Comaximal Ideas in a Principal Ideal Domain

• Mar 14th 2013, 02:43 AM
Bernhard
Comaximal Ideas in a Principal Ideal Domain
Prove that in a prinicpal ideal domain, two ideals (a) and (b) are comaximal if and only if a greatest common divisor of a and b (in which case (a) and (b) are said to be coprine or realtively prime)

Note: (1) Two ideals A and B of the ring R are said to be comaximal if A + B = R

(2) Let I and J be two ideals of R
The sum of I and J is defined as $I+J = \{ a+b | a \in I, b \in J \}$
• Mar 14th 2013, 04:26 AM
Gusbob
Re: Comaximal Ideas in a Principal Ideal Domain
Quote:

Prove that in a prinicpal ideal domain, two ideals (a) and (b) are comaximal if and only if a greatest common divisor of a and b is 1 (in which case (a) and (b) are said to be coprine or realtively prime)

Let $R$ be a principal ideal domain and suppose $(a)+(b)=R$. The following is a series of equivalences, let me know if you need clarification on any of the steps.

$(a)+(b)=R$
$\iff (a)+(b)=(1)$
$\iff (a)+(b)\ni 1$
$\iff 1=ax+by$ for some $x,y\in \mathbb{R}$
$\iff gcd(a,b)=1$
• Mar 14th 2013, 02:25 PM
Bernhard
Re: Comaximal Ideas in a Principal Ideal Domain
Thanks for the help

One immediate problem for me is that I cannot see why the following is true:

$(a) + (b) = R \Longleftrightarrow (a) + (b) = (1)$

Can you please show me explicitly why this is the case.

Thanks

Peter
• Mar 14th 2013, 02:33 PM
Siron
Re: Comaximal Ideas in a Principal Ideal Domain
Prove $(1) \subseteq R$ and $R \subseteq (1)$
• Mar 14th 2013, 11:58 PM
Bernhard
Re: Comaximal Ideals in a Principal Ideal Domain
Can clearly see that proving $(1) \subseteq R$ and $R \subseteq (1)$ is the way to go ... but ...

How do we even know that R has a 1?

Can someone help?

Peter
• Mar 15th 2013, 12:06 AM
Bernhard
Re: Comaximal Ideals in a Principal Ideal Domain
I think I just answered my most recent post.

Just checked definitions of PID and Integral domain

A PID is an integral domain in which every ideal is principal

and

An integral domain is a commutative ring with identity having no zero divisors

So by definition a PID has a 1!

Can someone confirm this is correct?

Peter
• Mar 15th 2013, 12:09 AM
Gusbob
Re: Comaximal Ideals in a Principal Ideal Domain
1 just denotes the multiplicative identity on the ring.

EDIT: Yes, you are right. A PID has to contain a multiplicative identity.