# Tagged Questions

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

39 views

### If ring $R$ is a principal ideal domain, then $R[x]$ is principal ideal domain. [duplicate]

If ring $R$ is a principal ideal domain, then $R[x]$ is principal ideal domain. I think that it is false. So I try to find some PID not satisfying above hypothesis. Could you help me?
83 views

### On describing a sort of “well-behaved” subgroups of a free abelian group.

I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finitely generated case. Let $M$ be a free abelian group and $N$ a ...
31 views

### I wonder if $K$ is a field then $K[X]$ is a PID then the ideals of it have formula that is $<f(X)>$ when $f(X)$ is a polynomial with coefficent in $K$ [on hold]

I see the solution of proving if K is a field then $K[X]$ is PID that is taking the minimal polynomial of ideal but I do not see they use the hypothesis K is field? Especially, I do not see the ...
368 views

### If two elements in a ED have the same Euclidean norm, are they associates?

Is it obvious that in a Euclidean Domain two elements $x$ and $y$ having the same Euclidean norm are associates? Can someone give me a proof of this?
142 views

### Can we characterize all infinite Euclidean-domains having exactly one invertible element?

$\mathbb Z_2$ and $\mathbb Z_2[x]$ are two euclidean-domains having exactly one invertible element ; my question is ; Can we characterize all euclidean domains $D$ having exactly one invertible ...
125 views

### In a P.I.D., if $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$ with $\gcd(m,n) = 1$, then $a=b$

Let $R$ be a principal ideal domain, and $a, b \in R$ with $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$, and $\gcd(m, n) = 1$. I now want to show that we then already have $a = b$. I think ...
31 views

### Prove that any finitely generated submodule of $R^+$ (the field of quotients) is free of rank $1$

I am working on the following problem: Let $R$ be a principal ideal domain and $R^+$ the field of quotients. Then $R^+$ is an $R$-module. Prove that any finitely generated submodule of $R^+$ is a ...
39 views

### ED,PID and UFD and the relation between them

Let R be a Commutative ring with unity, such that R[x] is UFD. If R[x] is a PID then it is a Eucledian Domain? Is the last statement about being eucledian domain correct?
24 views

### $R/\langle p^k\rangle$ is an associator (i.e. if $\langle a\rangle = \langle b\rangle,$ then $a$ and $b$ are associates) when $R$ is a PID.

As the title says, I want to show that when two principal ideals are equal in $R/\langle p^k\rangle,$ where $R$ is a principal ideal domain and $p\in R$ is a prime element, then their generators are ...
64 views

31 views

### quotient of P.I.D by a prime power a P.I.D? [closed]

If $R$ is a P.I.D. and $p\in R$ is prime, is it the case that $R/<p^k>$ will be a P.I.D for all k? If so how would one show this?
10 views

### if an element $q_1$ of a principal ideal domain is irreducible, then the ideal $q_1 A$ is a maximal ideal

In my lecture, my professor wrote within a proof: $q_1$ in the principal ideal domain $A$ is irreducible, therefore, the ideal $q_1A$ is a maximal ideal. I don't understand how that's true. I tried ...
57 views

### Shortest proof for showing $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID.

I'm looking for an easy proof for that $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID. One proof I know is to show that the field norm is a Dedekind-Hasse norm, but this proof is quite dirty( that it ...
93 views

### $\mathbb{Z}[x]$ doesn't have principal maximal ideals [closed]

Prove that $\mathbb{Z}[x]$ doesn't have principal maximal ideals. Please, I need help with this problem. Thanks!
39 views

48 views

292 views

22 views

### Examples of PID's with finitely many maximal ideals

Do you know some examples of principal ideal domains which have finitely many maximal ideals? More generally, do you know how to build such domains? I don't look for fields and discrete valuation ...
43 views

70 views

### Polynomial ring is not PID

I was trying to think of a proof of the following proposition: Let $K$ be a field. Then $R=K[x_1,...,x_n]$ is not a PID for $n>1$. So here's what I've written so far: Suppose $R$ is a PID ...