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 and

(ii) if are non-zero elements of R such that for all i, then there is a positive integer N such that is a unit times for all

Re: Integral Domaina and Principal Ideal Domains (PIDs)

Suppose is a (non-empty) ideal in . Given an element , we have: either or not. In the former case, we're done, so assume the latter. Choose and set gcd . Here, we have either 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.

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 , we have: either 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 and set . Here, we have either or not. ... ... "

When you say or not, I am assuming that there is some link between setting = gcd and the fact that is a possibility.

What is the link between setting = gcd and the fact that is a possibility

Can you clarify?

Would appreciate the help!

Peter

Re: Integral Domaina and Principal Ideal Domains (PIDs)

Re: Integral Domaina and Principal Ideal Domains (PIDs)

Quote:

Originally Posted by

**Gusbob** = gcd

implies that

for some

, so

. Now

by definition, so the question becomes whether

contains all elements of

, formulated as

or not.

The question of `

?' is always valid, the motivation being the gcd bit, which guarantees

. What you are doing in this proof is constructing bigger and bigger principal sub-ideals

of

, and argue that you must eventually get all of

(otherwise you get a contradiction). Essentially, the reason we want to ask whether

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:

" ... = gcd implies that for some ... ... "

But I think that = gcd implies that since

= gcd and and for some

Now and

So we have

So since you conclude 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 and set ... ... "

What motivated you to take . In other words in your proof strategy what made you want to not only exclude from consideration, but also to exclude all of the principal ideal generated by ie to exclude all of

Peter

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:

" ...

= gcd

implies that

for some

... ... "

But I think that

= gcd

implies that

since

= gcd

and

and

for some

Now

and

So we have

So since you conclude

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

and set

... ... "

What motivated you to take

. In other words in your proof strategy what made you want to not only exclude

from consideration, but also to exclude all of the principal ideal generated by

ie to exclude all of

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 to generate a principal ideal = I.

Gusbob's strategy was to choose an element from and I wished to know why he excluded elements in from consideration.

I now think that the reason is as follows:

If we choose where then

and so we are restricted to a subset of - 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 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 do we assume that we need to examine a larger ideal??

Peter

Re: Integral Domaina and Principal Ideal Domains (PIDs)

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 , we have: either ."

Then you conclude that if then

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

How do you know that ?

Peter

PS Sorry if I am being slow here!

Re: Integral Domaina and Principal Ideal Domains (PIDs)

Quote:

Originally Posted by

**Bernhard** How do you know that

?

Much apologies. I chose . It should have been . Sorry for all the confusion.