $JB$ is also principal in $B$.
0
votes
1answer
116 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 ...
0
votes
1answer
61 views

Is a ring generated by an irreducible polynomial a PID?

If $p(x)$ is an irreducible polynomial in $\mathbb R[X]$ (the set of polynomials with real coeffs), is the (sub)ring generated by $p(x)$ a PID? My guess is yes, at least if I have an ideal ...
2
votes
2answers
49 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?
0
votes
0answers
47 views

Invariant factors of torsion module under homomorphism

Let M be a torsion module for a PID D with invariant factors $(d_1) \supseteq (d_2) \supseteq...\supseteq (d_r) $ (which means $M = Dz_1 \bigoplus Dz_2 \bigoplus ... \bigoplus Dz_r $ s.t $ann(z_i) ...