# Principal ideal domain

• Jun 21st 2011, 07:08 AM
jishnuray
Principal ideal domain
R is an integral domain.Then prove that if the following two conditions hold then R is a principal ideal domain(PID)
1)any two non zero elements a, b in R have a greatest common divisor which can be written in the form ra+sb for some r,s belonging to R.
2)['a_1 denotes a suffix 1]
If a_1,a_2,a_3,...... are non zero elements of R such that a_(i+1) divides a_i for all i then there is a positive integer N such that a_n is unit times a_N for all n greater than or equal to N.

How to do the sum?
• Jun 21st 2011, 02:14 PM
NonCommAlg
Re: Principal ideal domain
Quote:

Originally Posted by jishnuray
R is an integral domain.Then prove that if the following two conditions hold then R is a principal ideal domain(PID)
1)any two non zero elements a, b in R have a greatest common divisor which can be written in the form ra+sb for some r,s belonging to R.
2)['a_1 denotes a suffix 1]
If a_1,a_2,a_3,...... are non zero elements of R such that a_(i+1) divides a_i for all i then there is a positive integer N such that a_n is unit times a_N for all n greater than or equal to N.

How to do the sum?

let $I$ be any non-zero ideal of $R$ and define $S = \{Ra : \ a \in I \}$. use condition 2) in your problem to show that $S$ has a maximal element, say $Rb$. the claim is that $I = Rb$. suppose our claim is not true and choose $a \in I \setminus Rb$. then by condition 1) we have $Ra+Rb=Rd$, where $d = \gcd(a,b)$. clearly $d \in I$ because $a,b \in I$. thus $Rd \in S$. we also have $a \in Rd \setminus Rb$ and so $Rb \subset Rd$, contradicting the maximality of $Rb$ in $S$. Q.E.D.
• Jun 21st 2011, 03:21 PM
Deveno
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:

$(a_i)$.

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

$I_1 = (a_1)$ where $a_1 \in I\setminus{0}$.

$I_2 = (a_2)$ where $a_2$ is chosen by finding $r_1 \in I\setminus I_1$ and setting $a_2 = gcd(r_1,a_1)$, so that $a_2|a_1$,

$I_3 = (a_3)$ where $a_3$ is chosen by finding $r_2 \in I\setminus I_2$ and setting $a_3 = gcd(r_2,a_2)$, so that $a_3|a_2$,

and so on, which by (2) terminates with some $I_N =(a_N)$.

now, it would be nice if we could conclude that $I = I_N$, but perhaps for some other choice of the $\{a_i\}$ 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 $I\setminus I_N$ an associate of $a_N$? (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 " $I_N$"s).

my apologies if i have stated this badly.
• Jun 21st 2011, 03:45 PM
NonCommAlg
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:

$(a_i)$.

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

$I_1 = (a_1)$ where $a_1 \in I\setminus{0}$.

$I_2 = (a_2)$ where $a_2$ is chosen by finding $r_1 \in I\setminus I_1$ and setting $a_2 = gcd(r_1,a_1)$, so that $a_2|a_1$,

$I_3 = (a_3)$ where $a_3$ is chosen by finding $r_2 \in I\setminus I_2$ and setting $a_3 = gcd(r_2,a_2)$, so that $a_3|a_2$,

and so on, which by (2) terminates with some $I_N =(a_N)$.

now, it would be nice if we could conclude that $I = I_N$, but perhaps for some other choice of the $\{a_i\}$ 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 $I\setminus I_N$ an associate of $a_N$? (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 " $I_N$"s).

my apologies if i have stated this badly.

yes, you're right. we don't need Zorn's lemma to show that $S$ has a maximal element. i removed it from my proof.
suppose that $S$ has no maximal element and choose $a_1 \in I$. since $Ra_1$ is not maximal in $S$, we may choose $a_2 \in I$ such that $Ra_1 \subset Ra_2$. we continue this way to find an ascending chain of elements of $S$ which does not stop, contradicting condition 2).