# Thread: Integral Domaina and Principal Ideal Domains (PIDs)

1. ## 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 $r, s \in R$ and

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

2. ## Re: Integral Domaina and Principal Ideal Domains (PIDs)

Suppose $I$ is a (non-empty) ideal in $R$. Given an element $a_1\in R$, we have: either $I=\langle a_1\rangle$ or not. In the former case, we're done, so assume the latter. Choose $b_1\in J- \langle a_1 \rangle$ and set $a_2 =$ gcd $(a_1,b_1)$. Here, we have either $I=\langle a_2\rangle$ 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.

3. ## 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 $a_1\in R$, we have: either $I=\langle a_1\rangle$ 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 $b_1\in I- \langle a_1 \rangle$ and set $a_2 = gcd (a_1,b_1)$. Here, we have either $I=\langle a_2\rangle$ or not. ... ... "

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

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

Can you clarify?

Would appreciate the help!

Peter

4. ## Re: Integral Domaina and Principal Ideal Domains (PIDs)

Originally Posted by Bernhard
What is the link between setting $a_2$ = gcd $( a_1, b_1 )$ and the fact that $I = \langle a_2 \rangle$ is a possibility
$a_2$ = gcd $( a_1, b_1 )$ implies that $a_2=r\cdot a_1$ for some $r\in R$, so $\langle a_1\rangle \subseteq \langle a_2 \rangle$. Now $\langle a_2 \rangle \subseteq I$ by definition, so the question becomes whether $\langle a_2 \rangle$ contains all elements of $I$, formulated as $I=\langle a_2 \rangle$ or not.

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

5. ## Re: Integral Domaina and Principal Ideal Domains (PIDs)

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

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

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

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

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

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

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

So since you conclude $( a_1 ) \subseteq ( a_2 )$ 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 $b_1 \in J - \langle a_1 \rangle$ and set $a_2 = gcd (a_1,b_1)$... ... "

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

Peter

6. ## Re: Integral Domaina and Principal Ideal Domains (PIDs)

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:

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

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

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

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

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

So since you conclude $( a_1 ) \subseteq ( a_2 )$ 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 $b_1 \in J - \langle a_1 \rangle$ and set $a_2 = gcd (a_1,b_1)$... ... "

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

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 $a_1$ to generate a principal ideal = I.

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

I now think that the reason is as follows:

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

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

and so we are restricted to a subset of $a_1$ - 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 $( a_1)$ 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 $I = ( a_1 )$ do we assume that we need to examine a larger ideal??

Peter

7. ## Re: Integral Domaina and Principal Ideal Domains (PIDs)

Originally Posted by Bernhard
Sorry to backtrack from my idea above - why when, say I fails to equal $( a_1)$ 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 $I = ( a_1 )$ do we assume that we need to examine a larger ideal??
You're right about my point about gcd, the $a_1$ and $a_2$ should have been switched. I'll assume the quote above reads $I \not= ( a_1 )$. The reason is that $\langle a_1\rangle \subseteq I$ always. If $I=\langle a \rangle$ is principal, then $a_1 = r\cdot a$ for some $r$, which means $\langle a_1\rangle \subseteq \langle a\rangle =I$. As such, it makes sense to take bigger ideals.

8. ## 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 $a_1\in R$, we have: either $I=\langle a_1\rangle or not$."

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

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

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

Peter

PS Sorry if I am being slow here!

9. ## Re: Integral Domaina and Principal Ideal Domains (PIDs)

Originally Posted by Bernhard
How do you know that $(a_1) \subseteq I$?
Much apologies. I chose $a_1 \in R$. It should have been $a_1 \in I$. Sorry for all the confusion.