Comaximal Ideals in a Principal Ideal Domain

[Math Amateur]

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

 

[Siron]

Re: Comaximal Ideas in a Principal Ideal Domain

I think you've found the answer on MHF.

 

[Math Amateur]

Re: Comaximal Ideas in a Principal Ideal Domain

Well ... I am still working on the problem ... but I will be using your guidance regarding the way to progress

At my day job at the moment ... but will use your hint when I return to the problem

Thanks again

Peter

 

[jakncoke1]

by def, two elements a,b in a PID are relatively prime if there exist, $m_1,m_2 \in $, such
that 1 = $m_1a+m_2b$

now if $a,b$ are relatively prime then

$\{r_1a + r_2b | r_1,r_2 \in R\}$, contains 1, if an ideal contains 1, then that ideal is identical to R.

Now <a> + <b> = $\{g_1a + g_2b | g_1,g_2 \in R\}$

 

[blue_raver22]

I would start by defining a principal ideal domain (PID) as a commutative ring in which every ideal is generated by a single element. This means that every ideal in a PID can be written as (a) = {ra | r \in R} where a is a single element in the ring R.

Now, to prove that in a PID, two ideals (a) and (b) are comaximal if and only if they have a greatest common divisor, we can use the definition of comaximal ideals given above. This definition states that two ideals A and B are comaximal if their sum A+B is equal to the whole ring R.

So, let us assume that (a) and (b) are comaximal ideals in a PID. This means that (a) + (b) = R. We can then choose an element c \in R such that c = ra + sb for some r,s \in R. Since R is a PID, we know that (a) and (b) are principal ideals and can be written as (a) = {ra | r \in R} and (b) = {sb | s \in R}. Therefore, c \in (a) and c \in (b), which means that c \in (a) \cap (b).

Now, let d be any common divisor of a and b. This means that d \in (a) \cap (b). Since c is also in this intersection, it must be a multiple of d, i.e. c = kd for some k \in R. This shows that d divides both a and b, making it a common divisor.

Conversely, let us assume that (a) and (b) have a greatest common divisor, denoted by d. This means that d \in (a) \cap (b). Since d is in both (a) and (b), it can be written as d = ra and d = sb for some r,s \in R. This shows that c = ra + sb \in (a) + (b), which means that c \in R. Therefore, (a) + (b) = R, making (a) and (b) comaximal ideals.

In conclusion, in a PID, two ideals (a) and (b) are comaximal if and only if they have a