Greatest Common Divisor for Integers
Given \(a, b \in \mathbb{Z}\), the greatest common denominator of \(a\) and \(b\) denoted by \(\mathrm{gcd}(a, b)\) is defined as the greatest integer which divides both \(a\) and \(b\).
This definition of the greatest common denominator depends on the fact that \(\mathbb{Z}\) is totally ordered. A modification is made to define the \(\mathrm{gcd}\) for general rings, however this generalisation is not entirely consistent with the above definition, which is why this is a different note.
This inconsistency is characterised in the following theorem.
If \(a = b = 0\) then \(\gcd(a, b)\) is not define in the context of integers, yet in the context of the general ring definition \(\gcd(a, b) = \{0\}\).
If either \(a\) or \(b\) is non-zero, and \(\gcd(a, b) = d\) by the integer definition, then \(\gcd(a, b) = \{-d, d\}\) by the ring definition.
Proof
By the definition of divisibility, because \(n \times 0 = 0\), \(n \mid 0\) for any integer \(n\). Hence there is no greatest such integer.
By the general ring definition, it is clear that \(0 \mid 0\), and if \(n \mid 0\), that is we have some other common divisor, then \(n \mid 0\) so \(0\) is the greatest such divisor.
Suppose either \(a\) or \(b\) are non-zero and let \(\gcd(a, b) = d\) by the integer definition. Then because all GCDs are associates in an integral domain, we have in the ring definition that \(\gcd(a, b) = \{-d, d\}\) because the only units of the integers are \(1\) and \(-1\).