1
vote
1answer
22 views

Help understanding statement relating to structure of modules over PIDs

Lemma IV.6.11 of Hungerford's Algebra is Lemma 6.11. Let $R$ be a principal ideal domain. If $r \in R$ factors as $r = p_1^{n_1} \cdots p_k^{n_k}$ with $p_1,\ldots,p_k \in R$ distinct primes and ...
1
vote
2answers
135 views

Is $\mathbb{C}[x,y] / (y^2-x^3)$ a PID?

First, I'd like to show $\mathbb{C}[x,y] / (y^2-x^3)$ is an integral domain. Then I need to find out whether or not it is a PID. For the first part, I want to show $y^2-x^3 \: | \: fg \implies ...
3
votes
2answers
57 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 ...
-2
votes
2answers
32 views

R is a PID, and a is a nonzero nonunit in R. How can we show R/Ra is an injective module over R?

If we use Baer's criterion then it suffices to show that if there exist a map from an ideal $I$ to $R/Ra$ we must find a map $g$ such that $g\circ i=f$.
2
votes
3answers
51 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 ...
15
votes
2answers
275 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 ...
10
votes
2answers
275 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
24 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
56 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 ...
5
votes
2answers
84 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?
7
votes
2answers
146 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
51 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
24 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 $Hom(F,T)=0$?
0
votes
1answer
45 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
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 $D=\mathbb{Z}$.
2
votes
0answers
61 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
62 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
62 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
1answer
73 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 ...
11
votes
3answers
385 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
40 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
94 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
191 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
123 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. ...
4
votes
1answer
145 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
237 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
131 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
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 ...
1
vote
1answer
102 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
52 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
99 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
38 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
56 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 ...
0
votes
0answers
139 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
61 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
1answer
126 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
63 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 ...
0
votes
0answers
48 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) ...
-2
votes
0answers
39 views

Quotient of a PID by a prime ideal [duplicate]

Prove that quotient of a PID by a prime ideal is PID.
2
votes
2answers
291 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 ...
1
vote
1answer
63 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 ...
3
votes
3answers
69 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 ...
0
votes
2answers
57 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?
0
votes
0answers
90 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 ...
3
votes
1answer
110 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?
6
votes
3answers
221 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 ...