# Integral Domaina and Principal Ideal Domains (PIDs)

• Mar 22nd 2013, 05:00 PM
Bernhard
Integral Domains and Principal Ideal Domains (PIDs)
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 $\displaystyle r, s \in R$ and

(ii) if $\displaystyle a_1, a_2, a_3, ...$ are non-zero elements of R such that $\displaystyle a_{i+1} | a_i$ for all i, then there is a positive integer N such that $\displaystyle a_n$ is a unit times $\displaystyle a_N$ for all $\displaystyle n \ge N$
• Mar 23rd 2013, 05:47 AM
Gusbob
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Suppose $\displaystyle I$ is a (non-empty) ideal in $\displaystyle R$. Given an element $\displaystyle a_1\in R$, we have: either $\displaystyle I=\langle a_1\rangle$ or not. In the former case, we're done, so assume the latter. Choose $\displaystyle b_1\in J- \langle a_1 \rangle$ and set $\displaystyle a_2 =$ gcd$\displaystyle (a_1,b_1)$. Here, we have either $\displaystyle I=\langle a_2\rangle$ or not. In the former case, we're done, so assume the latter... Déjà vu.

I think you see where this is going. This process terminates because of your second condition.
• Mar 23rd 2013, 05:40 PM
Bernhard
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Thanks Gusbob, most helpful; I can see the overall strategy ... but need a couple of details clarified.

You write:

"Suppose I is a (non-empty) ideal in R. Given an element $\displaystyle a_1\in R$, we have: either $\displaystyle I=\langle a_1\rangle$ or not. ... ... "

Now I follow that statement since it is just simple logic and is true independent of any conditions in the exercise - it says something is true or not!

But then you write:

"In the former case, we're done, so assume the latter. Choose $\displaystyle b_1\in I- \langle a_1 \rangle$ and set $\displaystyle a_2 = gcd (a_1,b_1)$. Here, we have either $\displaystyle I=\langle a_2\rangle$ or not. ... ... "

When you say $\displaystyle I = \langle a_2 \rangle$ or not, I am assuming that there is some link between setting $\displaystyle a_2$ = gcd $\displaystyle ( a_2 , b_2 )$ and the fact that $\displaystyle I = \langle a_2 \rangle$ is a possibility.

What is the link between setting $\displaystyle a_2$ = gcd $\displaystyle ( a_2 , b_2 )$ and the fact that $\displaystyle I = \langle a_2 \rangle$ is a possibility

Can you clarify?

Would appreciate the help!

Peter
• Mar 23rd 2013, 06:35 PM
Gusbob
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Quote:

Originally Posted by Bernhard
What is the link between setting $\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 )$ and the fact that $\displaystyle I = \langle a_2 \rangle$ is a possibility

$\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 )$ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$, so $\displaystyle \langle a_1\rangle \subseteq \langle a_2 \rangle$. Now $\displaystyle \langle a_2 \rangle \subseteq I$ by definition, so the question becomes whether $\displaystyle \langle a_2 \rangle$ contains all elements of $\displaystyle I$, formulated as $\displaystyle I=\langle a_2 \rangle$ or not.

The question of $\displaystyle I=\langle a_i \rangle$?' is always valid, the motivation being the gcd bit, which guarantees $\displaystyle \langle a_{i-1}\rangle \subseteq \langle a_i \rangle$. What you are doing in this proof is constructing bigger and bigger principal sub-ideals $\displaystyle \langle a_i\rangle$ of $\displaystyle I$, and argue that you must eventually get all of $\displaystyle I$ (otherwise you get a contradiction). Essentially, the reason we want to ask whether $\displaystyle I=\langle a_i \rangle$ at each step is because it is a principal ideal that is bigger (and contains) all previous principal ideals we looked at.

I hope that made sense, I'm not able to explain this eloquently.
• Mar 23rd 2013, 07:48 PM
Bernhard
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Quote:

Originally Posted by Gusbob
$\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 )$ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$, so $\displaystyle \langle a_1\rangle \subseteq \langle a_2 \rangle$. Now $\displaystyle \langle a_2 \rangle \subseteq I$ by definition, so the question becomes whether $\displaystyle \langle a_2 \rangle$ contains all elements of $\displaystyle I$, formulated as $\displaystyle I=\langle a_2 \rangle$ or not.

The question of $\displaystyle I=\langle a_i \rangle$?' is always valid, the motivation being the gcd bit, which guarantees $\displaystyle \langle a_{i-1}\rangle \subseteq \langle a_i \rangle$. What you are doing in this proof is constructing bigger and bigger principal sub-ideals $\displaystyle \langle a_i\rangle$ of $\displaystyle I$, and argue that you must eventually get all of $\displaystyle I$ (otherwise you get a contradiction). Essentially, the reason we want to ask whether $\displaystyle I=\langle a_i \rangle$ at each step is because it is a principal ideal that is bigger (and contains) all previous principal ideals we looked at.

I hope that made sense, I'm not able to explain this eloquently.

Thanks again Gusbob, things are getting clearer for me! Just a pedantic point - just to be sure of what is going on:

You write:

" ... $\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 )$ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$ ... ... "

But I think that $\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 )$ implies that $\displaystyle a_1 = r\cdot a_2$ since

$\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 ) \longrightarrow a_2 | a_1$ and $\displaystyle a_2 | b_1 \longrightarrow a_1 = r_1 \cdot a_2$ and $\displaystyle b_1 = r_2 \cdot a_2$ for some $\displaystyle r_1 , r_2 \in R$

Now $\displaystyle (a_1) = \{ r \cdot a_1 | r \in R \}$ and $\displaystyle (a_2) = \{ r \cdot a_2 | r \in R \}$

So we have $\displaystyle a_1 = r \cdot a_2 \longrightarrow a_1 \in ( a_2 ) \longrightarrow ( a_1 ) \subseteq ( a_2 )$

So since you conclude $\displaystyle ( a_1 ) \subseteq ( a_2 )$ all is well (Sorry if I am being pedantic)

A more important point:

In your earlier post on the proof you state at one point:

"Choose $\displaystyle b_1 \in J - \langle a_1 \rangle$ and set $\displaystyle a_2 = gcd (a_1,b_1)$... ... "

What motivated you to take $\displaystyle b_1\in J- \langle a_1 \rangle$ . In other words in your proof strategy what made you want to not only exclude $\displaystyle a_1$ from consideration, but also to exclude all of the principal ideal generated by $\displaystyle a_1$ ie to exclude all of $\displaystyle ( a_1 )$

Peter
• Mar 23rd 2013, 08:18 PM
Bernhard
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Quote:

Originally Posted by Bernhard
Thanks again Gusbob, things are getting clearer for me! Just a pedantic point - just to be sure of what is going on:

You write:

" ... $\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 )$ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$ ... ... "

But I think that $\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 )$ implies that $\displaystyle a_1 = r\cdot a_2$ since

$\displaystyle a_2$ = gcd $\displaystyle ( a_1, b_1 ) \longrightarrow a_2 | a_1$ and $\displaystyle a_2 | b_1 \longrightarrow a_1 = r_1 \cdot a_2$ and $\displaystyle b_1 = r_2 \cdot a_2$ for some $\displaystyle r_1 , r_2 \in R$

Now $\displaystyle (a_1) = \{ r \cdot a_1 | r \in R \}$ and $\displaystyle (a_2) = \{ r \cdot a_2 | r \in R \}$

So we have $\displaystyle a_1 = r \cdot a_2 \longrightarrow a_1 \in ( a_2 ) \longrightarrow ( a_1 ) \subseteq ( a_2 )$

So since you conclude $\displaystyle ( a_1 ) \subseteq ( a_2 )$ all is well (Sorry if I am being pedantic)

A more important point:

In your earlier post on the proof you state at one point:

"Choose $\displaystyle b_1 \in J - \langle a_1 \rangle$ and set $\displaystyle a_2 = gcd (a_1,b_1)$... ... "

What motivated you to take $\displaystyle b_1\in J- \langle a_1 \rangle$ . In other words in your proof strategy what made you want to not only exclude $\displaystyle a_1$ from consideration, but also to exclude all of the principal ideal generated by $\displaystyle a_1$ ie to exclude all of $\displaystyle ( a_1 )$

Peter

I have been reflecting on my own question which was concerned with selecting a new element from which to generate a new principal ideal when faced with the failure of $\displaystyle a_1$ to generate a principal ideal = I.

Gusbob's strategy was to choose an element from $\displaystyle I - ( a_1)$ and I wished to know why he excluded elements in $\displaystyle ( a_1 ) \ne a_1$ from consideration.

I now think that the reason is as follows:

If we choose $\displaystyle b_1 \in (a_1)$ where $\displaystyle b_1 \ne a_1$ then

$\displaystyle b_1 \in ( a_1 ) \longrightarrow (b_1) \subseteq (a_1)$

and so we are restricted to a subset of $\displaystyle a_1$ - BUT, we need a larger principal ideal domain and not a smaller one.

Can someone confirm that my thinking is correct.

FURTHER EDIT

Sorry to backtrack from my idea above - why when, say I fails to equal $\displaystyle ( a_1)$ do we assume that we need a larger ideal - all we are trying to show is that any ideal we examine is principal - why if $\displaystyle I = ( a_1 )$ do we assume that we need to examine a larger ideal??

Peter
• Mar 23rd 2013, 09:47 PM
Gusbob
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Quote:

Originally Posted by Bernhard
Sorry to backtrack from my idea above - why when, say I fails to equal $\displaystyle ( a_1)$ do we assume that we need a larger ideal - all we are trying to show is that any ideal we examine is principal - why if $\displaystyle I = ( a_1 )$ do we assume that we need to examine a larger ideal??

You're right about my point about gcd, the $\displaystyle a_1$ and $\displaystyle a_2$ should have been switched. I'll assume the quote above reads $\displaystyle I \not= ( a_1 )$. The reason is that $\displaystyle \langle a_1\rangle \subseteq I$ always. If $\displaystyle I=\langle a \rangle$ is principal, then $\displaystyle a_1 = r\cdot a$ for some $\displaystyle r$, which means $\displaystyle \langle a_1\rangle \subseteq \langle a\rangle =I$. As such, it makes sense to take bigger ideals.
• Mar 23rd 2013, 10:22 PM
Bernhard
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Hi Gusbob, I am still struggling with this - originally you say the following:

"Suppose I is a (non-empty) ideal in R. Given an element $\displaystyle a_1\in R$, we have: either $\displaystyle I=\langle a_1\rangle or not$."

Then you conclude that if $\displaystyle I \ne ( a_1 )$ then $\displaystyle (a_1) \subseteq I$

Why exactly does it follow: it seems to me that from $\displaystyle I \ne (a_1)$ all we can conclude is that either I is a subset of $\displaystyle (a_1)$ , I is a superset of $\displaystyle (a_1)$ or I and $\displaystyle (a_1)$ are disjoint.

How do you know that $\displaystyle (a_1) \subseteq I$?

Peter

PS Sorry if I am being slow here!
• Mar 23rd 2013, 11:12 PM
Gusbob
Re: Integral Domaina and Principal Ideal Domains (PIDs)
Quote:

Originally Posted by Bernhard
How do you know that $\displaystyle (a_1) \subseteq I$?

Much apologies. I chose $\displaystyle a_1 \in R$. It should have been $\displaystyle a_1 \in I$. Sorry for all the confusion.