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 $\displaystyle I+J = \{ a+b | a \in I, b \in J \} $

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

$\displaystyle (a)+(b)=R$

$\displaystyle \iff (a)+(b)=(1)$

$\displaystyle \iff (a)+(b)\ni 1$

$\displaystyle \iff 1=ax+by$ for some $\displaystyle x,y\in \mathbb{R}$

$\displaystyle \iff gcd(a,b)=1$

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:

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

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

Thanks

Peter

Re: Comaximal Ideas in a Principal Ideal Domain

Prove $\displaystyle (1) \subseteq R$ and $\displaystyle R \subseteq (1)$

Re: Comaximal Ideals in a Principal Ideal Domain

Can clearly see that proving $\displaystyle (1) \subseteq R $ and $\displaystyle R \subseteq (1) $ is the way to go ... but ...

How do we even know that R has a 1?

Can someone help?

Peter

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

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.