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

learn more… | top users | synonyms (1)

19
votes
2answers
657 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 ...
17
votes
2answers
2k 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 ...
16
votes
2answers
2k 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 ...
15
votes
4answers
823 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 ...
15
votes
2answers
4k 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 ...
14
votes
3answers
3k 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).
14
votes
5answers
789 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
1answer
739 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
3k 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 ...
10
votes
1answer
412 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
votes
3answers
601 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 ...
9
votes
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 ...
9
votes
1answer
884 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 ...
8
votes
5answers
5k 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 ...
8
votes
2answers
428 views

What information do we gain from PIDs

I am self-learning some algebraic number theory and my question is regarding the advantages to studying PIDs. I have seen that Euclidean Domains $\subseteq$ Principal Idea Domains $\subseteq$ Unique ...
8
votes
2answers
181 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
votes
2answers
1k 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
169 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 ...
7
votes
1answer
85 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
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
608 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
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
242 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
votes
1answer
1k 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, ...
6
votes
4answers
319 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 ...
6
votes
1answer
387 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 ...
6
votes
2answers
1k 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
59 views

Something wrong my proof that $\mathbb{Z}[x]$ is not a PID nor an Euclidean domain?

My proof goes as follows: Suppose for contradiction that $\mathbb{Z}[x]$ is a PID. Then the ideal generated by any irreducible element is maximal. We know that $x^2+1$ is irreducible in ...
6
votes
1answer
189 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 ...
6
votes
1answer
244 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 ...
6
votes
1answer
128 views

$\mathbb{Q}(\sqrt[3]{17})$ has class number $1$

Let $\alpha:=\mathbb{Q}(\sqrt[3]{17})$ and $K:=\mathbb{Q}(\alpha)$. We know that $$\mathcal{O}_K=\left\{\frac{a+b\alpha+c\alpha^2}{3}:a\equiv c\equiv -b\pmod{3}\right\}.$$ I have to show that $K$ has ...
6
votes
0answers
213 views

Reducing multivariate rational fractions to lowest terms

I wish to simplify multivariate rational fractions to a canonical form. Thanks to some very helpful mathematically inclined people who verified that my understanding of Wikipedia was correct, I'm now ...
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
3answers
3k 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 ...
5
votes
2answers
201 views

One-dimensional Noetherian UFD is a PID

I am looking for a reference which has a self-contained (elementary, that is, at the "undergraduate algebra level") proof of the the fact that any one-dimensional Noetherian UFD is a PID. Does anyone ...
5
votes
1answer
1k 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 ...
5
votes
1answer
404 views

Contraction of maximal ideals in polynomial rings over PIDs

Let $R$ be a principal ideal domain which is not a field, and let $M$ be a maximal ideal of the polynomial ring $R[X_1,\dots,X_n]$. If $n=1$ it is very easy to see that $M \cap R \neq 0$. Is this also ...
5
votes
2answers
64 views

Is $\mathbf{Z}[X]/(2,X^2+1)$ a field/PID?

I've been asked to determined whether the following are fields, PIDs, UFDs, integral domains: $$\mathbf{Z}[X],\quad \mathbf{Z}[X]/(X^2+1),\quad \mathbf{Z}[X]/(2,X^2+1)\quad \mathbf{Z}[X]/(2,X^2+X+1)$$ ...
5
votes
2answers
337 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
203 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 ...
5
votes
1answer
523 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
votes
2answers
97 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
1k views

For which $d$ is $\mathbb Z[\sqrt d]$ a principal ideal domain?

Is there any general idea about for which $d$, $\mathbb Z[\sqrt d]$ a principal ideal domain (PID)? As for example $\mathbb Z[\sqrt{-1}]$ and $\mathbb Z[\sqrt 2] $ are PIDs, but $\mathbb Z[\sqrt{-5}] ...
5
votes
1answer
127 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
2answers
436 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 ...
5
votes
1answer
483 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
votes
1answer
180 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
132 views

Applications of the Dedekind-Hasse criterion

It is a fact that an integral domain $R$ is a principal ideal domain if and only if there is a Dedekind-Hasse function $|R|\setminus\{0\}\xrightarrow{\ \ \delta\ \ }\mathbb{N}$ on $R$, i.e. a function ...
5
votes
0answers
83 views

Elementary divisors for chains of submodules

Given free modules $N \le M$ of finite rank over a PID $R$, it's well-known that there is a basis $\{x_1,\ldots,x_n\}$ of $M$ and there are $e_1,\ldots,e_n \in R$ such that $\{e_ix_i\mid e_i \neq ...
4
votes
3answers
100 views

Ideal of $\mathbb Q[x]$ which contains two polynomials

Suppose $I$ is an ideal of $\mathbb Q[x]$ which contains $x^2 + 2x +4$ and $x^3 - 3$. Prove $I =\mathbb Q[x]$. This is an exercise in my abstract algebra text book. I know the definition of an ...