# Questions tagged [principal-ideal-domains]

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

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### GTM73 Hungerford's Algebra. Problem IV.6.4(ii)

Problem IV.6.4 If $R$ is a principal ideal domain and $A$ is a cyclic $R$-module of order $r$, then (i) every submodule of $A$ is cyclic, with order dividing $r$; (ii) for every ideal $(s)$ containing ...
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### What does it mean for a submodule of a module over a PID to have invariant factors $1$ or $0$?

I will take a particular example for simplicity: suppose $D$ is a PID and $M$ is finitely generated submodule of $D^5$, say, with set of generators $x_i, i=1,2,3,4,5$. Suppose also that the smith ...
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### If $R$ is a principal ideal domain and $P \in Spec(R)$, why is $P^mR/P^n=0$ for $m \geq n$?

If $R$ is a principal ideal domain and $P \in Spec(R)$, why is $P^mR/P^n=0$ for $m \geq n$? This is a step on a proof I’m trying to understand. Answering to the question in the comments: The context ...
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### Sage: Constructing maximal ideal of a local ring, $M = \langle x - P_x \rangle$ gives invalid result, but $M = \langle x - P_x, y - P_y \rangle$ works

Anothony Knapp's book on Elliptic Curves, page 350 states: $M$ is principal given by $M = \langle t \rangle$ for any element $t$ of $M$ not in $M^2$. He later writes on that page, that $t$ is the ...
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### Let $R$ be a PID and $a \in R$ is non-invertible,then $\exists$ some prime element $p$ such that $p|a.$

I tried this and i don't know , if there is any fallacies in my arguement. If $a$ is prime then it is easy to show. If $a$ is not prime the $<a>$ is cannot be a maximal ideal of $R$, since $R$ ...
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### Prove that in a PID if $a$ is not writable as product of finitely many prime elements then $a$ is not prime

The statement I want to prove: Suppose $I$ is a principal ideal domain. An element $a \in I$ can not written as a product of finite number of prime elements. Then $a$ is not a prime element in $I$? My ...
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### Is it true that an uniformizing element of a discrete valuation ring $R$ is either zero or prime?

Let $R$ be a DVR. This means that it is a PID with a unique maximal ideal. Let us denote $m$ to be the maximal ideal. Then there is a unique irreducible element $\pi$ up to multiplication my a unit ...
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### Bezout Lemma in a PID [duplicate]

The Bezout Lemma in the integers states that For any $a, b \in\mathbb Z$, let $g = \gcd(a, b)$, There exists $x, y$ such that $ax+by = g$. This can be generalized to a commutative ring that is a ...
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### How can I finish the proof for the following proposition: every ideal in k[x] is a principal ideal? [duplicate]

Proposition : Every ideal in k[x] (polynomial ring) is a principal ideal Proof : Suppose that $I\subseteq K[x]$. Take $p(x)\in I$ such that $p(x)$ is monic and $deg(p(x))$ is minimal over all ...
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### On Showing that $\mathbb{Z}[x]$ is not a Principal Ideal Domain

I'm working on a question that asks to consider the ring $R = \mathbb{Z}[x]$. I need to show that $(x)$ is a prime ideal, show that $(x,7)$ is a prime ideal and use these facts to conclude that $R$ ...
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### Question on Definition of Canonical Epimorphism

Let $P$ be a finitely-generated $R$-module, where $R$ is a principal ideal domain. The text that I am reading refers to $$f:\bigoplus_{p \in P}R \rightarrow P$$ as the canonical epimorphism of $R$-...
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### What does the correspondence theorem tell us about the ideals of $\Bbb Z[x]$ that contain $x^2+1$?

What does the correspondence theorem tell us about the ideals of $\Bbb Z[x]$ that contain $x^2+1$? If I define $\varphi : \Bbb Z[x] \to \Bbb Z[i]$ as $x \longmapsto i$, then $\ker \varphi = (x^2+1)$. ...
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### Why are the direct summands obtained in the structure theorem for finitely generated modules over a PID indecomposable?

The structure theorem tells us that a finitely generated module $M$ over a principal ideal domain $R$ is isomorphic to a direct sum $\bigoplus _{i}R/(q_{i})$ where $( q_i ) ≠ R (q_{i})\neq R$ and ...
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### How does this quotient by a maximal ideal looks like?

I have the following situation: Let $R$ be a PID. Then let $\mathfrak{p}\subset R$ be a maximal ideal. We consider the set $$\left(R/\mathfrak{p}\right)[X]$$ which is clearly a quotient, but I would ...
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### Content of a polynomial and maximal ideals.

I have the following problem: Let $R$ be a PID. Let us denote for $P=a_0+a_1 X+\dots+a_n X^n\in R[X]$ that $c(P)=\gcd(a_0,...,a_n)$. Show that for $P\in R[X]$ nonzero $c(P)=1$ iff for every maximal ...
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Let $M$ be a finitely generated torsion module over the PID $R$. Suppose that $M$ has $s$ elementary divisors and $t$ invariant factors. Define S:=\{n\in\mathbb{N}\,|\,M\text{ is isomorphic to a ...