Questions tagged [principal-ideal-domains]

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

A PID is a type of integral domain where every proper ideal can be generated by a single element. Every Euclidean ring is a PID but the converse is not true. Every PID is a unique factorisation domain but not conversely.

Examples of PIDs are any field, $\mathbb{Z}$, rings of polynomials, Gaussian integers, and Eisenstein integers.

806 questions
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Non-zero ideals in ${\mathbb{Q}}_p$ are $p^n{\mathbb{Q}}_p$, $n\in\mathbb N_0$

How do I show that every non-zero ideal in ${\mathbb{Q}}_p$ is of the form $p^n{\mathbb{Q}}_p$ for some $n \in \mathbb{N}_0$, and investigate if ${\mathbb{Q}}_p$ is a principal ideal domain? If it is a field it has to be an PID, too right? But I…
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Determining ring type for the formal series ring

Given the following set of rings: $$\mathcal{C} = \{\mathbb{Z}[i\sqrt{3}], \ \mathbb{Z}[X], \ \mathbb{Z}, \ \mathbb{Q} , \ \mathbb{Q}[X]\}$$ I am told to determine, $\forall S\in\mathcal{C}$ what type of ring the formal series of $S$,…
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quotient of P.I.D by a prime power a P.I.D?

If $R$ is a P.I.D. and $p\in R$ is prime, is it the case that $R/$ will be a P.I.D for all k? If so how would one show this?
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Ring of polynomial functions on unit hyperbola is PID

Let $R=\mathbb{R}[X,Y]/(XY-1)$ be the ring of polynomial functions pn the unit hyperbola. How do I prove that $R$ is a principal ideal domain with unit group…
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how can i prove that every subring of $\mathbb{Q}$ is PID?

How can I prove that every subring of $\mathbb{Q}$ is PID?
user190974
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Quotient of a PID by a prime ideal

Prove that quotient of a PID by a prime ideal is PID.
EuReka
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Question about the relation between P.I.D. and quotient field.

Let $D$ be a P.I.D. and $F$ is a quotient field of $D$. How to prove that any subring $D'$ contained in $D$ in $F$ is a P.I.D., and $D'$ is a ring of fraction of a subset of $D$ closed under multiplication? Conversely, how to prove any ring of…
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If a is non zero non unit in a PID then a has at least one irreducible divisor?

Sources for the proof of this claim: If $a$ is non zero non unit in a PID then $a$ has at least one irreducible divisor. Definition of irreducible element: a non zero non unit element $p$ is irreducible if whenever $p=ab$, then $a$ or $b$ is a…
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Is PID with one divisible element a field?

Let $R$ be a PID and $r\in R$ be the divisible element i.e. for any $a\in R$ there exists $y\in R$ s.t. $r=ay$. Can we imply that $R$ is a field?
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Let $A$ be a PID, $M$ an injective finitely generated module. Prove that $M = 0$.

Help! Let $A$ be a PID, $M$ an injective finitely generated module. Prove that: $$ M = 0.$$
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Are $\mathbb C[x , y]$ and $ \frac {\mathbb R[x , y]}{ \left}$ PIDs?

Are $\mathbb C[x , y]$ and $ \frac {\mathbb R[x , y]}{ \left}$ PIDs? Can anyone please educate me how to handle two variable stuff? I know how to check this kind of stuff in one variable.
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