Question

Let p be an irreducible element in an integral domain. Prove that 1_Ris a gcd of p and a if and only if and only if p+a .

Short answer

It is proved that 1_Ris a gcd of p and a if and only if and only if p+a

Step-by-step solution

Defining greatest common divisor

The greatest common divisor (gcd) of two non-zero integers a, and b is the greatest positive integer that is a divisor of both a and b.

Proving p+a 

Given that 1_Ris a gcd of p and a.

Let’s assume p l a, and p is gcd of p and a.

As we know p is irreducible, so there must be no other divisor of p.

Since p l a $\Rightarrow$p is a gcd, so 1 is not a gcd

Vice versa, if 1 is the gcd $\Rightarrow$p+a.

Hence, p+a

 Step 3: Proving 1R is a gcd of p and a

Suppose p + a , x be gcd of p and a.

As we know that p is irreducible, x must be a unit or x=p* (where p* is an associate of p).

Suppose x is a common divisor, and it is not a unit.

Then, x=p*

This means p* + a

Since p and p* are associates, this implies p l a, which is a contradiction to our assumption.

Hence, x must be a unit.

This implies gcd of p and a must be a unit and 1 are associates.

From the above, we can conclude that is a gcd of p and a if and only if and only if p + a