What is a principal ideal domain?

An integral domain in which every ideal is generated by one element.

Why is every field a PID?

Its only ideals are the zero ideal and the whole field, both principal.

What does the ideal $(a,b)$ contain?

All ring-linear combinations $ra+sb$ with $r,s\in R$.

What factorization property does every PID have?

It is a unique factorization domain: nonzero nonunits factor uniquely into irreducibles, up to order and units.

Abstract algebraLesson 4 of 4 in the course Rings and ideals

Principal Ideal Domains: Ideals, Factorization, and Modules

5 flashcards · 5 quiz questions · 5 match pairs · 4 fill-in sentences · English

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Summary

A principal ideal domain (PID) is a type of commutative ring in which every ideal can be described using just one generator. PIDs connect the language of ideals to familiar arithmetic in the integers.

An integral domain is a nonzero commutative ring with identity and no nonzero zero divisors. In other words, if $ab=0$ , then $a=0$ or $b=0$ .

An ideal $I$ in a ring $R$ is an additive subgroup of $R$ that is closed under multiplication by elements of $R$ . Thus, if $a\in I$ and $r\in R$ , then $ra\in I$ .

All flashcards

What is a principal ideal domain?: An integral domain in which every ideal is generated by one element.
How is the principal ideal generated by $a$ in $R$ written?: $(a)=aR=\{ra:r\in R\}$ .
Why is every field a PID?: Its only ideals are the zero ideal and the whole field, both principal.
What does the ideal $(a,b)$ contain?: All ring-linear combinations $ra+sb$ with $r,s\in R$ .
What factorization property does every PID have?: It is a unique factorization domain: nonzero nonunits factor uniquely into irreducibles, up to order and units.

Quiz questions

Which statement correctly describes the relationship between Euclidean domains and PIDs?Answer: Every Euclidean domain is a PID, but some PIDs are not Euclidean domains.
In a PID, if $(a,b)=(d)$ , what can be concluded about $d$ ?Answer: $d$ is a greatest common divisor of $a$ and $b$ , up to a unit.
What does the structure theorem imply for a finitely generated module over a PID?Answer: It is a direct sum of cyclic modules of the form $R/xR$ .
What follows from the theorem about submodules of free modules over a PID?Answer: Every submodule of a free module over a PID is free.
Why is $K[x,y]$ not a PID when $K$ is a field?Answer: The ideal $(x,y)$ is not principal.

Based on the Wikipedia article Principal ideal domain, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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