Re: Principal ideal domain

Re: Principal ideal domain

can this be proved without the use of Zorn's lemma? perhaps i am not understanding the question, but condition two looks like an ascending chain condition on the principal ideals:

.

i was thinking about this, and i ran into the following difficulty:

given any (non-zero) ideal I, we can form a chain of ideals by taking:

where .

where is chosen by finding and setting , so that ,

where is chosen by finding and setting , so that ,

and so on, which by (2) terminates with some .

now, it would be nice if we could conclude that , but perhaps for some other choice of the we would wind up with "a different N". that is, how do we know that we can find an N that makes every element in an associate of ? (i think this is where Zorn's lemma comes in. if i understand what you wrote, each Ra in your answer, is one of my " "s).

my apologies if i have stated this badly.

Re: Principal ideal domain

Quote:

Originally Posted by

**Deveno** can this be proved without the use of Zorn's lemma? perhaps i am not understanding the question, but condition two looks like an ascending chain condition on the principal ideals:

.

i was thinking about this, and i ran into the following difficulty:

given any (non-zero) ideal I, we can form a chain of ideals by taking:

where

.

where

is chosen by finding

and setting

, so that

,

where

is chosen by finding

and setting

, so that

,

and so on, which by (2) terminates with some

.

now, it would be nice if we could conclude that

, but perhaps for some other choice of the

we would wind up with "a different N". that is, how do we know that we can find an N that makes every element in

an associate of

? (i think this is where Zorn's lemma comes in. if i understand what you wrote, each Ra in your answer, is one of my "

"s).

my apologies if i have stated this badly.

yes, you're right. we don't need Zorn's lemma to show that has a maximal element. i removed it from my proof.

suppose that has no maximal element and choose . since is not maximal in , we may choose such that . we continue this way to find an ascending chain of elements of which does not stop, contradicting condition 2).