Question

Let d be a gcd of a_1........a_k in an integral domain. Prove that every associate of d is also a gcd of a_1........a_k .

Short answer

It is proved that every associate of d is also a gcd of a_1........a_k .

Step-by-step solution

Defining greatest common divisor (gcd)

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 that every associate of d is also a gcd of a1........ak 

Let’s assume , d' be an associate of d.

Then, d' can be written as:

d' = du

d = d'v

Where u and v are unit such that, uv=1.

Since d I a, therefore, a_i = dx_i for some x_i .

Simplify a_i = dx_i as:

a_i = d (uv) x_i

= (du)(vx_i)

=d'vx_i

=d' (vx_i)

According to this, d' l a_i for every i.

Now, suppose c is another common divisor of a .

Since d is the gcd of alla_i , so c l d.

Therefore, for any y we can write d as:

d=cy

Multiplying both side by u as follows:

du=cyu

d_i=c (yu)

This implies c l d'.

As we know, d' is a common divisor of all a_i.

As c is another common divisor of a_i , therefore, c l d' .

Hence, d' is gcd of alla_1........a_k .