Principal Ideal Domains .... ....

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SUMMARY

This discussion centers on the proof of Lemma 7.2.13 from "The Basics of Abstract Algebra" by Paul E. Bland, specifically regarding Principal Ideal Domains (PIDs). The proof demonstrates that if two ideals generated by elements \(a\) and \(b\) are equal, then \(a\) and \(b\) are associates in an integral domain. The proof involves showing that if \((a) = (b)\), then both \(a\) and \(b\) can be expressed as multiples of each other by units in the domain, confirming their association.

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  • Understanding of integral domains
  • Familiarity with ideals in ring theory
  • Knowledge of units and associates in algebra
  • Basic concepts of Euclidean and Principal Ideal Domains
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  • Study the properties of Principal Ideal Domains (PIDs)
  • Explore the concept of Unique Factorization Domains (UFDs)
  • Learn about the structure of integral domains and their ideals
  • Review proofs related to ring theory and ideal equality
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Students and educators in abstract algebra, mathematicians focusing on ring theory, and anyone seeking to deepen their understanding of Principal Ideal Domains and their properties.

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I am reading The Basics of Abstract Algebra by Paul E. Bland ...

I am focused on Section 7.2 Euclidean Domains, Principal Ideal Domains, Unique Factorization Domains ... ...

I need help with the proof of Lemma 7.2.13 ... ... Lemma 7.2.13 reads as follows:https://www.physicsforums.com/attachments/8255Bland states Lemma 7.2.13 but does not prove it ... can someone please demonstrate a proof ...Peter
 
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Hi Peter,

If $(a)=(b)$ then $a\in(b)$ and $a=rb$ from some $r\in D$; a similar argument shows that $b=sa$ for some $s\in D$.
We have therefore $a = rb = (rs)a$ and $a(rs-1)=0$. As we are in an integral domain, this implies $rs=1$, which means that $r$ and $s$ are units and $a$ and $b$ are associates.

Conversely, if $a=rb$, we have $a\in(b)$ and $(a)\subset(b)$; a similar argument shows that $(b)\subset(a)$, and therefore $(a)=(b)$.
 
castor28 said:
Hi Peter,

If $(a)=(b)$ then $a\in(b)$ and $a=rb$ from some $r\in D$; a similar argument shows that $b=sa$ for some $s\in D$.
We have therefore $a = rb = (rs)a$ and $a(rs-1)=0$. As we are in an integral domain, this implies $rs=1$, which means that $r$ and $s$ are units and $a$ and $b$ are associates.

Conversely, if $a=rb$, we have $a\in(b)$ and $(a)\subset(b)$; a similar argument shows that $(b)\subset(a)$, and therefore $(a)=(b)$.
Thanks castor28 ... appreciate your help ...

Peter
 

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