Principal Ideal Domains .... ....

  • MHB
  • Thread starter Math Amateur
  • Start date
  • Tags
    domains
In summary, the conversation discusses Lemma 7.2.13 from The Basics of Abstract Algebra by Paul E. Bland and requests a proof for it. The lemma states that if two elements in a domain are associates, then their principal ideals are equal. The proof involves showing that if $(a)=(b)$, then $a$ and $b$ are associates. Conversely, if $a$ and $b$ are associates, then $(a)=(b)$. Thank you castor28 for providing the proof.
  • #1
Math Amateur
Gold Member
MHB
3,998
48
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
 
Physics news on Phys.org
  • #2
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 $(a)=(b)$.
 
  • #3
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
 

FAQ: Principal Ideal Domains .... ....

What is a Principal Ideal Domain (PID)?

A PID is a type of commutative ring in abstract algebra that shares properties with both integral domains and principal ideal rings. It is an integral domain, meaning it has no zero divisors, and every nonzero element has a multiplicative inverse. It is also a principal ideal ring, meaning every ideal in the ring can be generated by a single element.

What are the main properties of a PID?

There are three main properties of PIDs: every ideal is generated by a single element, every nonzero element can be factored into a product of irreducible elements, and it is a unique factorization domain (UFD). UFD means that every element can be written as a unique product of irreducible elements, up to order and multiplication by units.

How are PIDs related to other types of rings?

PIDs are a special type of ring that sits between integral domains and Euclidean domains. They are more general than Euclidean domains, which have a division algorithm, and more specific than integral domains, as they have additional properties such as unique factorization.

What are some examples of PIDs?

The most well-known example of a PID is the ring of integers, Z. Other examples include polynomial rings over a field (such as Q[x] or R[x]), the ring of Gaussian integers (Z[i]), and the ring of Hurwitz integers (Z[(1+√5)/2]).

What are some applications of PIDs?

PIDs have numerous applications in abstract algebra, number theory, and cryptography. They are used in the construction of public key encryption systems, such as the RSA algorithm. They also have applications in coding theory, as they can be used to construct error-correcting codes. Additionally, PIDs are used in algebraic geometry to study algebraic curves and surfaces.

Similar threads

Back
Top