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

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19
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2answers
521 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 ...
16
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2answers
1k 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 ...
15
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2answers
909 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 ...
14
votes
3answers
537 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 ...
14
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2answers
3k views

Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal

Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal. The forward direction is standard, and the reverse direction is giving me trouble. In particular, I can prove that ...
13
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5answers
706 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 ...
13
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4answers
2k 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).
11
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1answer
459 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
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1answer
384 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 $B$ ...
9
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1answer
1k 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 ...
8
votes
3answers
470 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 ...
8
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1answer
683 views

Proofs of the structure theorem for finitely generated modules over a PID

I'm looking for different proofs (references or sketch of main ideas) of the structure theorem for finitely generated modules over a PID. If possible, a comparison in terms of clarity, elegance or ...
7
votes
2answers
930 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 ...
7
votes
2answers
170 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 $M \simeq N.$
7
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1answer
71 views

Is this ring a PID? [closed]

Let $R$ be the $k$-subalgebra of $k(t)$ generated by the set $k[t]$, of all polynomials, and a pair of rational functions: ${1\over{t-1}}$ and ${1\over{t-2}}$. Is the ring $R$ a PID?
6
votes
5answers
3k 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 ...
6
votes
3answers
2k views

If F is a field, then $F[x,y]$ is a Principal Ideal Domain?

Let $F$ be a field, and $F[x,y]$ be a ring of polynomials in two variables. Is $F[x,y]$ a Principal Ideal Domain? Also show that $F[x,y]/(y^2-x)$ and $F[x,y]/(y^2-x^2)$ are not isomorphic for any ...
6
votes
3answers
401 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
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3answers
1k 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
3answers
200 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 ...
6
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2answers
951 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 ...
6
votes
2answers
148 views

Does there exist such an invertible matrix?

Let $n \geq 1$ and $A = \mathbb{k}[x]$, where $\mathbb{k}$ is a field. Let $a_1, \dots, a_n \in A$ be such that $$Aa_1 + \dots + Aa_n = A.$$ Does there exist an invertible matrix $\|r_{ij}\| \in ...
6
votes
1answer
211 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
1k 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 ...
5
votes
1answer
874 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
4answers
164 views

Why is $(2, 1+\sqrt{-5})$ not principal?

Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$? Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb ...
5
votes
1answer
79 views

Prove that $(2)$ is a prime ideal in $\mathbb Z[w]$

Let $w\in\mathbb C$ be such that $w^3=1$ and $w\neq1$. Prove that $(2)$ is a prime ideal in $\mathbb Z[w]$, and describe $\mathbb Z[w]/(2)$. What I wanted to do is to show that $\mathbb Z[w]$ is a ...
5
votes
1answer
255 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 ...
5
votes
2answers
267 views

Characterization of primary ideals in a principal ideal domain

On the commutative algebra wiki, a table of properties lists that "for a PID, the primary ideals coincide with the powers of prime ideals." I played around with it, couldn't produce a proof, ...
5
votes
1answer
163 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 ...
5
votes
2answers
91 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
231 views

Proof that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a PID

How would one prove that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a principal ideal domain (PID)? It isn't a Euclidean domain according to the Wikipedia article on PIDs.
5
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1answer
120 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 ...
5
votes
1answer
253 views

Proving a subring of $\mathbb{Q}$ containing $\mathbb{Z}$ is a PID

Let $S$ be a subring of $\mathbb{Q}$ containing $\mathbb{Z}$. Prove that it is a principal ideal domain. So here is what I tried. Take any ideal $I\subset S$. Take any two elements, say $a=p/q, ...
5
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1answer
175 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 ...
5
votes
0answers
31 views

Are the ring of integers of the constructible numbers a Euclidean domain?

I suspect that since Euclid uses the Euclidean Algorithm to perform division on constructible numbers in Elements, the ring of integers of the constructible numbers are a Euclidean Domain, but I have ...
4
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3answers
892 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 ...
4
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1answer
2k 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 ...
4
votes
2answers
70 views

In $\Bbb Z$, what element generates the ideal $(4,7)$?

I have a really silly question. $\mathbb{Z},+,\cdot$ is a HID, so all ideals are principal ideals. Now, $(4,7)$ is an ideal in $\mathbb{Z}$, so it must be a principal ideal, but which element is its ...
4
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1answer
71 views

Principal ideal domains that are not integral domains

In the usual definition, a principal ideal domain $R$ is also assumed to be an integral domain? However, the property that every ideal is generated by a single element does not seem to immediately ...
4
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3answers
90 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 ...
4
votes
1answer
139 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
2k 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 ...
4
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1answer
59 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 ...
4
votes
2answers
52 views

The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$)

The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$) Can I get walked through one of them so that I ...
4
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1answer
168 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
76 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 ...
4
votes
1answer
239 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 ...
4
votes
0answers
80 views

Property of free submodules for a module over a PID

It's possible to produce an example of an integral domain $R$ and a free $R$-module $M$ with free submodules $L, L'$ such that $L+L'$ is not free. We can take $R=M=K[x,y]$ , $L=<x>$ , ...
4
votes
2answers
207 views

Ring of formal power series over a principal ideal domain is a unique factorisation domain

An exercise in my algebra course book asks to prove that if $R$ is a PID, then $R[[x]]$ is a UFD, where $R[[x]]$ is the ring of formal power series over $R$. After some failed attempts at proving the ...