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

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26 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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0answers
170 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 ...
2
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1answer
307 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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2answers
70 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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1answer
162 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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2answers
791 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
319 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
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2answers
63 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
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2answers
171 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 ...
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1answer
40 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 ...
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2answers
77 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 ...
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1answer
128 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 ...
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1answer
44 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 ...
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1answer
87 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 ...
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1answer
66 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 ...
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1answer
69 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 ...
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1answer
222 views

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

Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PIR. I think I have proved this: Let $J$ be an ideal of $S$. Then $f^{-1}(J)=(a)$ is a principal ideal of ...
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1answer
75 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 ...
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1answer
215 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 ...
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1answer
65 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 ...
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2answers
59 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?
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49 views

Quotient of a PID by a prime ideal [duplicate]

Prove that quotient of a PID by a prime ideal is PID.
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2answers
490 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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1answer
67 views

Describe units and maximal ideals in these two PIDs

If $p$ is a fixed prime integer, let $R$ be the set of all rational numbers that can be written in the form $(a)$ $\frac{a}{b}$ with $b$ not divisible by $p$. $(b)$ $\frac{a}{b}$ with $b=p^k$ for a ...
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3answers
85 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 ...
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1answer
64 views

gcd & lcm in a PID

In a PID, $l={\rm lcm}(a,b)$ and $d=\gcd(a,b)$. Is it always true that the following product ideals are equal? $$<d><l> = <a><b>$$ Thanks in advance -- Mike
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1answer
70 views

Classifying torsion-free injective modules over a PID

So the question is as in the title: Let $R$ be a PID. Classify all torsion-free injective modules. I know that it is going to be divisible, and using torsion free, if we define ...
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2answers
65 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?
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0answers
124 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 ...
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1answer
127 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?
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3answers
335 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 ...
2
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1answer
45 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 ...
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2answers
220 views

surjective homomorphism between principal ideal rings

I asked this question before but did not get an answer, I tried solving it myself and I think I'm heading somewhere, I just need a push in the right direction. Let $\phi: R \to S$ be a homomorphism ...
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1answer
120 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 ...
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0answers
37 views

Inclusion-minimality of a lattice basis

An integer lattice is a subgroup of $\mathbb{Z}^n$. Since $\mathbb{Z}$ is PID, each lattice has a well-defined rank and a generating set of rank many elements is a basis. I wonder if there is a way ...
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2answers
1k views

Proving that a ring is not a Principal Ideal Domain

This is my first question on StackExchange. I'm taking a second semester course of Abstract Algebra. I have a general understanding of Principal Ideal Domains, but I am a bit confused on proving that ...
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1answer
55 views

Understanding this basic lemma proof in Ireland & Rosen

Lemma 2. Let $R$ be a PID and $p$ a prime element and $a \neq 0$ any element. Then there is an integer $n$ such that $p^n \ | \ a$ but $p^{n+1}$ doesn't divide $a$. Proof. If the lemma were ...
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1answer
30 views

In the proof that in a PID, every non-zero non-unit is the product of irreducibles…

In proving that all non-zero non-units of a PID are a product of irreducibles, theres: "We now show that $a$ is a product of irreducibles. If $a$ is irreducible, we are done. Otherwise let $p_1$ ...
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2answers
298 views

Integral domain in complex numbers.

Let $I = _{\mathbb {C} [X]} \langle X^2 + 1\rangle$ the principal ideal of $\mathbb{C}[X]$ generated by $X^2 + 1$. Is $\mathbb{C}[X]/I$ an integral domain? From my understanding $\mathbb{C}[X]/I$ ...
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94 views

Confusion regarding the proof of “Every PID satisfies the Ascending Chain Condition”.

I refer to this proof of the fact that Principal Ideal Domains satisfy the Ascending Chain Condition. It says Let $\bigcup\limits_{i=1}^{\infty}I_i=(a)$. As $a$ is present in ...
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5answers
2k views

Is $\mathbb{Z}[x]$ a principal ideal domain?

Is $ \mathbb{Z}[x] $ a principal ideal domain? Since the standard definition of principal ideal domain is quite difficult to use. Could you give me some equivalent conditions on whether a ring is a ...
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1answer
69 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
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1answer
111 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 ...
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1answer
110 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 ...
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1answer
199 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 ...
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1answer
130 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 ...
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1answer
48 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 ...
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0answers
54 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?
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1answer
226 views

Find a non free submodule of a free module over R which is not PID

I try to solve following question. Show that $R=\mathbf{Q}[x,x^{-1}]$ is not a PID. Construct a free module over $R$ having a non free submodule. One may give some examples for free modules but not ...