Welcome to our community

Be a part of something great, join today!

Integral Domains and Principal Ideal Domains (PIDs)

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Dummit and Foote, Section 8.2 (Principal Ideal Domains (PIDs) ) - Exercise 4, page 282.

Let R be an integral domain.

Prove that if the following two conditions hold then R is a Principal Ideal Domain:

(i) any two non-zero elements a and b in R have a greatest common divisor which can be written in the form ra + sb for some [TEX] r, s \in R [/TEX] and

(ii) if [TEX] a_1, a_2, a_3, ... [/TEX] are non-zero elements of R such that [TEX] a_{i+1} | a_i [/TEX] for all i, then there is a positive integer N such that [TEX] a_n [/TEX] is a unit times [TEX] a_N [/TEX] for all [TEX] n \ge N [/TEX]

I am somewhat overwhelmed by this exercise. I would appreciate it if someone could help me get started and indicate a strategy for formulating a proof.

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Dummit and Foote, Section 8.2 (Principal Ideal Domains (PIDs) ) - Exercise 4, page 282.

Let R be an integral domain.

Prove that if the following two conditions hold then R is a Principal Ideal Domain:

(i) any two non-zero elements a and b in R have a greatest common divisor which can be written in the form ra + sb for some [TEX] r, s \in R [/TEX] and

(ii) if [TEX] a_1, a_2, a_3, ... [/TEX] are non-zero elements of R such that [TEX] a_{i+1} | a_i [/TEX] for all i, then there is a positive integer N such that [TEX] a_n [/TEX] is a unit times [TEX] a_N [/TEX] for all [TEX] n \ge N [/TEX]

I am somewhat overwhelmed by this exercise. I would appreciate it if someone could help me get started and indicate a strategy for formulating a proof.
Start like this. Let $J$ be an ideal in $R$. We want to show that $J$ is a principal ideal. Choose $a_1\in J$. If $J = \langle a_1\rangle$ (the principal ideal generated by $a_1$), then $J$ is a principal ideal, and we are done. If not, choose $b_1\in J - \langle a_1\rangle$, and let $a_2 = \text{gcd}(a_1,b_1)\in J$. If $J = \langle a_2\rangle$, then $J$ is a principal ideal, and we are done. If not, ... .

Do you see how to continue? The idea should be to use (ii) to show that you can build up a sequence so that $J = \langle a_N\rangle$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Start like this. Let $J$ be an ideal in $R$. We want to show that $J$ is a principal ideal. Choose $a_1\in J$. If $J = \langle a_1\rangle$ (the principal ideal generated by $a_1$), then $J$ is a principal ideal, and we are done. If not, choose $b_1\in J - \langle a_1\rangle$, and let $a_2 = \text{gcd}(a_1,b_1)\in J$. If $J = \langle a_2\rangle$, then $J$ is a principal ideal, and we are done. If not, ... .

Do you see how to continue? The idea should be to use (ii) to show that you can build up a sequence so that $J = \langle a_N\rangle$.
=================================================================

Thanks Opalg ... the overall strategy for proving this is now clear but I do need some help on important aspects of the proof.

You write:

"Start like this. Let $J$ be an ideal in $R$. We want to show that $J$ is a principal ideal. Choose $a_1\in J$. If $J = \langle a_1\rangle$ (the principal ideal generated by $a_1$), then $J$ is a principal ideal, and we are done. ... ... "

I understand this since it is simple logic that something is the case or it is not - true and depends on nothing in the exercise conditions/givens.

Then you write:

"If not, choose $b_1\in J - \langle a_1\rangle$, and let $a_2 = \text{gcd}(a_1,b_1)\in J$. If $J = \langle a_2\rangle$, then $J$ is a principal ideal, and we are done. If not, ... . "


When you say " choose \(\displaystyle b_1\in J - \langle a_1\rangle \) and let \(\displaystyle a_2 = \text{gcd}(a_1,b_1)\in J \) " ... ...

and then go on to say "if \(\displaystyle J = \langle a_2\rangle \) ... ... " ... ... I am assuming that there is some logical link between setting \(\displaystyle a_2 = \text{gcd}(a_1,b_1) \) and then making the statement " if \(\displaystyle J = \langle a_2\rangle \) ... ...

Can you clarify? Can you make the link between the gcd and the principal ideal clear?

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
=================================================================

Thanks Opalg ... the overall strategy for proving this is now clear but I do need some help on important aspects of the proof.

You write:

"Start like this. Let $J$ be an ideal in $R$. We want to show that $J$ is a principal ideal. Choose $a_1\in J$. If $J = \langle a_1\rangle$ (the principal ideal generated by $a_1$), then $J$ is a principal ideal, and we are done. ... ... "

I understand this since it is simple logic that something is the case or it is not - true and depends on nothing in the exercise conditions/givens.

Then you write:

"If not, choose $b_1\in J - \langle a_1\rangle$, and let $a_2 = \text{gcd}(a_1,b_1)\in J$. If $J = \langle a_2\rangle$, then $J$ is a principal ideal, and we are done. If not, ... . "


When you say " choose \(\displaystyle b_1\in J - \langle a_1\rangle \) and let \(\displaystyle a_2 = \text{gcd}(a_1,b_1)\in J \) " ... ...

and then go on to say "if \(\displaystyle J = \langle a_2\rangle \) ... ... " ... ... I am assuming that there is some logical link between setting \(\displaystyle a_2 = \text{gcd}(a_1,b_1) \) and then making the statement " if \(\displaystyle J = \langle a_2\rangle \) ... ...

Can you clarify? Can you make the link between the gcd and the principal ideal clear?

Peter
The underlying idea is that you want to find an element that generates the whole of $J$. If you start with a random element $a_1\in J$, then $\langle a_1\rangle$ will be contained in $J$ but it might not be the whole of $J$. So you want to find an element $a_2\in J$ that generates a larger part of $J$. If you take $a_2 = \text{gcd}(a_1,b_1)$ then firstly $a_2\in J$ (by condition (i), because $a_2$ is a linear combination of $a_1$ and $b_1$), and hence $\langle a_2\rangle\subseteq J$; secondly, $\langle a_2\rangle$ contains $\langle a_1\rangle$ (because $a_1$ is a multiple of $a_2$); and thirdly, $\langle a_2\rangle$ is strictly larger than $\langle a_1\rangle$ (because $b_1$ is a multiple of $a_2$ and therefore $b_1\in\langle a_2\rangle$, whereas $b_1$ was not in $\langle a_1\rangle$).

Continuing in that way, you build up an increasing nest of sub-ideals of $J$, and then you need to use condition (ii) to show that this process cannot continue indefinitely without eventually encompassing the whole of $J$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
The underlying idea is that you want to find an element that generates the whole of $J$. If you start with a random element $a_1\in J$, then $\langle a_1\rangle$ will be contained in $J$ but it might not be the whole of $J$. So you want to find an element $a_2\in J$ that generates a larger part of $J$. If you take $a_2 = \text{gcd}(a_1,b_1)$ then firstly $a_2\in J$ (by condition (i), because $a_2$ is a linear combination of $a_1$ and $b_1$), and hence $\langle a_2\rangle\subseteq J$; secondly, $\langle a_2\rangle$ contains $\langle a_1\rangle$ (because $a_1$ is a multiple of $a_2$); and thirdly, $\langle a_2\rangle$ is strictly larger than $\langle a_1\rangle$ (because $b_1$ is a multiple of $a_2$ and therefore $b_1\in\langle a_2\rangle$, whereas $b_1$ was not in $\langle a_1\rangle$).

Continuing in that way, you build up an increasing nest of sub-ideals of $J$, and then you need to use condition (ii) to show that this process cannot continue indefinitely without eventually encompassing the whole of $J$.
Thanks Opalg

Your post was extremely helpful


Peter